St.distance Formula Xi

Distance FormulaDistance between two points P(x,y) and Q(x,y) is|PQ| = [(x2 -x1)² +(y2 -y1)²].Hence the distance of the point P (x, y) from the origin (0, 0) = [(x -0)² +(y -0)²] = [x² +y²]

Remarks

To prove that a quadrilateral is a(i) rhombus, show that all the sides are equal.(ii) square, show that all the sides are equal and the diagonals are also equal.(iii) parallelogram, show that the opposite sides are equal.(iv) rectangle, show that the opposite sides are equal and the diagonals are also equal.

Illustrative Examples

Example

Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right angled triangle.

Solution

Let the points be A (7, 10), B (-2, 5) and C (3, -4), then|AB| = [(- 2 -7)² +(5 -10)²] = [81 +25] = 106,|BC| = [(3 +2)² +(11 -3)²] = [25 +81] = 106 and|CA| = [(7- 3)² +(10 -(-4))²] = [16 +196] = 212=> AB² = 106, BC² = 106 and CA² = 212,Hence AB² +BC² = 106 +106 = 212 = CA²=> ABC is right angled and it is right angled at B.Also |AB| = 106 = |BC| => ABC is isosceles.

Example

Show that the points (-1, -1), (2, 3) and (8, 11) are collinear.

Solution

Let the points be A (-1, -1), B (2, 3) and C (8, 11), then|AB| = [(2 -(-1))² +(3 -(-1))²] = [32 +42] = [9 +16] = 25 =5,|BC| = [(8 -2)² +(11-3)²] = [36 +64] = 100 = 10 and|CA| = [(8 -(-1))² +(11 -(-1))²] = [92 +122] = [81 +144] = 225 =15.Hence |AB| +|BC| = 5 +10 = 15 = |CA|=> the given points are collinear.

Example

The vertices of a triangle are A (1, 1), B (4, 5) and C (6, 13). Find cos A.

Solution

If a, b, c are the sides of ABC, thena = |BC| = [(6 -4)² +(13 -5)²] = [4 +64] = 68 = 2. 17,b = |CA| = [(6 -1)² +(13 -1)²] = [25+144] = 169 = 13, andc = |AB| = [(4 -1)² +(5 -1)²] = [9 +16] = 25 = 5By cosine formula, we havecos A = (b² +c² -a²)/ 2bc = (169 +25 -68)/(2.13.5) = 126/130 = 63/65

Exercise

1. A is a point on y-axis whose ordinate is 5 and B is the point (-3, 1). Compute the length of AB.

2. The distance between A (1, 3) and B (x, 7) is 5. Find the values of x.3. What point (or points) on the y-axis are at a distance of 10 units from the point (8, 8)?4. Find point (or points) which are at a distance of 10 from the point (4, 3) given that the

ordinate of the point (or points) is twice the abscissa.5. Find the abscissa of points whose ordinate is 4 and which are at a distance of 5 units from

(5, 0).6. What point on x-axis is equidistant from the points (7, 6) and (-3, 4)?7. Find the value of x such that |PQ| = |QR| where P, Q, R are (6, -1), (1, 3) and (x, 8)

respectively. Are the points P, Q, R collinear?8. Show that the points (4, 2), (7, 5) and (9, 7) are collinear.9. Show that the points (2, 3), (-4, -6) and (1, 3/2) do not form a triangle.

[Hint. Show that the given points are collinear.]10. Show that the points A (2, 2), B (-2, 4) and C (2, 6) are the vertices of a triangle. Prove that

ABC is an isosceles triangle.11. Show that the points:

(i) (-2, 2), (8, -2) and (-4, -3) are the vertices of a right-angled triangle.(ii) (0, 0), (5, 5) and (-5, 5) are the vertices of a right-angled isosceles triangle.(iii) (1, 1), (-1, -1),(- 3 , 3)are the vertices of an equilateral triangle.(iv) (2 a, 4 a), (2 a, 6 a), (2 a + 3 a, 5 a) are the vertices of an equilateral triangle.

12. Show that the points:(i) (2, 1), (5, 4), (4, 7), (1, 4) are the angular points of a parallelogram.(ii) (7, 3), (3, 0), (0, -4), (4, -1) are the vertices of a rhombus.(iii) (2, -2), (8, 4), (5, 7), (-1, 1) are the vertices of a rectangle.(iv) (3, 2), (0, 5), (-3, 2), (0, -1) are the vertices of a square.

13. The points A (0, 3), B (-2, a) and C (-1, 4) are the vertices of a right-angled triangle at A, find the value of a.

14. Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of equal sides is 3 units.

Answers

1. 5 2. 4 or -2 3. (0, 2), (0, 14)4. (1, 2), (3, 6) 5. 2 or 8 6. (3, 0)

7. 5 or -3; No 13. 1

14. (2 +( 11)/2, 5/2) or (2 -( 11)/2, 5/2)

Section Formula

When the Point divides the line segment Internally

Let P (x1, y1) and Q (x2, y2) be two given points in the co-ordinate plane, and R (x, y) be the point which divides the segment [PQ] internally in the ratio m1 : m2 i.e. PR/RQ = m1 / m2, where m1 0, m2 0, m1 + m2 0Then the coordinates of R are (m1 x2 +m2 x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2)Note. [PQ] stands for the portion of the line PQ which is included between the points P and Q including the points P and Q. [PQ] is called segment directed from P to Q. It may be observed that [QP] is the segment directed from Q to P. If a point R divides [PQ] in the ratio m1 : m2 then it divides [QP] in the ratio m2 : m1.

When the Point divides the line segment Externally

Let P (x1, y1) and Q (x2, y2) be two given points in the co-ordinate plane, and R (x, y) be the point which divides the segment [PQ] externally in the ratio m1 : m2 i.e. PR/RQ = m1 / m2, where m1 0, m2 0, m1 - m2 0Then the co-ordinates of R are m1 x2 -m2 x1)/(m1 -m2), (m1y2 -m2y1)/(m1 -m2)

Mid-point formula

The co-ordinates of the mid-point of [PQ] are ((x1 +x2)/2, (y1 +y2)/2)


Example

Find the co-ordinates of the point which divides the line segment joining the points P (2, -3) and Q (-4, 5) in the ratio 2 : 3(i) internally(ii) externally.

Solution

i. Let (x, y) be the co-ordinates of the point R which divides the line segment joining the points P (2, -3) and Q (-4, 5) in the ratio 2 : 3 internally, thenx = [2.(-4) +3.2]/(2+3) = - 2/5 andy = [2.5 +3.(-3)]/(2+3) = 1/5Hence the co-ordinates of R are (-2/5, 1/5)

ii. Let (x, y) be the co-ordinates of the point R which divides the line segment joining the points P (2, - 3) and Q (-4, 5) in the ratio 2 : 3 externally i.e.internally in the ratio 2 : -3.

x = [2.(-4) + (-3).2]/[2 +(-3)] = -14/1 = 14and y = [2.5 + (-3)(-3)]/[2 +(-3)] = 19/(-1) -19Hence the co-ordinates of R are (14, -19).

Example

In what ratio is the line segment joining the points (4, 5) and (1, 2) divided by the y-axis? Also find the co-ordinates of the point of division.

Solution

Let the line segment joining the points A (4, 5) and B (1, 2) be divided by the y-axis in the ratio k : 1 at P.By section formula, co-ordinates of P are ((k +4)/(k+1), (2k +5)/(k+1)).But P lies on y-axis, therefore, x-coordinate of P = 0=> (k +4)/(k+1) = 0 => k +4 = 0 => k = -4The required ratio is -4 : 1 or 4 : 1 externally.Also the co-ordinates of the point of division are (0, (2.(-4) +5)/(-4+1)) i.e (0, 1)

Exercise

1. Find the co-ordinates of the point which divides the join of the points (2, 3) and (5, -3) in the ratio 1 : 2(i) internally(ii) externally.

2. Find the co-ordinates of the point which divides the join of the points (2, 1) and (3, 5) in the ratio 2 : 3(i) internally(ii) externally.

3. Find the co-ordinates of the point that divides the segment [PQ] in the given ratio:(i) P (5, -2), Q (9, 6) and ratio 3 : 1 internally.(ii) P (-7, 2), Q (-1, -1) and ratio 4 : 1 externally.

4. Find the co-ordinates of the points of trisection of the line segment joining the points (3, - 1) and (-6, 5).

5. Find point (or points) on the line through A (- 5, -4) and B (2, 3) that is twice as far from A as from B.

6. Find the point which is one-third of the way from P (3, 1) to Q (-2, 5).7. Find the point which is two third of the way from P(0, 1) to Q(1, 0).8. Find the co-ordinates of the point which is three fifth of the way from (4, 5) to (-1, 0).

9. If P (1, 1) and Q (2, -3) are two points and R is a point on PQ produced such that PR = 3 PQ, find the co-ordinates of R.

10. In what ratio does the point P (2, -5) divide the line segment joining the points A (- 3, 5) and B (4, -9)?

11. In what ratio is the line joining the points (2, - 3) and (5, 6) divided by the x-axis? Also find the co-ordinates of the point of division.

12. In what ratio is the line joining the points (4, 5) and (1, 2) divided by the x-axis? Also find the co-ordinates of the point of division.

13. In what ratio is the line joining the points (3, 4) and (- 2, 1) divided by the y-axis? Also find the co-ordinates of the point of division.

14. Point C (-4, 1) divides the line segment joining the points A (2, - 2) and B in the ratio 3 : 5. Find the point B.

15. The point R (-1, 2) divides the line segment joining P (2, 5) and Q in the ratio 3 : 4 externally, find the point Q.

16. Find the ratio in which the point P whose ordinate is 3 divides the join of (-4, 3) and (6, 3), and hence find the co-ordinates of P.

17. By using section formula, prove that the points (0, 3), (6, 0) and (4, 1) are collinear.18. Points P, Q, R are collinear. The co-ordinates of P, Q are (3, 4), (7, 7) respectively and

length PR = 10 unit, find the co-ordinates of R.19. The mid-point of the line segment joining (2 a, 4) and (-2, 3 b) is (1, 2 a +1). Find the values

of a and b.20. The center of a circle is (-1, 6) and one end of a diameter is (5, 9), find the co-ordinates of

the other end.21. Show that the line segments joining the points (1, - 2), (1, 2) and (3, 0), (-1, 0) bisect each

other.22. Show that the points A(-2, -1), B (1, 0), C (4, 3) and D (1, 2) from a parallelogram. Is it a

rectangle?23. The vertices of a quadrilateral are (1, 4), (- 2, 1), (0, -1) and (3, 2). Show that the diagonals

bisect each other. What does quadrilateral become?24. Three consecutive vertices of a parallelogram are (4, - 11), (5, 3) and (2, 15). Find the fourth

vertex.

Answers1. (i) (3, 1) (ii) (-1, 9) 2. (i) (12/5, 13/5) (ii) (0, - 7)3. (i) (4, 8) (ii) (1, - 2) 4. (0, 1) and (-3, 3)5. (-1/3, 2/3) and (9, 10) 6. (4/3, 7/3)7. (2/3, 1/3) 8. (1, 2)9. (4, -11) 10. 5 : 2 internally11. 1 : 2 internally; (3, 0) 12. 5 : 2 externally; (-1 , 0)13. 3 : 2 internally 14. (- 14, 6)15. (3, 6) 16. 3 : 2 internally; (2, 3)18. (11 , 10) 19. a = 2, b = 220. (-7 , 3) 22. No23. Parallelogram 24. (1, 1)

Centroid and IncenterThe point which divides a median of a triangle in the ratio 2 : 1 is called the centroid of the triangle. Thus, if AD is a median of the triangle ABC and G is its centroid, then AG/GD = 2/1

By section formula, the co-ordinates of G are

The symmetry of the co-ordinates of G shows that it also lies on the medians through B and C. Hence the medians of a triangle are concurrent.Incenter of a triangleThe point of the intersection of any two internal bisectors of the angles of a triangle is called the incenter of the triangle. It is usually denoted by I.

If the internal bisector of A of a ABC meets the side BC in D, then BD/DC = AB/ACBy section formula, the co-ordinates of I are

The symmetry of the co-ordinates of I shows that it also lies on the internal bisector of C. Hence the internal bisectors of the angles of a triangle are concurrent.


Example

Find the co-ordinates of the incenter of the triangle whose vertices are(-2,4), (5,5) and (4,-2).

Solution

Let A (-2,4), B (5,5) and C (4,-2) be the vertices of the given triangle ABC, thena = | BC| = [(4 -5)² +(-2 -5)²] = [1 +49] = 50 = 5 2,b = |CA| = [(4 -2)² +(-2 -4)²] = [36 +36] = 72 = 6 2 andc = |AB| = [(5 -2)² +(5 -4)²] = [49 +1] = 50 = 5 2.The co-ordinates of the incenter of ABC are

Exercise

1. Find the centroid of the triangle whose vertices are (-1,4), (2,7) and (-4,-3).2. Find the point of intersection of the medians of the triangle whose vertices are (3,-5), (-7,4)

and (10,-2). [Hint. Find centroid.]3. Find the third vertex of a triangle if two of its vertices are (3,-5) and (-7,4), and the medians

meet at (2,-1).4. Find the centroid of the triangle ABC whose vertices are A(9,2), B(1,10) and C(-7,-6). Find

the co-ordinates of the middle points of its sides and hence find the centroid of the triangle formed by joining these middle points. Do the two triangles have same centroid?

5. If (-1,5), (2,3) and (-7,9) are the middle points of the sides of a triangle, find the co-ordinates of the centroid of the triangle.

6. If A(1, 5), B (-2,1) and C(4,1) are the vertices of ABC, and internal bisector of A meets side BC at D, find |AD|. Also find the incenter of ABC.

7. Find the co-ordinates of the center of the circle inscribed in a triangle whose angular points are (-36,7), (20,7) and (0,-8).

Answers1. (-1,8/3) 2. (2,-1) 3. (10,-2)4. (1,2); mid-points of BC, CA, AB are (-3,2), (1,-2), (5,6); (1,2); Yes5. (-2,17/3) 6. 4 units; (1,5/2) 7. (-1,0)

Slope of a Straight Line

Slope (or gradient) of a straight line

If ( 90°) is the inclination of a straight line, then tan is called its slope (orgradient). The slope of a line is usually denoted by m.Remark. Since tan is not defined when = 90°, therefore, the slope of a vertical line is not defined.

Slope of the line joining two points

The slope m of a non-vertical line passing through the points P(x1 , y1) and Q(x1, y1) is given by slope = m = (y2 -y1)/(x2 -x1)

Remarks

Two (non-vertical) lines are parallel iff their slopes are equal. Two (non-vertical) lines are perpendicular iff the product of their slopes = -1. Slope of a perpendicular line is the negative reciprocal of the slope of the given line.


Example

Without using Pythagoras theorem, show that the points A (1, 2), B (4, 5) and C (6, 3) are the vertices of a right-angled triangle.

Solution

In ABC, we havem1 = slope of AB = (5 -2)/(4-1) = 3/3 = 1 andm2 = slope of BC = (3 -5)/(6-4) = 2/2 = -1m1 m2 = 1. (-1) = -1 => AB BCHence, the given points are the vertices of a right-angled triangle.

Example

Using slopes, show that the points A (6, -1), B (5, 0) and C (2, 3) are collinear.

Solution

Slope of AB = [0 -(-1)]/(5-6) = 1/(-1) = -1and slope of AC = [3 -(-1)]/(2-6) = 4/(-4) = -1=> slope of AB = slope of AC=> AB and AC are parallel.But AB and AC have point A is common, therefore, the given points A, B and C are collinear.

Exercise

1. Find the slope of a line whose inclination is:(i) 30° (ii) 2 /3 (iii) /3

2. Find the inclination of a line whose gradient is:(i) 1/ 3 (ii) -1 (iii) - 3

3. Find the gradient of the line containing the points(i) (-2, 3) and (5, -7) (ii) (3, -7) and (0, 2)

4. A line passes through the points (4, -6) and (-2, -5). Does it make an acute angle with the positive direction of x-axis?

5. Find the equation of the locus of all points P such that the slope of the line joining origin and P is - 2.

6. Show that the line joining (2, -3) and (-5, 1) is(i) parallel to the line joining (7, -1) and (0, 3)(ii) perpendicular to the line joining (4, 5) and (0, - 2)

7. State, whether the two lines in each of the following problems are parallel, perpendicular or neither:(i) through (2, -5) and (-2, 5); through (6, 3) and (1, 1)(ii) through (5, 6) and (2, 3); through (9, -2) and (6, - 5)(iii) through (8, 2) and (-5, 3); through (16, 6) and (3, 15)

8. Find y if the slope of the line joining (-8, 11), (2, y) is - 4/3.9. Find the value of x so that the line through (x, 9) and (2, 7) is parallel to the line through (2, -

2) and (6, 4).10. Without using Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the

vertices of a right angled triangle.

Answers

1. (i) 1/ 3 (ii) - 3 (iii) not defined2. (i) 30° (ii) 135° (iii) 120°3. (i)- 10/7 (ii) -34. No 5. 2x +y = 07. (i) Perpendicular (ii) parallel (iii) neither8. -7/3 9. 10./3

Equation of a Straight Line

Important Definitions, Results and Formulae

1. The angle which a line makes with the positive direction of x-axis measured in the anti-clockwise direction is called the inclination of the line.

2. If ( 90°) is the inclination of a line, then tan is called its slope (or gradient).3. Slope of a line.

(i) If the inclination of a line is , its slope = m = tan .(ii) Slope of a line through (x1, y1) and (x2, y2) is given by m = (y2 -y1)/(x2 -x1)

4. Equation of a straight line.(i) Equation of a line parallel to x-axis is y = b.(ii) Equation of a line parallel to y-axis is x = a.(iii) Equation of a line with slope m and y-intercept c is y = m x +c.(iv) Equation of a line through (x1, y1) and with slope m is y -y1 = m (x -x1).

5. Conditions of parallelism and perpendicularity.Two lines with slopes m1 and m2 are(i) parallel if and only if m1 = m2

(ii) perpendicular if and only if m1m2 = -16. Reflection of P( , ) in the line y = x is P'( , ).

Exercise

1. Find the inclination of a line whose gradient is(i) 1 (ii) 3 (iii) 1/ 3

2. Find the equation of a straight line parallel to x-axis and passing through the point (2, -7).

3. Find the equation of a straight line whose(i) gradient = 3 , y-intercept = -4/3(ii) inclination = 30°, y-intercept = -3

4. Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.5. The equation to the line PQ is 3y -3x +7 = 0.

(i) Write down the slope of the line PQ.(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.

6. The given figure represents the lines y = x +1 and y = 3 x -1. Write down the angles which the lines make with the positive direction of x-axis. Hence determine .[Hint. Ext. = sum of two opposite int. s; 60° = +45°]

7. Given that (a, 2a) lies on the line& y/2 = 3x -6, find the value of a.8. The graph of the equation y = mx +c passes through the points (1, 4) and (-2, -5). Determine

the values of m and c.9. Find the equation of a straight line passing through (-1, 2) and having y-intercept 4 units.10. Find the equation of a st. line whose inclination is 60° and passes through the point (0, -3).11. Given that the line y/2 = x -p and the line ax +5=3y are parallel, find the value of a.12. Find the value of m, if the lines represented by 2mx -3y = 1 and y = 1 -2x are perpendicular

to each other.13. If the lines 3x +y = 4, x -ay +7 = 0 and bx +2y +5 = 0 form three consecutive sides of a

rectangle, find the values of a and b.14. Find the equation of a straight line perpendicular to the line 2x +5y +7 = 0 and with y-

intercept - 3 units.15. Find the equation of a straight line parallel to the line 2x +3y = 5 and having the same y-

intercept as x +y +4 = 0.16. Find the equation of the line which is parallel to 3x -2y -4 = 0 and passes through the point

(0, 3).17. Write down the equation of the line perpendicular to 3x +8y = 12 and passing through the

point (-1, -2).18. The co-ordinates of two points E and F are (0, 4) and (3, 7) respectively. Find

(i) the gradient of EF(ii) the equation of EF(iii) the co-ordinates of the point where the line EF intersects the x-axis.

19. Find the equation of the line passing through the points (4, 0) and (0, 3). Find the value of k, if the line passes through (k, 1·5).

20. If A (-1, 2), B (2, 1) and C (0, 4) are the vertices of a ABC, find the equation of the median through A.

21. Find the equation of a line passing through the point (-2, 3) and having x-intercept 4 units.[Hint. Since x-intercept is 4, the line passes through (4, 0)]

22. Find the equation of the st. line containing the point (3, 2) and making positive equal intercepts on the axes.

23. The intercepts made by a st. line on the axes are -3 and 2 units. Find:(i) the gradient of the line.(ii) the equation of the line.(iii) the area of the triangle enclosed between the line and the co-ordinate axes.

24. A line through the point P (2, 3) meets the co-ordinate axes at points A and B. If PA = 2 PB, find the co-ordinates of A and B. Also find the equation of the line AB.

25. Calculate the co-ordinates of the point of intersection of the lines represented by x +y = 6

and 3x -y = 2.26. The line joining the points P (4, k) and Q (-3, -4) meets the x-axis at A (1, 0) and y-axis at B.

Find(i) the value of k. (ii) the ratio of PB : BQ.

27. Find the equations of the diagonals of a rectangle whose sides are x = -1, x = 2, y = -2 and y = 6.

28. Find the co-ordinates of the image of (3, 1) under reflection in x-axis followed by reflection in the line x =1.

29. If P' (-4, -3) is the image of the point P under reflection in the origin, find(i) the co-ordinates of P.(ii) the co-ordinates of the image of P under reflection in the line y = -2.

30. Find the co-ordinates of the image of the point P (4, 3) under reflection in the x-axis followed by reflection in the line x = -2.

Answers1. (i) 45° (ii) 60° (iii) 30° 2. y +7 = 0

3. (i) 3 3x -3y -4 = 0 (ii) x - 3y -3 3 = 0. 4. -2/3; 45. (i) 1 (ii) 45° 6. 45°, 60°; 15° 7. 3

8. m = 3, c = 1 9. y = 2x +4 10. 3x -y -3 = 011. 6 12. 3/4 13. a = 3, b = 614. 5x -2y -6 = 0 15. 2x +3y +12 = 0 16. 3x -2y +6 = 017. 8x -3y +2 = 0 18. (i) 1 (ii) x -y +4 = 0 (iii) (-4, 0)19. 3x +4y -12 = 0; 2 20. x -4y +9 = 0 21. x +2y -4 = 022. x +y -5 = 023. (i) 2/3 (ii) 2x -3y +6 = 0 (iii) 3 sq. units24. A (6, 0), B(0,9/2); 3x +4y -18 = 025. (2, 4). 26. (i) 3 (ii) 4 : 327. 8x -3y +2 = 0, 8x +3y -10 = 0 28. (-1, -1)29. (i) (4, 3) (ii) (4, -7) 30. (-8, -3)

Angle between two Lines

The angle between two non-vertical and non-perpendicular lines

Let l1 and l2 be the two non-vertical and non-perpendicular lines with slopes m1and m2 respectively. Let 1 and 2 be their inclinations, then m1 = tan 1 and m2= tan 2. There are two angles and - between the lines l1 and l2, given bytan = ± (m1-m2)(1+m1m2)


Example

Find the angle between the lines joining the points(-1,2), (3,-5) and (-2,3), (5,0).

Solution

Here, m1 = slope of the line joining (-1,2) and (3,-5) = (-5-2)/(3+1) = -7/4 and m2 = slope of the line joining (-2,3) and (5,0) = (0-3)/(5+2) = -3/7Let be the acute angle between the given lines, then

tan = .

=Hence the acute angle between the lines is given by tan = 37/49

Exercise

1. Find the angle between the following pairs of lines:(i) 3 x -7 y +5 = 0 and 7 x +3 y -11 = 0(ii) 3 x +y -7 = 0 and x +2 y +9 = 0(iii) y = (2 - 3) x +9 and y = (2 + 3) x +1(iv) 2 x -y +3 = 0 and x +y -2 = 0

[Hint. (iv) It will be found that acute angle is given by tan = 3which gives as 71° 34', by using tables of natural tangents]

2. Find the angle between the lines joining the points (0,0), (2,3) and (2,-2), (3,5).3. If A(-2,1), B(2,3) and C(-2,-4) are three points, find the angle between the lines AB and BC.4. Find the angles between the lines x +1 = 0 and 3 x +y -3 = 0.5. Find the angle between the lines which make intercepts on the axes a,-b and b,-a

respectively.6. Find the measures of the angles of the triangle whose sides lie along the lines x +y -5 = 0, x

-y +1 = 0 and y -1 = 0.7. Find the equations of the two straight lines passing through the point (4,5) which make an

acute angle of 45° with the line 2 x-y +7 = 0.8. Find the equations of the two straight lines passing through the point (1,-1) and inclined at an

angle of 45° to the line 2 x -5 y +7 = 0.9. A vertex of an equilateral triangle is (2,3) and the equation of the opposite side is x +y +2 =

0. Find the equations of the other two sides.10. One diagonal of a square lies along the line 8 x -15 y = 0 and one vertex of the square is at

(1,2). Find the equations of the sides of the square passing through this vertex.11. If (1,2) and (3,8) are a pair of opposite vertices of a square, find the equations of the sides

and the diagonals of the square.

Answers

1.(i) 90° (ii) 45° (iii) 60° (iv) 71° 34'2. 25° 34' 3. 33° 42' 4. 30°

5. The acute angle is given by tan =6. 45°, 45°, 90°

7. 3 x +y -17 = 0, x -3 y +11 = 08. 7 x -3 y -10 = 0, 3 x +7 y +4 = 0

9. (2 + 3) x -y -1 -2 = 0, (2 - 3) x -y -1 +2 = 010. 23 x -7y -9 = 0, 7x +23 y -53 = 0.

11. Sides are 2 x +y -4 = 0, x -2 y +3 = 0, 2 x +y -14 = 0, x -2 y +13 = 0 and

diagonals are 3 x -y -1 = 0, x +3 y -17 = 0

Position of two Points relative to a LineThe points P (x1, y1) and Q (x2 , y2) lie on the same side or on opposite sides of the line a x +b y +c = 0 according as the expressions a x1 + b y1 +c and a x2 + by2 +c have same sign or opposite signs.


Example

The sides of a triangle are given by the equations 3 x +4 y = 10, 4 x -3 y = 5 and 7 x +y +10 = 0. Show that the origin lies with in the triangle.

.

Solution

The given lines are3 x +4 y -10 = 0 ...(i)4 x -3 y -5 = 0 ...(ii)and 7 x + y +10 = 0 ...(iii)Let ABC be the triangle formed by these lines.Solving these equations simultaneously, taking two at a time, the vertices of the triangle areA (-2, 4), B (2, 1) and C (-1, -3)On substituting (-2, 4) in L.H.S. of (ii), we get -8 -12 -5 = -25and substituting (0, 0) in L.H.S. of (ii), we get 0 -0 -5 = -5Since both have same sign, therefore, origin and A lie on the same side of BC.Similarly origin and B lie on the same side of CA; and origin and C lie on the same side of AB. (Please check it)From these results, it follows that the origin lies with in the triangle formed by the given lines.

Exercise

1. Are the points (2, 3) and (-1, 5) on the same side or on opposite sides of the line y = 2 x +5.2. Show that the points (3, 5) and (-3, -2)lie on the same side of the line 3 x = 7 y + 8.3. Which of the points (1, 1), (-1, 2), (2, 3) and (-3, 0) lie on the same side of the line 4 x +3 y =

5 on which the origin lies?4. Find by calculation whether the points (13, 8), (26, -4) lie in the same, adjacent or opposite

angles formed by the straight line 5 x + 6 y -112 = 0 and 10 x +11 y - 217 = 0.5. Prove that the point P (x1, y1) and the origin lie on the same side or on opposite sides of the

line ax +by +c = 0 according as ax1 + by1 + c and c have same sign or opposite signs.

Answers1. Opposite sides

3. (-1, 2) and (-3, 0)4. Opposite

Distance of a Point from a LineThe perpendicular distance d of a point P (x 1, y 1) from the line ax +by +c = 0 is given by d =| ax1 +by1 +c|/[ (a² +b²)]

Rule to find the distance between parallel lines

i. Choose a point on one of the given parallel lines.ii. Find the perpendicular distance from this point to the other line.


Example

Find the equation of a straight line, with a positive gradient, which passes through the point (-5, 0) and is at a perpendicular distance of 3 units from the origin.

Solution

Let m (> 0) be the gradient of the line, then any line through (-5, 0) and with gradientm is y -0 = m(x +5) i.e. mx -y +5m = 0 ...(i)It will be the required line if its perpendicular distance from origin (0, 0) is 3 units=> |m.0 -0 +5 m| /[ (m² +(-1)²)] = 3 => | 5 m | = 3 [m² +1]=> 25 m² = 9 (m² +1) => 16 m² = 9 => m = 3/4 (m > 0)Substituting this value of m in (i), the equation of the required line is (3/4) x -y + 5.3/4 = 0 or 3 x -4 y +15 = 0

Example

Find the distance between the lines 3 x -4 y +7 = 0 and 6 x -8 y = 18.

Solution

The given lines are 3 x -4 y +7 = 0 ...(i)and 6 x -8 y -21 = 0 ...(ii)We note that the slope of (ii) = - 6/(-8) = 3/4 = the slope of (i)=> the given lines are parallel.To find distance between these lines, we choose a point on (i).On putting x = 0 in (i), we get -4y +7 = 0 => y = 7/4

Thus (0, 7/4) is a point on (i).Required distance between given parallel lines = perpendicular distance from (0, 7/4) to the line (ii)

= = |-35 | = units

Exercise

1. Find the distance of the point P from the line AB in the following cases:(i) P (2, -3), line AB is 2 x -3 y -25 = 0(ii) P (4, 1), line AB is 3 x -4 y -9 = 0(iii) P (0, 0), line AB is h (x +h) +k (y +k) = 0

2. Find the distance of the point (0, - 1) from the line joining the points (1, 3) and (-2, 6).3. Calculate the length of the perpendicular from (7, 0) to the straight line 5 x +12 y -9 = 0 and

show that it is twice the length of perpendicular from (2, 1).4. Find the value (s) of k, given that the distance of the point (4, 1) from the line 3 x -4y +k = 0 is

4 units.5. The points A (0, 0), B (1, 7), C (5, 1) are the vertices of a triangle. Find the length of

perpendicular from A to BC and hence the area of ABC.6. Find the lengths of altitudes of the triangle whose sides are given by x -4 y = 5, 4 x +3 y = 5

and x +y = 1.7. Find the length of perpendicular from the point (4, -7) to the line joining the origin and the

point of intersection of the lines 2 x -3 y +14 = 0 and 5 x +4 y -7 = 0.8. A vertex of a square is at the origin and its one side lies along the line x -4 y -10 = 0. Find the

area of the square.

Answers

1. (i) 12/ 13 units (ii) 1/5 units (iii) [h² +k²] units

2. 5/ 2 units 3. 2 units 4. 12, -28

5. 17/ 13 units, 17 sq. units

6. 1 unit, 1/7 units, 1/[5 2] units7. 1 unit 8. 4 sq. units

Families of Lines

One parameter family of lines

It may seem that the equation of a straight line ax +by +c = 0 contains three arbitrary constants. In fact, it is not so. On dividing it by a (or b, which ever is non-zero), we get x + y(b/a) + (c/a),which can be written as x + By + C = 0 where B =b/a and C = c/a.It follows that the equation of a straight line contains two arbitrary constants, and the number of these arbitrary constants cannot be decreased further. Thus, the equation of every straight line

contains two arbitrary constants, consequently, two conditions are needed to determine the equation of a straight line uniquely.One condition yields a linear relation among two arbitrary constants and hence each arbitrary constant determines the other. Therefore, the lines which satisfy one condition contain a single arbitrary constant. Such a system of lines is called one parameter family of lines and the unknown arbitrary constant is called theparameter.

Examples of one parameter families

i. The equation y = m x +2, for different real values of m, represents a family of lines with y-intercept 2 units. A few members of this family are shown in figure below.

ii. The equation 2 x +3 y + k = 0, for different real values of k, represents a family of lines with

slope -2/3.A few members of this family are shown in figure below.

iii. The equation y -y1 = m (x -x1), for different real values of m, represents a family of lines which

pass through the fixed point (x1, y1) except the vertical line x = x1.iv. The equation x = a, for different real values of a, represents the family of lines parallel to y-

axis (including the y-axis itself).v. The equation a x +b y +k = 0, for different real values of k, represents a family of lines

parallel to the line ax +by +c = 0.vi. The equation b x -a y +k = 0, for different real values of k, represents a family of lines

perpendicular to the line ax +by +c = 0.vii. If l1 = a1 x +b1 y +c1 = 0 and l2 = a2 x +b2 y +c2 = 0 then l1 +k l2 = 0, for different real values of

k, represents a family of lines passing through the point of intersection of the lines l1 and l2.


Example

Find the equation of the family of lines with x-intercept -4.

Solution

Since the x-intercept of the family is given to be -4, therefore, each member of the family passes through the point (-4, 0).By using point-slope form, the equation of such a family of lines is y -0 = m (x -(-4)) i.e. y = m (x +4), where m is a parameter.Note. The above equation of the family does not give the vertical line through the point (-4, 0). However, the equation of this line is x = -4 i.e. x +4 = 0.

Example

Find the equation of the straight line which is parallel to 3 x -7 y = 11 and makes x-intercept 3 units.

Solution

The given line is 3x -7 y -11 = 0 ...(i)The equation of the family of lines parallel to (i) is 3 x -7 y+k = 0 ...(ii) where k is a parameter.To find x-intercept, put y = 0, we get 3 x +k = 0 => x =k/3For the required member of the family which makes x-intercept 3 units, k/3 = 3 => k = -9Substituting this value of k in (ii), the equation of the required line is 3 x -7 y -9 = 0

Exercise

1. Write the equations of the family of lines:(i) with y-intercept -3(ii) with slope 2(iii) parallel to 2 x +3 +3 y -5 = 0(iv) perpendicular to 3 x +7 +7 y = 8

2. Find the equation of the line through the intersection of the lines 4 x -3 y +7 = 0 and 2 x +3 y +5 = 0 and the point (-4, 5).

3. Find the equation of the straight line parallel to 2 x -5 y +3 = 0 and having x-intercept -4.4. Find the equations of two straight lines which are parallel to the line x +7 y +2 = 0 and at a

unit distance from the point (2, -1).5. Find the equations of straight lines which are perpendicular to the line 3 x +4 y -7 = 0 and are

at a distance of 3 units from (2, 3).6. Find the equations of the two straight lines drawn through the point (0, a) on which the

perpendiculars dropped from the point (2 a, 2) are each of length a.7. Find the equation of the line which lies mid-way between the lines 2 x +3 y +7 = 0 and 2 x +3

y +5 = 0.8. Find the equations of straight lines parallel to the lines 3 x - y -3 = 0 and 3 x - y +5 = 0 and

whose distances from these lines are in the ratio 3 : 5. Point out the line which lies between the given lines.

9. Find the equation of a straight line parallel to 2 x +3 y = 10 and which is such that the sum of its intercepts on the axes is 15.

10. A line is drawn perpendicular to 5 x = y +7. Find the equation of the line if the area of the triangle formed by this line with co-ordinate axes is 5 sq. units.

Answers1. (i) y = m x -3, m parameter (ii) y = 2 x +c, c parameter (iii) 2x + 3 + 3y + k = 0, k parameter (iv) 7 x -3 y + k = 0, k parameter

2. 8 x +3 +3 y +17 = 0 3. 2 x-5 y +8 = 0

4. x +7 y +5 +5 ( 2 +1) = 0, x +7 y -5 ( 2 -1) = 05. 4 x -3 y +16 = 0, 4x + 16 = 0, 4 x -3 y -14 = 06. y -a = 0, 4 x -3 y +3 +3 a = 07. 2 x +3 +3 y +6 = 08. 3 x - y = 0, 3 x - y -15 = 0 ; 3 x - y = 09. 2 x +3 +3 y -18 = 0

10. x +5 y -5 2 = 0, x +5 y +5 2 = 0

Equation of a Straight Line in different forms

Slope-Intercept form

The equation of a straight line having slope m and making an intercept c on y-axis is y = m x +c

Point-slope form

The equation of a straight line passing through the fixed point (x1, y1) and having slope m is y -y1 = m (x -x1)Corollary. The equation of a line passing through origin and having slope m is y = m x.

Two-point form

The equation of the line passing through two fixed points A (x1, y1) and B (x2,y2) isy - y1 = [(y2 - y1)/(x2 - x1)](x -x1)

Intercept form

The equation of the line cutting off intercepts a and b on the axes is x/a + y/b = 1

Normal (or perpendicular) form

The equation of a straight line in terms of the length of perpendicular from the origin upon it and the angle which this perpendicular makes with the positive direction of x-axis is given by x cos + y sin = p

General form

Every straight line can be represented by an equation of the first degree in x and y, and conversely every first degree equation in x, y represents a straight line.The equation A x +B y +C = 0 (where at least one of A and B is non-zero) is called the general form.


Example

Find the equation of a straight line whose inclination is 5 /6 and which cuts off an intercept of 4 units on negative direction of y-axis.

Solution

Let m be the slope of the line, thenm = tan 5 /6 = tan 150° = tan (180° -30°) = -tan 30° = 1/ 3Also c = y-intercept = -4So the equation of the line is y =(-1/ 3) x +(-4) (As y = m x +c)i.e. x + 3y + 4 3 = 0

Example

Show that the points (a, 0), (0, b) and (3 a, -2 b) are collinear. Also find the equation of the line containing them.

Solution

The equation of the line through (a, 0) and (0, b) isy -0 =[(b-0)/(0-a)](x -a) y - y1 = [(y2 - y1)/(x2 - x1)](x -x1)i.e. -ay = bx -ab i.e. bx +ay -ab = 0The third point (3 a, -2 b) lies on it if b. 3 a + a. (-2 b) - ab = 0 i.e. if 0 = 0, which is true.Hence the given points are collinear and the equation of the line containing them is bx +ay -ab = 0.

Example

In what ratio is the line joining the points (2, 3) and (4, -5) divided by the line joining the points (6, 8) and (-3,-2)?

Solution

The equation of the straight line joining the points (6, 8) and (-3, -2) isy -8 = [(-2 -8)/(-3 -6)](x -6) (Two-point form)=> y -8 = (10/9) (x -6) => 9 y -72 = 10 x -60=> 10 x -9 y +12 = 0 ...(i)Let the line joining the points (6, 8) and (-3, -2) i.e. the line (i) divide the line segment joining the points (2, 3) and (4, -5) at the point P in the ratio k : 1, then the co-ordinates of point P are((k.4 +1.2)/(k+1), (k (-5) +1 .3)/(k+1)) i.e. ((4k +20)/(k+1), (-5k +3)/(k+1))Since P lies on (i), we get

10[(4k +20)/(k+1)] -9[(-5k +3)/(k+1)] + 12 = 0=> 40 k +20 +45 k -27 +12 k +12 = 0=> 97 k +5 = 0 => k = -5/97Hence the required ratio is -5/97 i.e. 5 : 97 externally.

Example

Prove that the locus of the point which is equidistant from the points (-3, 7) and (2, -5) is a straight line.

Solution

Let A (-3, 7) and B (2, -5) be the given points and P (x, y) be any point on the locus, then | AP | = | BP | (given)=> (x +3)² +(y -7)² = (x -2)² +(y +5)²=> x² +6 x +9 + y² -14 y +49 = x² -4 x +4 + y² +10 y +25=> 10 x -24 y +29 = 0, which is a first degree equation in x and y, and so it represents a straight line.Hence the locus is a straight line. In fact, this Is the perpendicular bisector of the line segment joining the two given points.

Exercise 1

1. Find the equation of a straight line parallel to x-axis at a distance(i) 3 units above it(ii) 3 units below it.

2. Find the equation of a straight line parallel to y-axis at a distance(i) 2 units to the right(ii) 2 units to the left of it.

3. Find the equation of a horizontal line passing through (5, -2).4. Find the equation of a vertical line passing through (-7, 3).5. Find the equation of a line which is equidistant from the lines y +5 = 0 and y -2 = 0.6. Find the equation of a line whose

(i) gradient = -1, y-intercept = 3(ii) slope = - 2/7, y-intercept = -3(iii) inclination = 3 /4, y-intercept = -5

7. Find the equations of the straight lines cutting off an intercept of 3 units from the negative direction of y-axis and equally inclined to the axes.[Hint. The lines which are equally inclined to the axes have slope = ± 1]

8. Find the equation of the line passing through the point (2, -5) and making an intercept of -3 on the y-axis.

9. If the straight line y = mx +c passes through the points (2, 4) and (-3, 6) find the values of m and c.

10. Find the equation of a straight whose y-intercept is -5 and which is(i) parallel to the line joining the points (3, 7) and (-2, 0)(ii) perpendicular to the line joining the points (-1, 6) and (-2, -3).

11. Find the equation of a straight line passing through origin and making an angle of 120° with the positive direction of x-axis.

12. Find the equations of the bisectors of the angles between the co-ordinate axes.13. Find the equation of the straight line passing through the point (5, 7) and inclined at 45° to x-

axis. If it passes through the point P whose ordinate is -7, what is the abscissa of P?

14. Find the equation of the line through the point (-5, 1) and parallel to the line joining the points (7, -1) and (0, 3).

15. Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining (1, 4), (2, 3).

Answers1. (i) y -3 = 0 (ii) y +3 = 02. (i) x -2 = 0 (ii) x +2 = 03. y +2 = 0 4. x +7 = 05. 2y +3 = 06. (i) x +y - 3 = 0 (ii) 2 x +7 y +21 = 0 (iii) x +y +5 = 07. x +y +3 = 0, x -y -3 = 08. x +y +3 = 0 9. m = -2/5, c = 24/510. (i) 7 x -5 y -25 = 0 (ii) x +9 y +45 = 0

11. 3 x +y = 0 12. x - y = 0, x +y = 013. x - y +2 = 0; -9 14. 4 x +7 y +13 = 015. x -y +3 = 0

Exercise 2

1. State the geometrical meaning of the constants involved in(i) x/a + y/b = 1(ii) x cos + y sin = p(iii) (x -x1)/cos = (y -y1)/ sin = r

2. Find the equation of the line which cuts off intercepts 3 and 4 from the axes.3. Prove that the straight line whose intercepts on the axes are 2 and -3 respectively passes

through the point (4, 3).4. Find the equation of a straight line which passes through the point (1, -3) and makes an

intercept on y-axis twice as long as on x-axis.5. Find the equation of the straight line which passes through the point (3, -4) and makes

(i) equal intercepts on the axes(ii) intercepts equal in magnitude but opposite in sign on the axes.

6. Find the equation of the straight line which passes through the point (3, -2) and cuts off positive intercepts on the x-axis and y-axis which are in the ratio 4 : 3.

7. A straight line passes through the point (2, 3) and the portion of the line intercepted between the axes is bisected at this point, find its equation.

8. A straight line is such that the portion of it intercepted between the axes is bisected at the point (x1 , y1). Find its equation.

9. A straight line passes through the point (1, 1) and the portion of the line intercepted between the axes is divided at this point in the ratio 3 : 4, find its equation.

10. Find the equation of a line which passes through the point (-5, 2) and whose segment between the axes is divided by this point in the ratio 2 : 3.

11. If the straight line x/a + y/b = 1passes through the points (12, -15) and (8, -9), find the values of a and b.

12. Find the equation of a line which passes through the point (22, -6) and whose intercept on the x-axis exceeds the intercept on y-axis by 5.

13. Find the equation of a line which passes through (-3, 10) and sum of its intercepts on the axes is 8.

14. A straight line passes through the points (a, 0) and (0, b), the length of the line segment contained between the axes is 13 and the product of the intercepts on the axes in 60. Calculate the values of a and b and find the equation of the straight line.

15. The area of a triangle formed by a line and the co-ordinate axes is 6 sq. units and the length of the segment intercepted between the axes is 5 units. Find the equation (s) of line.

Answers

1. (i) a and b are the intercepts made by the line on x-axis and y-axis respectively. (ii) p is the length of the perpendicular drawn from origin on the line and the angle which this perpendicular makes with the x-axis. (iii) (x1, y1) is the point through which the line passes, is the inclination of the line and r is the directed distance of any point (x, y) on the line from the point (x1, y1).2. 4 x +3 y = 124. 2 x +y +1 = 05. (i) x +y +1 = 0 (ii) x -y = 76. 3 x +4 y = 17. 3 x +2 y -12 = 08. x/x1 + y/y1 = 29. 4 x +3 y = 710. 3 x -5 y +25 = 011. a = 2, b = 312. 6 x +11 y -66 = 0, x +2 y -10 = 013. 2 x - y +16 = 0, 5 x +3 y -15 = 014. a = 12, b = 5, 5 x +12 y = 60; a = 5, b = 12, 12 x +5 y = 60; a = -12, b = -5, 5 x +12 y +60 = 0; a = -5, b = -12, 12 x +5 y +60 = 015. 3 x +4 y = 12, 4 x +3 y = 12; 3 x +4 y +12 = 0, 4 x +3 y +12 = 0

Forms for the equation of a straight line

Suppose that we have the graph of a straight line and that we wish to find its equation. (We will assume that the graph has x and y axes and a linear scale .) The equation can be expressed in several possible forms. To find the equation of the straight line in any form we must be given either:

two points, (x1, y1) and (x2, y2), on the line; or one point, (x1, y1), on the line and the slope, m; or the y intercept, b, and the slope, m.

In the first case where we are given two points, we can find m by using the formula:

Once we have one form we can easily get any of the other forms from it using simple algebraic manipulations. Here are the forms:

1. The slope-intercept form:y = m x + b.The constant b is simply the y intercept of the line, found by inspection. The constant m is the slope, found by picking any two points (x1, y1) and (x2, y2) on the line and using the formula:

2. The point-slope form:y − y1 = m (x − x1).(x1, y1) is a point on the line. The slope m can be found from a second point, (x2, y2), and using the formula:

3. The general form:a x + b y + c = 0.a, b and c are constants. This form is usually gotten by manipulating one of the previous two forms. Note that any one of the constants can be made equal to 1 by dividing the equation through by that constant.

4. The parametric form:x = x1 + ty = y1 + m tThis form consists of a pair of equations; the first equation gives the x coordinate and the second equation gives the y coordinate of a point on the line as functions of a parameter t. (x1, y1) is a known point on the line and m is the slope of the line. Each value of t gives a different point on the line.

For example when t = 0 then we get the pointx = x1

y = y1

or the ordered pair (x1, y1), and when t = 1 then we get the pointx = x1 + 1y = y1 + mor the ordered pair (x1 + 1, y1 + m), and so on.

Example: Show all of these forms for the straight line shown to the right.

Solution: Two points on this line are (x1, y1) = (0, 15) and (x2, y2) = (3, 0). Thus the y intercept isb = 15and the slope is

= −51. To get the slope-intercept form, we simply substitute in the two values m = −5 and b = 15:y = −5 x + 15.2. To get the point-slope form, we could use the point (0, 15) as “the” point together with m = −5:y − 15 = −5 (x − 0) or simplifying: y − 15 = −5 x.Or we could instead use the other point, (3, 0) and get:y − 0 = −5 (x − 3) or simplifying: y = −5 (x − 3).3. To get the general form, take any of these three forms found so far and distribute and collect all terms on the left-hand-side. The result is the same for all:5 x + y −15 = 0.Note that dividing both sides by, say 5, results in the equationx + 0.2 y − 3 = 0,which is also in general form and is equivalent in every way to the previous one.

4. To get the parametric form, we could use the point (0, 15) as “the” point together with m = −5:x = 0 + ty = 15 −5 tWith this choice, when t = 0 we are at the point (0, 15) and when t = 3 we are at the point (3, 0). We could instead use the other point, (3, 0) and get another parametric form:x = 3 + ty = 0 −5 t

With this choice, when t = 0 we are at the point (3, 0) and when t = −3 we are at the point

(0, 15). With either choice we will get all the points on the line as we let t range through all

values.

Line through two points

The line through two distinct points (x1, y1) and (x2, y2) is given by

(1) y = y1 + [(y2 - y1) / (x2 - x1)]·(x - x1),

where x1 and x2 are assumed to be different. In case they are equal,

the equation is simplified to

x = x1

and does not require a second point.

Equation (1) can also be written as

y - y1 = [(y2 - y1) / (x2 - x1)]·(x - x1),

or even as

(x2 - x1)·(y - y1) = (y2 - y1)·(x - x1),

where one does not have to worry whether x1 = x2 or not. However,

the simplest for me to remember is this

(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)

which is not as universal is the one before.

General equation

A straight line is defined by a linear equation whose general form is

Ax + By + C = 0,

where A, B are not both 0.

The coefficients A and B in the general equation are the components

of vector n = (A, B) normal to the line. The pair r = (x, y) can be

looked at in two ways: as a point or as a radius-vector joining the

origin to that point. The latter interpretation shows that a straight

line is the locus of points r with the property

r·n = const.

That is a straight line is a locus of points whose radius-vector has a

fixed scalar product with a given vector n, normal to the line. To see

why the line is normal to n, take two distinct but otherwise arbitrary

points r1 and r2 on the line, so that

r1·n = r2·n.

But then we conclude that

(r1 - r2)·n = 0.

In other words the vector r1 - r2 that joins the two points and thus

lies on the line is perpendicularto n.

Normalized equation

The norm ||n|| of a vector n = (A, B) is defined via ||n||2 = A2 +

B2 and has the property that, for any non-trivial vector n, n/||n|| is a

unit vector, i.e., || n/||n|| || = 1.

Note that the line defined by a general equation would not change if

the equation were to be multiplied by a non-zero coefficient. This

property can be used to keep the coefficient A non-negative. It can

also be used to normalize the equation by dividing it by ||n||. As a

result, in a normalized equation

Ax + By + C = 0,

A2 + B2 = 1.

(In the applet, the coefficients of the normalized equation are

rounded to up to 6 digits, for which reason the above identity may

only hold approximately.)

The normalized equation is conveniently used in determining

the distance from a point to a line.

Intercept-intercept

Assume a straight line intersects x-axis at (a, 0) and y-axis at (0,

b). Then it is defined by the equation

x/a + y/b = 1,

which also can be written as

xb + ya = ab.

The latter form is somewhat more general as it allows either a or b

to be 0. a and b are defined as x-intercept and y-intercept of the

linear function. These are signed distances from the points of

intersection of the line with the axes.

Point-slope

The equation of a straight line through point (a, b) with a given slope

of m is

y = m(x - a) + b, or y - b = m(x - a).

As a particular case, we have

Slope-intercept equation

The equation of a line with a given slope m and the y-intercept b is

y = mx + b.

This is obtained from the point-slope equation by setting a = 0. It

must be understood that the point-slope equation can be written for

any point on the line, meaning that the equation in this form is not

unique. The slope-intercept equation is unique because if the

uniqueness for the line of the two parameters: slope and y-intercept.

General Form of a CircleThe equation x² +y² +2 gx +2 fy +c = 0 represents a circle iff g² +f² -c > 0.Its center is (-g, -f) and radius = [g² +f² -c].Concentric circles. Circles having same center are called concentric circles.Equal circles Circles having equal radius are called equal circles.


Example

One end of a diameter of the circle x² +y² -6 x +5 y -7 = 0 is (-1, 3). Find the co-ordinates of the other end.

Solution

The given equation is x² + y² -6 x +5 y -7 = 0It is easy to see that this represents a circle with center C (3, - 5/2)Let A(-1,3) and B( , ) be the ends of the diameter.Since C is mid-point of [AB], we get ( - 1)/2 = 3 and ( +3) / 2 = - 5/2 = 7 and = -8Hence the other end of the diameter is (7, -8).

Example

Show that the points (7, 5), (6, -2), (-1, -1) and (0, 6) are concyclic. Also find the radius and the center of the circle on which they lie.

Solution

Let us find the equation of the circle passing through the points(7, 5), (6, - 2) and (-1, -1).Let the equation of this circle be x² + y² + 2gx + 2fy + c = 0 ...(i)As the points (7, 5), (6, -2) and (-1, -1) lie on it, we get

49 +25 +14 g +10 f +c => 14 g +10 f +c +74 = 0 ...(ii)36 +4 +12 g -4 f +c = 0 => 12 g -4 f + c +40 = 0 ...(iii)1 -2 g -2 f +c = 0 => 2 g +2 f -c -2 = 0 ...(iv)Adding (ii) and (iv), we get16 g +12 f +72 = 0 => 4 g +3 f +18 = 0 ...(v)Adding (iii) and (iv), we get14 g -2 f +38 = 0 => 7 g - f +19 = 0 ...(vi)Solving (v) and (vi) simultaneously, we get g = -3, f = -2. ...(vi)From (ii), we get c = -14 (-3) -10 (-2) -74 = -12.Substituting these values of g, f and c in (i), we getx² +y² -6 x -4 y -12 = 0 ...(vii)The fourth point (0, 6) will lie on (vii) if 0 +36 -0 -24 -12 = 0 i.e. if 0 = 0, which is true.Hence the given points are concyclic.Also, (vii) is the equation of the circle on which these points lie.Its center is (3, 2) and radius = [9 +4 -(-12)] = 5.

Exercise

1. Which of the following equations represent a circle? If so, determine its center and radius:(i) x² + y² +4 x -4 y -1 = 0(ii) 2 x² +2 y² = 3 x -5 y +7(iii) x² +y² +4 x +2 y +14 = 0(iv) 2 x² +2 y² = 5 x +7 y + 3(v) (x +3)² +(y -2)² = 0(vi) x² + y² - a x - b y = 0

2. Find the value of p so that x² + y² +8 x; +10 y +p = 0 is the equation of a circle of radius 7 units.

3. The radius of the circle x² +y² -2 x +3 y +k = 0 is 2. Find the value of k. Find also the equation of the diameter of the circle which passes through the point.

4. (i) Find the equation of the circle the end points of whose one diameter are the centers of the circles x² +y² +6 x -14 y +5 = 0 and x² + y² -4 x +10 y +7 = 0.(ii) One end of a diameter of the circle x² + y² -3 x +5 y -4 = 0 is (2, 1), find the co-ordinates of the other end.

5. Find the equation of the circle concentric with the circle x² +y² -8 x +6 y -5 = 0 and passing through the point (-2, -7).

6. Find the equation of the circle which passes through the center of the circle x² +y² -4 x -8 y -41 = 0 and is concentric with the circle x² +y² -2 y +1 = 0.

7. Find the equation of the circle concentric with the circle 2 x² +2 y² +8 x +10 y -35 = 0 and with area 16 square units.

8. Find the equation of the circle which is concentric with the circle x² + y² -4 x +6 y -3 = 0 and of double its (i) circumference (ii) area.

9. Prove that the centres of three circles x² + y² -4 x -6 y -14 = 0, x² + y² +2 x +4 y -5 = 0 and x² + y² -10 x -16 y +7 = 0 are collinear.

10. Prove that the radii of the circles x² + y² = 1, x² + y² -2 x -6 y -6 = 0 and x² + y² -4 x -12 y -9 = 0 are in A.P.

11. Find the equation of the circle which has its center on the line y = 2 and which passes through the points (2, 0) and (4, 0).

12. Find the equation of the circle which passes through the points (1, - 2), (4, -3) and has its center on the line 3 x +4 y +10 = 0.

13. Find the equation of the circle passing through the three points(i) (0, 0), (0, 1) and (2, 3)

(ii) (0, 2), (3, 0) and (3, 2);Also find its center and radius.

Answers

1. (i) circle; (- 2 , 2), 3 (ii) circle; (3/4, 5/4), (3 10)/4 (iii) empty set (iv) circle; (5/4, 7/4), (v) point circle; (-3, 2), zero (vi) circle;(a/2, b/2)2. -8 3. -3; 2 x -2 y = 54. (i) x² + y² +x -2 y -41 = 0 (ii) (1, -6)5. x² +y² -8 x +6 y -27 = 06. x² + y² -2 y -12 = 07. 4 x² +4 y² +16 x +20 y -23 = 08. (i) x² + y² -4 x +6 y -51 = 0 (ii) x² + y² -4 x +6y -19 = 011. x² +y² -6 x -4 y +8 = 012. x² +y² -4 x +8 y +15 = 0

13. (i) x² +y² -5 x - y = 0 -y = 0 ; (5/2, 1/2), 26 /2

(ii) x² +y² -3 x -2 y = 0 ; (3/2, 1), 13 /2

Orthogonal Circles

Angle of intersection of two circles

An angle between the two tangents to the two circles at a point of intersection is called an angle of intersection between two circles.

Orthogonal circles

Two circles are said to cut orthogonally iff angle of intersection of these circles at a point of intersection is a right angle i.e. iff the tangents to these circles at a common point are perpendicular to each other.If r, r' are radii of circles S and S' respectively and d is the distance between their centers and is an angle of intersection of these circles, then cos =(r² +r'² -d²)/(2 r r')If the circles S and S' cut orthogonally, then = 90°,cos = cos 90° = 0 => (r² +r'² -d²)/(2r r') = 0 => r² +r'² = d²Two circles S = x² +y² +2gx +2 f y +c = 0 andS' = x² +y² +2 g' x +2 f'y +c' = 0 cut orthogonally if 2 (g g' +f f') = c +c'.


Example

Find the equation of the circle which intersects the circles x² +y² +2 x -2 y +1 = 0 and x² +y² +4 x -4 y +3 = 0 orthogonally and whose center lies on the line 3 x -y -2 = 0.

Solution

Let the equation of the required circle be x² +y² +2 g x +2 f y + c = 0 ...(i)Since this circle cuts the circles x² +y² +2 x -2 y +1 = 0 and x² + y² +4 x -4 y +3 = 0 orthogonally, we get 2 (g. 1 + f. (-1)) = c +1=> 2 g -2 f - c -1 = 0 ...(ii)and 2 (g.2 + f.(-2)) = c +3=> 4 g -4 f -c -3 = 0 ...(iii)As center of (i) i.e. (-g, -f) lies on 3 x -y -2 = 0, we get -3 g + f -2 = 0=> 3 g - f +2 = 0 ...(iv)Subtracting (ii) from (iii), we get 2 g -2 f -2 = 0=> g -f -1 = 0 ...(v)Solving (iv) and (v) simultaneously, we get g = -3/2 and f = -5/2From (ii), we get c = 2 g -2 f -1 = -3 +5 -1 = 1Substituting these values of g, f and c in (i), we getx² +y² -3 x -5 y +1 = 0, which is the equation of the required circle.

Exercise

1. Determine the angle of intersection of the two circles (x -3)² +(y -1)² = 8 and (x -2)² +(y +2)² = 2.

2. Find the angle at which the circles x² + y² = 16 and x² +y² -2 x -4 y = 0 intersect each other.3. Show that the circles x² + y² -4 x -6 y +4 = 0 and x² +y² -10 x -14 y +58 = 0 cut orthogonally.4. Show that the circles x² +y² -2 a x +2 b y + c = 0 and x² + y² +2 bx +2 a y -c = 0, a² + b² > | c

|, cut orthogonally.5. For what value of do the circles x² + y² +5 x +3 y +7 = 0 and x² +y² -8 x +6 y + = 0 cut

orthogonally?

Answers1. /2

2. The acute angle is given by cos = 2/ 55. -18

Equation of a Circle in different formsA circle is the locus of a point which moves in a plane so that it remains at a constant distance from a fixed point. The fixed point is called the center and the constant distance is called radius. Radius is always positive.

Standard (or simplest) form

The equation of a circle with O(0,0) as center and r (>0) as radius is x² +y² = r²

Central form

The equation of a circle with C(h,k) as center and r (>0) as radius is given by (x -h)² +(y -k)² = r²

Diameter form

Let A(x1,y1) and B(x2,y2) be the extremities of a diameter of the circle.Then the equation of the circle is(x -x1)(x -x2) + (y -y1)(y -y2) = 0


Example

Find the equation of a circle whose center is (3,-2) and which passes through the intersection of the lines 5x +7y = 3 and 2x -3y = 7.

Solution

Given lines are5x +7y -3 = 0 ...(i), and2x -3y -7 = 0 ...(ii)Solving (i) and (ii) simultaneously, we get x = 2, y = -1.So the point of intersection, say P, of the given lines is (2,-1).Since the center of the circle is C(3,-2) and it passes through the point P(2,-1), radius = |CP| = [(2 -3)² +(-1 +2)²] = [1 +1] = 2Hence the equation of the circle is

(x -3)² +(y +2)² = ( 2)² (central form)or x² +y² -6x +4y +11 = 0.

Example

Find the equation of a circle which touches

i. the y-axis at origin and whose radius is 3 unitsii. both the co-ordinate axes and the line x = 3.

Solution

i. There are two circles satisfying given conditions. As the circles touch y-axis at the origin, their centers lie on x-axis. Since radius is 3 units, centers of the circles are (3, 0) and (-3, 0) and hence the equations of the circles are (x ±3)² +(y -0)² = 3² or x² +y² ±6x = 0

ii. There are two circles satisfying the given conditions. From above figure, clearly, the centers of these circles are and (3/2,3/2) and (3/2, -3/2) radius of each circle is 3/2. So the required equation is x² +y² -3x ±3y + (9/4) = 0or 4x² +4y² -12x ±12y +9 = 0

Exercise

1. Find the equation of a circle whose(i) center is at the origin and the radius is 5 units.(ii) center is (-1, 2) and radius is 5 units.

2. Determine the equation of a circle whose center is (8, -6) and which passes through the point (5, -2).

3. Prove that the points (7, -9) and (11, 3) lie on a circle with center at the origin. Also find its equation.

4. Find the equation of the circle whose(i) center is (a, b) and passes through origin.(ii) center is (2, -3) and passes through the intersection of the lines 3x -2y -1 = 0 and 4x +y -27 = 0.

5. Find the equation of a circle whose center is the point of intersection of the lines 2x +y = 4 and x -y = 5 and passes through the origin.

6. Find the equation of a circle whose two diameters lie along the lines 2x -3y +12 = 0 and x +4y + 12 = 0 and x +4y -5 = 0 and has area 154 square units.

7. Find the equation of the circle whose center lies on the negative direction of y-axis at a distance 3 units from origin and whose radius is 4 units.

8. Find the equations of the circles of radius 5 whose centers lie on the x-axis and pass through the point (2,3).

9. Find the equation of the circle(i) whose center is (0, -4) and which touches the x-axis.(ii) whose center is (3, 4) and touches the y-axis.

10. Find the equations of circles which touch both the axes and(i) has radius 3 units(ii) touch the line x = 2a.

11. Find the equations of circles which pass through two points on x-axis at distances of 4 units from the origin and whose radius is 5 units.

12. Find the equations of circles(i) which touch the x-axis on the positive direction at a distance 5 units from the origin and has radius 6 units.(ii) passing through the origin, radius 17 and ordinate of the center is -15.(iii) which touch both the axes and pass through the point (2, 1).

13. Find the equations of circles which touch the y-axis at a distance of 4 units from the origin and cut off an intercept of 6 units along the positive direction of x-axis.

14. Find the equations to the circles touching axis of y at the point (0, 3) and making an intercept of 8 units on x-axis.

15. Find the equations to the circles which touch the x-axis at a distance of 4 units from the origin and cut off an intercept of 6 from the y-axis.

Answers1. (i) x² +y² = 25 (ii) x² +y² -2x +4y = 02. x² +y² -16x + 12y +75 = 03. x² +y² -130 = 04. (i) x² +y² -2ax -2by = 0 (ii) x² +y² -4x +6y -96 = 05. x² +y² -6x +4y = 0.6. x² +y² +6x -4y -36 = 07. x² +y² +6y -7 = 0.8. x² +y² -12x +11 = 0, x² +y² +4x -21 = 09. (i) x² +y² +8y = 0 (ii) x² +y² -6x -8y +16 = 010. (i) x² +y² ±6x ±6y +9 = 0 (ii) x² +y² -2ax ±2ay +a² = 011. x² +y² ±6y -16 = 012. (i) x² +y² -10x ±12y +25 = 0 (ii) x² +y² ±6x +30y = 0 (iii) x² +y² -2x -2y +1 = 0, x² +y² -10x -10y +25 = 013. x² +y² -10x ±8y +16 = 014. x² +y² ±10x -6y +9 = 015. x² +y² ±8x ±10y +16 = 0

Parametric form of Circle

x = r cos , y = r sin ,0 < 2 represent the circle x² +y² = r², where is calledparameter and the point P (r cos , r sin ) is called the point " " on the circle x² +y² = r².

Parametric form of the circle (x -h)² +(y -k)² = r²

Every point P on the circle can be represented as x = h + r cos , y = k + r sin , 0 < 2 Thus, x = h + r cos , y = k + r sin , 0 < 2 , represent the circle (x -h)² +(y -k)² = r². is called parameter and the point (h +r cos , k +r sin ) is called the point " " on this circle.


Example

Find the parametric equations of the circle x² +y² = 5

Solution

The given circle is x² + y² = 5We know that the parametric equations of the circle x² +y² = r² are x = r cos , y = r sin , 0 < 2 The given circle is comparable with x² +y² = r², here r = 5Therefore, the parametric equations of the given circle x² +y² = 5 are x = 5cos , y = 5 sin , 0 < 2

Example

Find the cartesian equations of the curves x = p +c cos , y = q +c sin , where is parameter. Do these equations represent a circle? If so, find center and radius.

Solution

Given x = p +c cos , y = q + c sin => x -p = c cos , y -q = c sin To eliminate the parameter , on squaring and adding these equations, we get (x -p)² + (y -q)² = c² (cos² +sin² )=> (x -p)² +(y -q)² = c²,which represents a circle with center (p, q) and radius = | c |.

Exercise

1. Find the parametric equations of the following circles :(i) x² +y² = 13(ii) (x -2)² +(y +3)² = 36(iii) x² + y² +4 x - 6 y -12 = 0(iv) 2 x² +2 y² = 5 x +7 y +3(v) x² + y² -2 a x - 2 a y = 0(vi) x² + y² + p x +q y = 0

2. Find the cartesian equations of the following curves:(i) x = 2 cos , y = 2 sin (ii) x = 1 +5 cos , y = 2 +5 sin (iii) x = -3 + 7cos , y = 4 + 7 sin ,where is parameter. Do these equations represent circles? If so, find center and radius.

Answers

1. (i) x = 13 cos , y = 13 sin , 0 < 2 (ii) x = 2 +6 cos , y = -3 +6 sin , 0 < 2 (iii) x = -2 +5 cos , y = 3 +5 sin , 0 < 2 (iv) x = cos , y = sin , 0 < 2 (v) x = a +| a | cos , y = a +| a | sin , 0 < 2

(vi) x = -p/2 +(1/2) (p² +q²) cos , y = -p/2 + (1/2) (p² +q²) sin , 0 < 2 2. (i) x² +y² = 4; circle, (0, 0), 2 (ii) (x -1)² +(y -2)² = 25; circle, (1, 2), 5

(iii) (x +3)² +(y -4)² = 7; circle, (-3 , 4), 7

Point and CircleLet S be a circle with center C and radius r (> 0) and P be any point in the plane of the circle S, then

i. P is called exterior to S iff | C P | > r,ii. P is called interior to S iff |CP| < r andiii. P is said to lie on S iff |CP| = r.

If P is exterior to S then we say that P lies outside S, and if P is interior to S then we say that P lies inside SNotation. Let S = x² + y² +2 g x +2 f y + c = 0, g² + f² - c > 0, be a circle and P (x1, y1) be a point in the plane of S, then S1 = x1² +y1² +2 g x1 +2 f y1 + c.Let S be a circle and P (x1, y1) be a point in the plane of S, then

i. P is exterior to S iff S1 > 0ii. P is interior to S iff S1 < 0iii. P lies on S iff S1 = 0

Corollary.Let S be a circle and P (x1, y1), Q (x2, y2) be two points in the plane of S then they lie

i. on the same side of S iff S1 and S2 have same sign

ii. on the opposite sides of S iff S1 and S2 have opposite signs.

Exercise

1. Among the points given below, point out which of these are interior, exterior or lie on the circle S with center (1, 2) and radius 3?(i) (2, 3) (ii) (-1, -3) (iii) (4, 2)

2. Among the points given below, point out which of these are exterior, interior or lie on the circle 3 x 2 +3 y 2 -2 x -11 = 0?

(i) (-1, 0) (ii) (1, -2) (iii) 3. Do the following pairs of points lie on the same side or on opposite sides of the circle with

center (- 1, 2) and radius?(i) (2, -3) and (1, 2) (ii) (0, 4) and (1, 3)

Answers1. (i) Interior (ii) exterior (iii) lies on2. (i) Interior (ii) exterior (iii) lies on3. (i) Opposite sides (ii) both lie on the circle

Line and CircleThe condition that the line y = mx +c may intersect the circle x² + y² = a² is given by a²(1 + m²) c²

Remark

The line y = m x + c will intersect the circle x² + y² = a² in two distinct points iff a² (1 +m²) > c², and the line will intersect the circle in one and only one point i.e. the line will be atangent to the circle iff a²(1 +m²) = c², and the line will not intersect the circle iff a²(1 + m²) < c².

Corollary 1. Condition of the tangency

The line y = mx + c will touch the circle x² +y² = a²

iff a²(1 + m²) = c² i.e. iff c = ± a [1 +m²]

Corollary 2. Equations of tangents in slope form

Substituting the values of c = ± a [1 +m²] in equation y = mx + c, we get

y = mx ± a [1 +m²]Thus, there are two parallel tangents to the circle x² +y² = a² having m as their slope.

Length of intercept made by a circle on a line

Let a line l meet a circle S with center C and radius r in two distinct points. If d is the distance of C

from l then the length of intercept = p [r² -d²]

Length of tangent

Let S be a circle and P be an exterior point to S, and PT1, PT2 be two tangents to S through P, then the distance |PT1| or |PT2| is called the length of tangent from P to the circle S.

The length of tangent =


Example

Find the locus of the point of intersection of perpendicular tangents to the circle x² +y² = a²

Solution

he given circle is x² + y² = a² ...(i)The equation of any tangent to the circle (i) in the slope form is y = mx +a [1 +m²] .... (ii)Let (ii) pass through the point P ( , ), then = m + a [1 +m²] - m = a [1 +m²]=> ( - m )² = a²(1 + m²)=> ² + m 2 ² - 2 m - a² - a² m² = 0=> ( ² - a²) m² - 2 m + ( ² - a²) = 0,which is a quadratic in m having two roots, say m1, m2; and these represent slopes of two tangents passing through P ( , ).Since the tangents are at right angles, m1 m2 = -1=> ² - a² = -1 => ² - a² = - ² + a² ² -a²=> ² + ² = 2 a²The locus of P ( , ) is x² +y² = 2 a²

Thus, the locus of point of intersection of perpendicular tangents to the circle x² +y² = a² is x² +y² = 2 a², which is a circle concentric with the given circle.This is known as director circle of the circle x² +y² = a².

Exercise

1. Determine the number of points of intersection of the circle x² +y² + 6x -4y +8 = 0 with each of the following lines:(i) 2 x + y -1 = 0(ii) x +1 = 0(iii) 4x +3y -12 = 0

2. Determine the points of intersection (if any) of the circle x² +y² +5 x = 0 with each of the following lines:(i) x = 0(ii) 3x - y +1 = 0(iii) 3x -4 y = 7

3. Find the points in which the line y = 2 x +1 cuts the circle x² + y² = 2. Also find the length of the chord intercepted.

4. (i) Find the points of intersection of the circle 3 x² +3 y² -29 x -19 y -56 = 0 and the line y = x +2. Also find the length of the chord intercepted.(ii) If y = 2 x is a chord of the circle x² + y² -10 x = 0, find the equation of the circle with this chord as diameter. Hence find the length of the chord intercepted.

5. Find the lengths of intercepts made by the circle x² + y² -4 x -6 y - 5 = 0 on the co-ordinate axes.

6. Find the length of the chord intercepted by the circle x² +y² -8 x -6 y = 0 on the line x -7 y -8 = 0.

7. Find the length of the chord intercepted by the circle x² +y² = 9 on the line x +2 y = 5. Determine also the equation of the circle described on this chord as diameter.[Hint. The center of the circle described on the chord x +2 y = 5 as diameter is the point of intersection of this line and the line through (0, 0) and perpendicular to this line.]

8. (i) Prove that the lines x = 7 and y = 8 touch the circlex² + y² -4 x -6 y -12 = 0. Also find points of contact.(ii) Find the co-ordinates of the center and the radius of the circle x² + y² -4 x +2 y -4 = 0. Hence, or otherwise, prove that x +1 = 0 is a tangent to the circle. Calculate the co-ordinates of the point of contact. If this point of contact is A, find the co-ordinates of the other end of the diameter through A.

9. Prove that the line y = x +a 2 touches the circle x² +y² = a². Also find the point of contact.10. Prove that the line 4 x +y -5 = 0 is a tangent to the circle x² + y² +2 x -y -3 = 0, also find the

point of contact.11. Find the condition that the line l x +m y + n = 0 may touch the circle x² +y² = a².12. Find the condition that the line l x + m y +n = 0 may touch the circle x² +y² +2 g x +2 f y + c =

0.13. If the circle 2 x² +2 y² = 5 x touches the line 3 touches the line 3 x + 4 y = k, find the values of

k.14. (i) Find the equation of the circle with center (3, 4) and which touches the line 5x +12y -1 = 0.

(ii) Find the equation of the circle whose center is (4, 5) and touches the x-axis. Find the co-ordinates of the points at which the circle cuts y-axis.

15. Find the equation to the circle concentric with x² +y² -4 x -6 y -3 = 0 and which touches the y-axis.

16. Find the equation to the circle which is concentric with x² +y² -6 x +7 = 0 and touches the line x +y +3 = 0.

17. Find the length of the chord made by the x-axis with the circle whose center is (0, 3 a) and which touches the straight line 3 x +4 y = 37.

18. Show that 3 x -4 y +11 = 0 is a tangent to the circle x² + y² -8y +15 = 0 and find the equation of the other tangent which is parallel to the line 3 x = 4 y.

19. Find the equations of the tangents to the circle x² +y² = 25 which are parallel to the line y = 2 x +4.

20. Find the equations of the tangents to the circle x² +y² -2 x -4 y = 4 which are perpendicular to the line 3 x - 4 y -1 = 0.

Answers1. (i) one point (ii) two distinct points (iii) none2. (i) (0, 0) (ii) (-1, -2), (-1/10, 7/10) (iii) none

3.(-1, -1),

4. (i) (1, 3), (5, 7) ; 4 2 (ii) x² + y² -2 x -4 y = 0 ; 2 5

5. Intercept on x-axis = 6, intercept on y-axis = 2 14

6. 5 2 7. 4; x² +y² -2 x -4 y +1 = 08. (i) (7, 3), (2, 8) (ii) (2, -1), 3; point of contact (-1, -1), other end of diameter (5, -1)

9. (-a/ 2, a/ 2) 10. (1, 1)

11. n = ± a [l² +m²]12. (l g + m f -n)² = (l² + m²)(g² + f² -c)13. 10, 5/214. (i) 169 (x² + y² -6 x -8 y) +381 = 0 (ii) x² + y² -8 x -10 y +16 = 0; (0, 2), (0, 8)15. x² + y² -4 x -6 y +9 = 016. x² + y² -6 x -9 = 017. 8 | a | 18. 3 x -4 y +21= 0

19. 2 x - y ± 5 5 = 020. 4 x + 3 y +5 = 0, 4 x +3 y -25 = 0

Tangent and Normal to a Circle at a Point

The equation of the tangent at a point on a circle

The equation of the tangent to the circle x² +y² +2 g x +2 f y +c = 0 at the point P (x1 , y1) is xx1 +yy1 +g (x +x1) +f(y +y1) +c = 0

The equation of the normal at a point on the circle

The equation of the normal to the circle x² +y² +2 g x +2 f y +c = 0 at the point P (x1, y1) is (y1 +f) x -(x1 +g) y +(g y1 -f x1) = 0Normal at a point on the circle passes through the center of the circle.


Example

Find the equations of tangent and normal to the circle x² +y² -5 x +2 y +3 = 0 at the point (2, -3).

Solution

The given circle is x² +y² -5 x +2 +2 y +3 = 0 ... (i)The equation of the tangent to the circle (i) at the point P (2, - 3) isx. 2 + y. (-3) -5.(1/2).(x +2) +2.(1/2).(y/3) +3 = 0=> 4 x -6 y -5 x -10 +2 y -6 +6 = 0=> -x -4 y -10 = 0 => x +4 y +10 = 0The slope of the tangent at P = - 1/4=> the slope of the normal at P = 4The equation of the normal to the circle (i) at P (2, -3) isy +3 = 4 (x - 2) i.e. 4x - y -11 = 0

Exercise

1. (i) Find the equation of the tangent to the circle x² + y²= a² at the point P (x1, y1) on it.(ii) Find the equation of the normal to the circle x² + y² = a² at the point P (x1, y1) on it.

2. Find the equations of the tangent and the normal to the following circles at the given points.(i) x² +y² = 169 at (12, - 5)(ii) 4 x² +4 y² = 25 at (3/2, -2)(iii) x² + y² -4 x +2 y +3 = 0 at (1, -2)(iv) 3 x² +3 y² -4 x -9 y = 0 at the origin.

3. Find the equations of the tangent and the normal to the following circles:(i) x² +y² = 10 at the points whose abscissa is 1.(ii) x² +y² -8 x -2 y +12 = 0 at the points whose ordinate is -1.

4. Show that the tangents drawn at the points (12, - 5) and (5, 12) to the circle x² + y² = 169 are perpendicular to each other.

Answers1. (i) x x1 +y y1 = a² (ii) y1 x -x1 y = 02. (i)12 x -5 y -169 = 0; 5x +12 y = 0 (ii) 6x +8 y +25 = 0; 4x -3y = 0 (iii) x +y +1 = 0; x -y -3 = 0 (iv) 4 x + 9 y = 0; 9 x -4 y = 03. (i) x +3 y -10 = 0, x -3 y -10 = 0, 3 x -y = 0, (ii) x -2 y -7 = 0, x +2 y -1 = 0, 2 x +y -9 = 0, 2 x - y -7 = 0

Relative position of two CirclesLet S, S' be two (non-concentric) circles with centers A, B and radii r1 , r2 and d be the distance between their centers, then(i) One circle lies completely inside the other iff d < | r1 -r2 |(ii) The two circles touch internally iff d = | r1 -r2 |

(iii) The two circles intersect in two points iff d > |r1 -r2| and d < r1 +r2

(iv) The two circles touch externally iff d = r1 +r2

(v) One circle lies completely outside the other iff d > r1 +r2

Note

1. If two circles intersect, then we can solve the equations of the circles simultaneously to find the points of intersection. In particular, the equation of the common chord is given by S -S' = 0.2. If the two circles touch (internally or externally), then the equation of their common tangent is given by S -S' = 0.


Example

Show that the circles x² +y² -2 x = 0 and x² +y² +6 x -6 y +2 = 0 touch each other. Do these circles touch externally or internally? Find the point of contact and the common tangent.

Solution

The equations of the given circles areS = x² + y² -2 x = 0 ...(i)and S = x² + y²+6x -6y +2 = 0 ...(ii)Their centers are A (1, 0) and B (-3, 3), and their radii arer1 = [1² +0² -0] = 1 and r2 = [9 +9 -2] = 4 respectivelyThe distance between their centers= d = [(- 3 -1)² +(3 -0)²] = 5 = 1 +4=> d = r1 + r2

=> the given circles (i) and (ii) touch externally and the point of contact P divides [AB] internally in the ratio r1 : r2 i.e. in the ratio 1 : 4The co-ordinates of the point of contact are (1.(-3) +4.1)/(1+4), (1.3 +4.0)/(1+4) i.e. (1/5, 3/5)The equation of the common tangent is S -S' = 0=> -8 x +6 y -2 = 0 => 4 x -3 y +1 = 0

Exercise

1. Prove that the circle x² + y² -6 x -2 y +9 = 0 lies entirely inside the circle x² + y² = 18.2. Prove that the circles x² + y² -4 x +6 y +8 = 0 and x² + y² -10 x -6 y +14 = 0 touch each other

externally. Find their point of contact and also the common tangent.3. Show that the circles x² + y² +2 x -6 y +9 = 0 and x² + y² +8 x -6 y +9 = 0 touch internally.

Find their point of contact and also the common tangent.

4. Prove that the circles x² + y² -6x -2 y +1 = 0 and x² + y² +2 x -8 y +13 = 0 touch one another and find the equation of the tangent at their point of contact.

5. Show that the circles x² + y² = 2 and x² + y² -6x -6 y +10 = 0 touch each other. Do these circles touch externally or internally? Also find their point of contact.

6. Find the equation of the circle whose radius is 3 and which touches the circle x² + y² -4 x -6 y -12 = 0 internally at the point (-1, -1).

Answers3. (3, -1, x +2 y -1 = 0

3. (0, 3), x = 04. 4 x -3 y +6 = 05. Externally; (1, 1)6. 5 (x² + y²) -8x -14 y +32 = 0

Families of CirclesA collection of circles is called a family or a system of circles.Let S and S' be two intersecting (or touching) circles, then S +k S' = 0, k -1, represents a family of circles through their points (or point) of intersection.

Remarks

1. If k = -1, then the equation S +kS' = 0 reduces to S -S' = 0 which represents the common chord in case of intersecting circles or common tangent in case of touching circles.

2. The equation S +kS = 0 represents all members of the family except the member S'. If we need the member S', then take the equation of the family as S' +kS = 0.


Example

Find the equation of the family of circles passing through the points A(0,4) and B(0,-4).

Solution

Let the equation of the desired family of circles bex² +y² +2gx +2fy +c = 0 ...(i)As all these circles pass through the points A(0,4) and B(0,-4), we get0 +16 +0 +8f +c = 0 => 8f +c +16 = 0 ...(ii),0 +16 +0 -8f +c = 0 => -8f +c +16 = 0 ...(iii)On solving (ii) and (iii), we get f = 0, c = -16.Substituting these values in (i), we getx² +y² +2gx -16 =0 ...(iv)Note that for every real value of g,g² +f² -c = g² +16 > 0, therefore, (iv) represents a circle.Hence the equation x² +y² +2gx -16 = 0 for different real values of g represents the desired family of circles. It is one-parameter family of circles where g is the parameter.

Example

Find the equation of the circle which passes through the points of inter-section of x² +y² -4 = 0 and x² +y² -2x -4y +4 = 0 and touches the line x +2 y = 0.

Solution

The equations of the given intersecting circles areS = x² +y² -4 = 0, andS' = x² +y² -2 x -4y +4 = 0The equation of the common chord of these circles isS -S' = 0 => l = 2 x +4 +4 y -8 = 0The equation of the family of circles passing through the intersection of the given circles isx² +y² -4 +k (2x +4 +4y -8) = 0 ...(i)Its center is (-k,-2k) andradius = [k² +4k² +4 +8k] = [5k² +8k +4].For the particular member of the family which touch the line x +2y = 0, we have|-k +2 (-2k)|/ [1² +2²] = [5k² +8k +4]=> 5|k|/ 5 = [5k² +8k +4]=> 5 k² = 5 k²+8 k +4=> 8 k +4 = 0 => k = -1/2Substituting this value of k in (i), the equation of the required circle isx² +y² -4 - (1/2)(2x +4y -8) = 0 i.e. x² + y² -x -2y = 0.

Exercise

1. Find the equation of the family of circles passing through the origin.2. Find the equation of the family of circles passing through the origin and the point (0,1).3. Find the equation of the family of circles passing through the points A(5,0) and B(-5,0).4. Find the equation of the family of circles with radius 3 and whose centers lie on the x-axis.5. Find the equation of the family of concentric circles with center as (-4,2). Also find a member

of the family which touches the line x -y = 3.6. Show that the equation of the family of circles which touch both the co-ordinate axes can be

put into the form x² +y² ±2rx ±2ry +r² = 0, where r is radius.[Hint. Let ( , ) be center of the family. Since it touches both the axesi.e. y = 0 and x = 0, so | | = | | r => = ±r, = ±r.]

7. Find the equation of the circle which passes through the origin and the points of intersection of the circles x² +y² +2x +2y -2 = 0 and x² +y² +4x -8y +4 = 0.

8. Find the equation of the circle through the points of intersection of the circles x² +y² +2x +3y -7 = 0 and x² +y² +3x -2y -1 = 0 and through the point (1,2).

9. Find the equation of the circle which passes through the point (1,-1) and through the points of intersection of the circles x² +y² +2x -2y -23 = 0 and 3x² +3y² +12x -4y -9 = 0.

10. Find the equation of the circle passing through the point (2,3) and through the points of intersection of the circle x² +y² +3x -4y +5 = 0 and the line x -y +2 = 0.

11. Find the equation of the circle through the intersection of the circles x² +y² -8x -2y +7 = 0 and x² +y² -4x +10y +8 = 0 and having its center on the x-axis.

Answers1. x² +y² +2 gx +2 +2fy = 0, where g, f are any real numbers2. x² +y² +2gx -y = 0, where g is any real number

3. x² +y² +2fy -25 = 0, where f is any real number4. (x -h)² +y² = 9, where h is any real number5. x² +y² +8x -4y +20 -r² = 0, where r is radius; 2(x² +y²) +16x -8y -41 = 07. 3(x² +y²) +8x -4y = 08. x² +y² +4x -7y +5 = 09. 32(x² +y²) +115x -47y -226 = 010. x² +y² -9x +8y -19 = 011. 6(x² +y²) -44x +43 = 0

ConicsThe curves known as conics were named after their historical discovery as the intersection of a plane with a right circular cone.

Applonius (before 200 B.C.) realized that a conic (or conic section) is a curve of intersection of a plane with a right circular cone of two nappes, and the three curves so obtained are parabola, hyperbola and ellipse.

Let l be a fixed line and F be a fixed point not on l, and e > 0 be a fixed real number. Let |MP| be the perpendicular distance from a point P (in the plane of the line l and point F) to the line l, then the locus of all points P such that |FP| = e |MP| is called aconic.

The fixed point F is called a focus of the conic and the fixed line l is called thedirectrix associated with F. The fixed real number e (> 0) is called eccentricity of the conic.In particular, a conic with eccentricity e is called(i) a parabola iff e = 1 (ii) an ellipse iff e < 1 (iii) a hyperbola iff e > 1.

Four standard forms of the parabola

Main facts about the parabola

Equationsy²= 4ax (a>0)Right hand

y² = -4axa>0

Left hand

x² = 4aya>0

Upwards

x² = -4aya>0

Downwards

Axis y=0 y = 0 x = 0 x = 0

Directrix x +a = 0 x -a = 0 y +a = 0 y -a = 0

Focus (a, 0) (-a, 0) (0,a) (0, -a)

Vertex (0,0) (0,0) (0,0) (0,0)

Length of Latus-rectum

4a 4a 4a 4a

Equation of Latus-rectum

x -a = 0 x +a = 0 y -a = 0 y +a = 0

Focal distance of the point(x,y)

x +a a -x y +a a -y

Another definition of ellipse

An ellipse is the locus of a point in a plane, sum of whose distances from two given points F and F' (in the plane) is a constant and greater than |FF'|.Remark. The focal property of ellipse gives us a practical method of drawing an ellipse. Fasten the ends of a string of length 2a > |FF'| at two distinct points F and F'. Keep the string taught by means of a pencil placed against the string and slide it along the string, the curve thus traced is an ellipse with F and F' as its foci.

Main facts about the ellipse

Equationx²/a² + y²/b² = 1

a > b > 0x²/b² + y²/a² = 1

a > b> 0

Equation of major axis y = 0 x = 0

Length of major axis 2a 2a

Equation of minor axis x = 0 y = 0

length of minor axis 2b 2b

Vertices (a,0),(-a,0) (0, a),(0, -a)

Foci (ae, 0), (-ae,0) (0, ae), (0, -ae)

Directrices x - a/e = 0, x + a/e = 0 y - a/e = 0, y + a/e = 0

Length of Latus -rectuum 2b²/a 2a²/b

Equation of a latera-recta Centerx-ae = 0, x +ae = 0

(0,0)y -ae = 0,y +ae = 0

Focal distance of any point (x,y) a -ex, a +ex a -ey, a +ey

Another definition of hyperbola

A hyperbola is the locus of a point in a plane, the difference of whose distances from two given points F and F' is 2a (constant), and 0 < a < (1/2)|FF'.

Main facts about the hyperbola

Equationx²/a² - y²/b²= 1

a > 0,b > 0y²/a² - x²/b² = 1

a > 0,b > 0

Length of transverse axis y = 0 x = 0

Equation of transverse axis 2a 2a

Length of conjugate axis 2b 2b

Equation of conjugate axis x = 0 y = 0

Vertices (a, 0), (-a, 0) (0, a), (0, -a)

Foci (ae, 0), (-ae, 0) (0, ae), (0, -ae)

Directrices x - a/e = 0, x + a/e=0 y - a/e= 0, y + a/e =0

Length of lactus-rectum 2b²/a 2a²/b

Equation of latera-rectacenter

x -ae = 0, x + ae = 0(0,0)

y -ae = 0,y + ae = 0(0,0)

Focal Distance point(x,y) |ex -a|, |ex +a| |ey -a|,|ey +a|


Example

Find the equation of the parabola with focus at (-2, 0) and whose directrix is the line x +2y -3 = 0.

Solution

The focus of the parabola is at F(-2, 0) and directrix is the line x +2y -3 = 0.Let P(x, y) be any point on the parabola and |MP| be the perpendicular distance from P to the directrix, then by def. of parabola |FP| = |MP| (As e = 1 for parabola)So ((x +2)² +y²) = |x +2y -3|/ (1² +2²)On squaring,=> 5 ((x +2)² +y²) = (x +2 y -3)² (since |x|² = x²)=> 5 (x² +4 +4 x +y²) = x² +4 y² +9 +4 x y -6 x -12 y=> 4 x² -4 xy +y² +26 x +12 y +11 = 0,which is the required equation of the parabola.

Example

Find the equation of the ellipse whose focus is (1, -2), the corresponding directrix x -y +1 = 0 and eccentricity is 2/3.

Solution

The focus of the ellipse is at F(1, -2), the corresponding directrix is the line x -y +1 = 0 and e = 2/3.Let P (x, y) be any point on the ellipse and | MP | be the perpendicular distance from P to the directrix, then by def. of ellipse |FP| = e |MP|

=> => 9 [(x -1)² +(y +2)²] = 2 (x -y +1)²=> 9 [x² -2 x +1 +y +1 +y² +4 y +4] = 2 [x² +y² +1 -2 x y +2 x -2 y]=> 7 x² +4 x y +7 y² -22 x +40 y +43 = 0,which is the required equation of the ellipse.

Example

Find the focus, directrix and eccentricity of the conic represented by the equation 3y² = 8x.

Solution

The given equation is 3 y² = 8 x i.e. y² = (8/3) x ...(i)which is the same as y² = 4ax, so (i) represents a standard (right hand) parabola, and hence its eccentricity is 1, as e = 1 for parabola.

Also 4a = 8/3 => a = 2/3, therefore, focus is (a, 0) i.e. and the equation of directrix isx +2/3 = 0 i.e. 3 x +2 = 0.

Example

Find the locus of a point P, the sum of whose distances from the points F(-2, 3) and F'(2, 0) is constant equal to 4 units.

Solution

Here, we note that |FF'| = [(2+2)² +(y -2)²] = 5and |PF| +|PF'| = 4 (given)=> |PF| +|PF'| < |FF'|, which is not possible wherever P may be (since sum of two sides of a triangle cannot be less than third side)Therefore, the locus of P is the empty set.

Exercise

1. Write the equation of a conic with eccentricity e, focus (a, 0) and directrix y-axis.2. Write the equation of a conic with eccentricity e, focus (0, a) and directrix x-axis.3. Write the equation of the parabola with the line x +y = 0 as directrix and the point (1, 0) as

focus.4. Find the equation of the parabola whose focus is (2, -1) and directrix is x +2 y -1 = 0.5. Find the equation of the ellipse whose focus is (1, -1), the corresponding directrix is x -y +3 =

0 and e = 1/2.6. Find the equation of the hyperbola with directrix x + 2 y = 1, focus at (0, 0) and eccentricity

2.7. Find the equation to the parabola with the focus (a, b) and directrix x/a + y/b = 1.8. Find the equation to the parabola whose focus is (-2, 1) and directrix is 6 x -3 y = 8.9. Find the equation of the parabola whose focus is (5, 3) and the directrix is given by 3x -4 y

+1 = 0.also find the equation of axis.10. Find the equation to the conic section whose focus is (1, -1), eccentricity is 1/2 and the

directrix is the line x -y = 3. Is the conic section an ellipse?11. Find the eccentricity of the ellipse if:

(i) the latus-rectum is one half of its minor axis.(ii) the latus rectum is one half of its major axis.(iii) the distance between foci is equal to the length of latus-rectum.

12. Find the eccentricity, co-ordinates of the foci and the length of the latus-rectum of the ellipse 4 x² +3 y² = 36.

13. Find the co-ordinates of the foci and the ends of the latera-recta of the ellipse 16x² +9y² = 144.

14. Find the vertices, eccentricity, foci and the equations of the directrices of the hyperbola x² -y² = 1.

15. Find the lengths of axes, co-ordinates of foci, the eccentricity and the length of latus-rectum of the hyperbola 25 x² -9 y² = 225.

16. In each of the parabolas(i) y² = 3 x (ii) y² = -4 x (iii) 3 x² = 4 y (iv) x² = -12 y,find the length of latus-rectum, coordinates of focus and the equation of directrix.

17. If the parabola y² = 4 p x passes through the point (3, -2), find the length of latus-rectum and the co-ordinates of the focus.

18. Find the equation of the parabolas with vertices at the origin and satisfying the following conditions:(i) Focus at (-4, 0) (ii) Directrix y -2 = 0(iii) Passing through (2, 3) and axis along x-axis.

19. Find the equation to the ellipse referred to its axes as co-ordinates axes(i) whose major axis = 8, eccentricity = 1/2(ii) which passes through the points (-3, 1) and (2, -2).

(iii) which passes through the point (-3, 1) and has eccentricity .(iv) whose minor axis is equal to the distance between foci and whose latus-rectum is 10.

20. Find the equation of the ellipse whose eccentricity is 1/2 and whose foci are (±2, 0).

Answers1. (1 -e²) x² +y² -2 a x +a² = 02. x² +(1 -e²) y² -2 ay +a² = 03.(x -y)² -4 x +2 = 04. 4 x² -4x y +y² -18 x +14 y +24 = 05. 7 x² +2 x y +7 y² -22 x +22y +7 = 0

6. x² +8 2x y +5 y² -8 x -8y +4 = 07. a² x² -2 a b x y +b² y² -2 a³ x -2 b³ y +(a 4 - a² b² +b4) = 08. 9 x² +36 x y +36 y² +276 x -138 y -169 = 09.16 x² +24 x y +9 y² -256 x -142 y +849 = 0; 4x +3 y -29 = 010. 7 x² +2 x y +7 y² -10 x +10 y +7 = 0; Yes

11. (i) ( 3)/2 (ii) 1/ 2 (iii) ( 5 1)/2

12. e = 1/2, foci are (0, 3), (0, - 3), length of latus-rectum = 3 3

13. (0, 7), (0, - 7);

14. (1, 0), (-1, 0); 2 ; ( 2, 0), (- 2 , 0); 2 x -1 = 0, 2x +1 = 0

15. 6, 10; ( 34 , 0), (- 34, 0); ( 34)/3; 10

16. (i) 2 3 (( 3)/2, 0) ; 2 x + 3 = 0 (ii) 4; (-1, 0); x -1 = 0 (iii) 4/3 ; (0, 1/3) ; 3 y +1 = 0 (iv) 12; (0, -3); y -3 = 017. 4/3, (1/3, 0)18. (i) y² = -16 x (ii) x² = -8y (iii) 2 y² = 9 x19. (i) 3 x² +4 y² = 48 (ii) 3x² +5 y² = 32 (iii) 3 x² +5 y² = 32 (iv) x² +2y² = 10020. 3 x² +4 y² = 48

St.distance Formula Xi

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