Modul Sal Chap 6 Coordinate Geometry Final

8/2/2019 Modul Sal Chap 6 Coordinate Geometry Final

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MODUL SAL : FASA 1

ADDITIONAL MATHEMATICS

CHAPTER 6 : COORDINATE GEOMETRY

1Distance, s =

2 Midpoint (x,y) = , 3 Area of triangle

= | |

4 A point dividing a segment of a line

)

Examples :

Q1:

A point P(x, y) moves such that it is always equidistant from points M(1, 4) and N(2,3).

Find the equation of the locus ofP.

Ans:

PM= PN

The equation of the locus ofP is

x22x + 1 +y

28y + 16 =x

24x + 4 +y

2+ 6y + 9

2x14y + 4 = 0

x

7y + 2 = 0

2. Diagram below shows three points,A, B and C, on a straight line.

Diagram

Given 4AB =AC, find the coordinates of point C.

Ans :


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The coordinates of point C are (14, 9).

3.

Diagram

In Diagram above, the straight lineAB intersects the y-axis and thex-axis at pointsA and B

respectively.

Find the equation of the perpendicular bisector ofAB.

Ans :


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EXERCISE SET 1 :

1.

Diagram

Diagram above shows two straight lines, PQ and PR, which are perpendicular to each other at

point P. Given the equation of the straight line PQ is 3x+ 4y 12 = 0.

Find the equation of the straight line PR.

[ans : 2. Solutions by scale drawing will not be accepted.

In Diagram below, the straight lineAB has an equation 4y= 5x

8.AB intersects the y-axis at point B. Point P lies onAB such

that BP : PA = 2 : 1.

Diagram

Find

(a) the coordinates ofP.

(b) the equation of the straight line that passes through P and


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is perpendicular toAB

[ans : a.P(4,3), b. 5y + 4x = 31]

3. Solution by scale drawing is not accepted.

Diagram below shows a straight line KL which is perpendicular to a straight line ML at point

L. Given the equation ofKL is 3y+ 2x 6 = 0.

Diagram

(a) Find the equation ofML.

(b) IfML is produced to meet thex-axis at point N, where NL = ML, find the coordinates of

point M [ans : y = 3/2 x + 2, b. M(4/3,4)]


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EXERCISE SET 2 :


Diagram below shows a straight lineAB.

Diagram

(a) Write the equation of the straight lineAB.

(b) Find the coordinates of point C

(c) Point CdividesAB in the ratio m : n.

Find the ratio m : n

[ans : a. y = -x + 2, b. (0,2), c . 4:3]

5. The straight line y+ 2x 4 = 0 is a tangent to the curve y=x

3

+ 3x

2

11x+ 9 at a point P.

(a) Find the gradient of the tangent at point P.

(b) Find the coordinates of point P.

(c) Find the equation of another tangent which is parallel to the tangent at point P.

[ans : a. -2 , b. (1,2), c. y = -2x +36]

6. Solution by scale drawing will not be accepted.

In Diagram below, the straight lineAB has an equation y+ 2x+ 8 = 0.

AB intersects thex-axis at pointA and intersects the y-axis at point B.


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Diagram

Point P lies onAB such thatAP : PB = 1 : 3.

Find

(a) the coordinates ofP.

(b) the equation of the straight line that passes through P and is perpendicular toAB.

[ans : a. (-3,-2), b. 2y = x 1]


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EXERCISE SET3 :

7. Solution by scale drawing will not be accepted.

Diagram below shows the straight lineACwhich intersects the y-axis at point B.

Diagram

The equation ofACis 3y= 2x 15.

Find

(a) the equation of the straight line which passes through pointA and is perpendicular to

AC.

(b) (i) the coordinates ofB.(ii) the coordinates ofC, givenAB : BC= 2 : 7.

[ans : a. 3x + 2y + 23, b. i. (0, -5), ii. (21/2,2)]

8. A point A moves such that it is always at a constant distance from the point R(0, 5). The

locus of pointA passes through points P(12, 0) and Q(13, k).

(a) (i) Calculate the distance of pointA from the point R(0, 5).

(ii) Find the equation of the locus of pointA.

(iii) Find the value ofk.

(b) Show that RQ is parallel to thex-axis.

(c) Calculate the area of triangle PQR.

[ans : a. i. 13 units, ii. X2+y2 10y -144=0, iii.5, c. 32.5 unit2]


Diagram below shows a trapezium PQRS. Line PS is perpendicular to line PQ, which


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intersects the y-axis at pointA. The equation ofPS is and the equation ofPQ is 4y

= kx+ 26, where kis constant.

Diagram

(a) (i) Find the value ofk.

(ii) Find the coordinates of pointA.

(b) Given PA :AQ = 1 : 3, find

(i) the coordinates of point Q.

(ii) the equation of the straight lineQR.

(c) A point T(x, y) moves such that 2TP = TQ.

Find the equation of the locus ofT.

[ans : a. i. k= 6, ii. (0, 6.5), b. i. (9, 20), ii. Y = -2/3 x +26, c. 3x2 + 3y2 +42x + 24y 429 = 0]


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EXERCISE SET 4 :


Diagram

Diagram above shows a right-angled triangleABC.

(a) Find the value ofk.

(b) A point P moves such thatAPC= 90.Find the equation of the locus of point P.

(c) If the locus of point P intersect the y-axis at points B and D, find the coordinates of point

D.

(d) Calculate the area of the quadrilateralABCD.

[ans : a. k = 7, b. x2 +y2 + 7x 15y + 26 =0, c. D(0,13), d. 71.5 unit2]


Diagram below shows a triangleABC.

Diagram


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(a) Given the area of triangleABCis 37.5 unit2.

Find the value ofh.

(b) Find the coordinates of point N which lies on the line BCsuch that 2BN = NC.

(c) Determine whether the straight linesAN and BCare perpendicular to each other.

(d) A point P moves such that its distance from point B is always twice its distance from

point C.

Find the equation of the locus ofP

[ans: a. h = 7, b. (-1,-1), d. x2 +y2 -22x +20y +121 =0]


Diagram

Diagram above shows a straight lineAB which is perpendicular to a straight line BCat point

B.

(a) Find the value oft.

(b) The straight lineAB is produced to a point D such thatAB : BD = 1 : 2. Find the

coordinates of point D.

(c) A point P moves such that it passes through pointsA, B and C.

Find the equation of the locus ofP.

[ans : a. 5, b. (-5,13), c. x2 + y2 +4x -4y 2 =0]


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EXERCISE SET 5 :


Diagram

Diagram above shows a triangle PQR.

(a) Find the equation of the straight line QR.

(b) Find the coordinates of point R.

(c) Show that PQR is an isosceles triangle.

[ans : a. 2y = x + 6, b. (0,3), c. QR=PQ = ]14. Solution by scale drawing is not accepted.

Diagram below shows a triangle PQR. Point S lies on PR such that PS : SR = 2 : 3. The

equation of the straight line PSR is 2y 3x+ 6 = 0.

Diagram

(a) Find the coordinates of points P and S.

(b) Given the area of triangleQRS is 49.5 unit2.

Find the value ofk.

(c) Find the equation of the straight line PQ in the intercept form.


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(d) A point M(x, y) moves such that MQ = 2MS.

Find the equation of the locus ofM.

[ans : a. P(0, -3), S(4,3), b. k = 13, c. d. 3x2 +3y2 6x -24y 69 =0]


Diagram below shows a triangle OAB.

Diagram

(a) A point Clies on OB such that OC:CB = 1 : 2.

Find the values ofp and q.

(b) Calculate the area, in unit2, of the triangle OAB.

(c) A point P moves such that it is always equidistant from points B and C.

(i) Find the equation of the locus of point P.(ii) Hence, determine whether the locus of point P intersects the y-axis.

[ans: a. p = -6, q = 3, b. 10 unit2 , c. i. 2y 3x + 26 =0, ii. Y = -13, yes]


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ANSWER MODUL SAL FASA 1 : coodinate goemetry

1.

2. (a) P(4, 3)

(b) 5y+ 4x= 31

3. (a)

(b)


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4. (a) The equation of the straight line AB is

(b) Equation of AB: y =x + 2

At point C, x = 0.

y = 0 + 2 = 2

The coordinates of point C are (0, 2).

(c)


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Thus, the ratio is 4 : 3.

5. (a) y+ 2x 4 = 0

y=2x+ 4

Gradient of the tangent =2

(b) y=x3 + 3x2 11x+ 9

= 3x + 6x 11 =2

3x2 + 6x 9 = 0

x2 + 2x 3 = 0

(x 1)(x+ 3) = 0

x= 1 or3

Whenx= 1,

y= 13 + 3(1)2 11(1) + 9

= 2

Whenx=3,

y= (3)3 + 3(3)2 11(3) + 9

= 42

Substitute (1, 2) into y=2x+ 4.

2 =2(1) + 4

2 = 2

Substitute (3, 42) into y=2x+ 4.

42 =2(3) + 4


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42 10

Thus, the coordinates of point P are (1, 2).

(c) The equation of another tangent which is parallel to the tangent at point P is

y 42 =2(x+ 3)

y 42 =2x 6

y=2x+ 36

6. (a) y+ 2x+ 8 = 0

On thex-axis, y= 0.

2x+ 8 = 0

2x=8

x=4

The coordinates ofA are (4, 0).

On the y-axis,x= 0.

y+ 8 = 0

y=8

The coordinates ofB are (0,8).

AP : PB = 1 : 3

Coordinates ofP

(b)Gradient ofAB = =2

Gradient of the line perpendicular toAB

=

The equation of the line passes through P and is perpendicular toAB is


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y+ 2 = (x+ 3)

2y+ 4 =x+ 3

2y=x 1

7. (a) Equation ofAC: 3y= 2x 15

y=

x 5

Gradient ofAC=

Gradient of the line that is perpendicular to

AC=

The equation of the line that passes throughA(3,7) and is perpendicular toACis

y+ 7 = (x+ 3)

2y+ 14 =3x 9

3x+ 2y+ 23 = 0

(b) (i) The equation ofACis y= x 5.

At point B,x= 0.

y= (0) 5 =5

Hence, the coordinates ofB are (0,5).

(ii)

Let the coordinates ofCbe (h, k).


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= 0

2h 21 = 0

2h = 21

h =

=5

2k 49 =45

2k= 4

k= 2

Hence, the coordinates ofCare ( , 2).

8. (a) (i)

=

(ii) Let the coordinates of point A be (x, y).

AR = 13 units

The equation of the locus of

point A is

(x 0)2 + (y 5)

2 = 132

x2 + y2 10y+ 25 = 169

x2 + y2 10y 144 = 0

(iii) The locus of point A passes through Q(13, k).

Hence,

132 + k2 10k 144 = 0

k2 10k+ 25 = 0


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(k 5)(k 5) = 0

k= 5

(b) Gradient of RQ = = 0

Hence, RQ is parallel to the x-axis.

(c) Area of PQR

9. (a) (i) k= 6 (ii)

(b) (i) Q(9, 20)

(ii)

(c) 3x2 + 3y2 + 42x+ 24y 429 = 0

10. (a)

(b)


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(c) At points B and D, x = 0.

y2 15y + 26 = 0

(y 13)(y 2) = 0

y = 13 or y = 2

The coordinates of point D are (0, 13).

(d)

11. (a)

From the diagram, h = 7.

(b) Given 2BN = NC

(c)

Thus, the straight lines AN and BC are perpendicular to each other.

(d)Let the coordinates of point P be

(x, y).


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Given PB = 2PC

12. (a)

(b) Let the coordinates of point D

Thus, the coordinates of point D

are (5, 13).

(c) The locus of point P is a circle with AC as its diameter.

APC forms a right-angled triangle.

Let the coordinates of point P be (x, y).

CP is perpendicular to AP.

Gradient of CP Gradient of AP =1

(y 3)(y 1) = (x + 5)(x 1)

y2 4y + 3 = (x2 + 4x 5)

x2 + y2 + 4x 4y 2 = 0

Thus, the equation of the locus

of P is x2 + y2 + 4x 4y 2 = 0.


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13.(a) Gradient ofPQ = =2

PQ is perpendicular to QR.

Gradient ofQR =

The equation ofQR is

y 5 = (x 4)

2y 10 =x 4

2y=x+ 6

(b) At point R,x= 0

2y= 0 + 6

y= 3

The coordinates of point R are (0, 3).

(c)

14. (a) 2y 3x+ 6 = 0

At point P,x= 0

2y 3(0) + 6 = 0

2y=6

y=3

The coordinates of point P are (0,3).


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Let the coordinates of point S be (x, y).

The coordinates of point S are (4, 3).

(b)

Based on the diagram, k= 13.

(c) The equation ofPQ is

(d) MQ = 2MS

The equation of the locus ofM is

x2 26x+ 169 + y

2 = 4(x2 8x+ 16 + y2 6y+ 9)

= 4x2 32x+ 4y2 24y+ 100

3x2 + 3y2 6x 24y 69 = 0


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15.

The equation of the locus of point P is

2y 3x+ 26 = 0.

(ii) At the y-axis,x= 0.

2y 3(0) + 26 = 0

2y=26

y=13

The locus of point P intersects the y-axis at y=13.

Modul Sal Chap 6 Coordinate Geometry Final

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