Class- X-CBSE-Mathematics Coordinate Geometry Practice more on Coordinate Geometry Page - 1 www.embibe.com CBSE NCERT Solutions for Class 10 Mathematics Chapter 7 Back of Chapter Questions 1. Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a, b), (−a, −b) Solution: (i) We know that the distance between two points (x 1 ,y 1 ) and (x 2 ,y 2 ) is given by, �(x 1 − x 2 ) 2 + (y 1 − y 2 ) 2 Hence, the distance between (2, 3) and (4, 1) is given by, Distance = �(2 − 4) 2 + (3 − 1) 2 = �(−2) 2 + (2) 2 = √4+4 = √8 =2√2 (ii) We know that the distance between two points (x 1 ,y 1 ) and (x 2 ,y 2 ) is given by, �(x 1 − x 2 ) 2 + (y 1 − y 2 ) 2 Hence, the distance between (−5, 7) and (−1,3) is given by, Distance = � �−5 − (−1)� 2 + (7 − 3) 2 = �(−4) 2 + (4) 2 = √16 + 16 = √32 =4√2 (iii) We know that the distance between two points (x 1 ,y 1 ) and (x 2 ,y 2 ) is given by, �(x 1 − x 2 ) 2 + (y 1 − y 2 ) 2 Hence, the distance between (a, b) and (−a, −b) is given by, Distance = � �a − (−a)� 2 + �b − (−b)� 2 = �(2a) 2 + (2b) 2 = � 4a 2 + 4b 2 =2 � a 2 +b 2 2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
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Class- X-CBSE-Mathematics Coordinate Geometry
Practice more on Coordinate Geometry Page - 1 www.embibe.com
CBSE NCERT Solutions for Class 10 Mathematics Chapter 7 Back of Chapter Questions
1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (−5, 7), (−1, 3)
(iii) (a, b), (−a,−b)
Solution:
(i) We know that the distance between two points (x1, y1) and (x2, y2) isgiven by,
�(x1 − x2)2 + (y1 − y2)2
Hence, the distance between (2, 3) and (4, 1) is given by,
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Solution:
We know that the distance between two points (x1, y1) and (x2, y2) is given by,
�(x1 − x2)2 + (y1 − y2)2
Now, the distance between the points 𝐴𝐴(0, 0) and 𝐵𝐵(36, 15) is given by,
AB = �(36 − 0)2 + (15 − 0)2 = �362 + 152
AB = √1296 + 225 = √1521 = 39
Now, we are told to find distance between two towns A & B in section 7.2
It is given that B is located 36 km east and 15 km north of town A
Let us take A as origin Therefore, A will be A (0, 0) & B will be B (36, 15)
Since, we have already calculated above the distance between the points 𝐴𝐴(0, 0) and 𝐵𝐵(36, 15). Hence, we can say that the distance between the two towns A and B discussed in section 7.2 is 39 km.
3. Determine if the points (1, 5), (2, 3) and (−2,−11) are collinear.
Solution:
Three points are collinear if they lie on a same line i. e., one point lies in between the line joining any other two points.
Since the two sides are equal in length, therefore,ABC is an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees.
Using distance formula, find which of them is correct.
We can observe that the opposite sides of this quadrilateral are of equal length, but diagonals are of different lengths. Therefore, ABCD is a parallelogram and given points are vertices of a parallelogram.
7. Find the point on the 𝑥𝑥 −axis which is equidistant from(2,−5) and(−2, 9).
Solution:
We know that, for a point on the 𝑥𝑥 − axis, its 𝑦𝑦 − coordinate will be 0.
So, let the point on 𝑥𝑥 −axis be (𝑘𝑘, 0)
Distance between (𝑘𝑘, 0) and (2,−5) = �(𝑘𝑘 − 2)2 + �0 − (−5)�2
=
�(𝑘𝑘 − 2)2 + (5)2
Distance between (𝑘𝑘, 0) and (−2, 9) = �(𝑘𝑘 − (−2))2 + �0 − (9)�2
=
�(𝑘𝑘 + 2)2 + (9)2
By given condition, these distances are equal in measure.
�(𝑘𝑘 − 2)2+(5)2 = �(𝑘𝑘 + 2)2 + (9)2
⇒ (𝑘𝑘 − 2)2 + 25 = (𝑘𝑘 + 2)2 + 81
⇒ 𝑘𝑘2 + 4 − 4𝑘𝑘 + 25 = 𝑘𝑘2 + 4 + 4𝑘𝑘 + 81
⇒ 8𝑘𝑘 = 25 − 81
⇒ 8𝑘𝑘 = −56
⇒ 𝑘𝑘 = −7
Hence, the required point on the 𝑥𝑥 − axis is (−7, 0).
8. Find the values of 𝑦𝑦 for which the distance between the points P(2,−3) and Q(10,𝑦𝑦) is 10 units.
Solution:
As per the question, it is given that the distance between (2,−3) and (10, 𝑦𝑦) is 10.
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We know the section formula, which gives coordinates of point P, dividing the line segment joining A(𝑥𝑥1,𝑦𝑦1) & B(𝑥𝑥2,𝑦𝑦2) internally in the ratio m ∶ n
�m𝑥𝑥2+n𝑥𝑥1m+n
,m𝑦𝑦2+n𝑦𝑦1m+n
�,
Upon substitution of values, we get
𝑥𝑥 =2 × 4 + 3 × (−1)
2 + 3=
8 − 35
=55
= 1
𝑦𝑦 =2 × (−3) + 3 × 7
2 + 3=−6 + 21
5=
155
= 3
Hence, the coordinates of the required point are given by (1, 3).
2. Find the coordinates of the points of trisection of the line segment joining (4,−1) and (−2,−3).
Solution:
Trisection means division into three equal parts.
So, we need to find two points such that they divide the line segment in three equal parts.
Let P(𝑥𝑥1,𝑦𝑦1) and Q(𝑥𝑥2,𝑦𝑦2) are the points of trisection of the line segment joining the given points i. e. AP = PQ = QB
Therefore point P divides AB internally in ratio 1 ∶ 2
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3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in figure. Niharika runs 1
4𝑡𝑡ℎ the distance AD on the 2nd line and posts a green flag. Preet runs
15
th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Solution:
According to the question, it is given that Niharika posted the green flag (G) at 14
th the distance AD i. e. 14
× 100 = 25 m from the starting point of 2nd line.
So, coordinates of this point G are(2, 25).
Similarly, Preet posted red flag at 15th the distance AD i. e. 1
5× 100 = 20 m from
the starting point of 8th line.
So, coordinates of this point Rare(8, 20).
Now, below is the figure, which simplifies the given question by considering G as a position of green flag posted by Niharika and similarly R as a position of red flag posted by Preet.
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Now distance between these flags by using distance formula = GR
GR = �(8 − 2)2 + (20 − 25)2 = √36 + 25 = √61 m
Since it is given that Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags. Now the point at which Rashmi should post her blue flag is the midpoint of line joining these points.
Let this point be B(𝑥𝑥,𝑦𝑦).
𝑥𝑥 =2 + 8
2,𝑦𝑦 =
25 + 202
𝑥𝑥 =102
= 5, 𝑦𝑦 =452
= 22.5
So, B(𝑥𝑥,𝑦𝑦) = (5, 22.5)
So, Rashmi should post her blue flag on 5th line at a distance of 22.5 m.
4. Find the ratio in which the line segment joining the points (−3, 10) and (6,−8) is divided by (−1, 6).
Solution:
Let the ratio in which line segment joining A(−3, 10) and B(6,−8) is divided by point P(−1, 6) is k ∶ 1.
Let P(𝑥𝑥,𝑦𝑦) be the required point
We know the section formula, which gives coordinates of point P, dividing the line segment joining A(𝑥𝑥1,𝑦𝑦1)&B(𝑥𝑥2,𝑦𝑦2) internally in the ratio k ∶ 1
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So, −1 = 6k−3k+1
−k − 1 = 6k − 3
7k = 2
k =27
Similarly, we can get k using 𝑦𝑦 as follows:
6 =−8k + 10
k + 1
6k + 6 = −8k + 10
14k = 4
k =27
Hence, the required ratio is 2 ∶ 7.
5. Find the ratio in which the line segment joining A(1,−5) and B(−4, 5) is divided by the 𝑥𝑥 −axis. Also find the coordinates of the point of division.
Solution:
If the ratio in which P divides AB is 𝑘𝑘 ∶ 1, then the co-ordinates of the point P will be
�𝑘𝑘𝑥𝑥2 + 𝑥𝑥1𝑘𝑘 + 1
,𝑘𝑘𝑦𝑦2 + 𝑦𝑦1𝑘𝑘 + 1
�
Let the ratio in which line segment joining A(1,−5) and 𝐵𝐵(−4, 5) is divided by 𝑥𝑥 axis be k ∶ 1.
So, coordinates of the point of division is �−4𝑘𝑘+1𝑘𝑘+1
, 5𝑘𝑘−5𝑘𝑘+1
�
We know that 𝑦𝑦 coordinate of any point on 𝑥𝑥 −axis is 0.
Thus,5𝑘𝑘−5𝑘𝑘+1
= 0
⇒ 𝑘𝑘 = 1
Hence, 𝑥𝑥 −axis divides it in the ratio 1 ∶ 1.
Coordinates of point of division are given by,
�−4(1) + 1
1 + 1,5(1) − 5
1 + 1� = �
−4 + 12
,5 − 5
2� = �
−32
, 0�
6. If (1, 2), (4,𝑦𝑦), (𝑥𝑥, 6) and (3, 5) are the vertices of a parallelogram taken in order, find 𝑥𝑥 and 𝑦𝑦.
3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are(0,−1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:
Let vertices of the triangle be A (0,−1), B (2, 1), C (0,3)
Let D, E, F are midpoints of the sides of this triangle. Coordinates of D, E, and F are given by-
Area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD
= 212
+ 352
= 28 square units
5. You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ ABC whose vertices are A(4,−6), B(3,−2) and C(5, 2)
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⇒ 𝑥𝑥 = 1
We know that in a square all interior angles are of 90o
So, in ∆ABC
AB2 + BC2 = AC2
��(1 + 1)2 + (𝑦𝑦 − 2)2�2
+ ��(1 − 3)2 + (𝑦𝑦 − 2)2�2
= ��(3 + 1)2 + (2 − 2)2�2
⇒ 4 + 𝑦𝑦2 + 4 − 4𝑦𝑦 + 4 + 𝑦𝑦2 − 4𝑦𝑦 + 4 = 16
⇒ 2𝑦𝑦2 + 16 − 8𝑦𝑦 = 16
⇒ 2𝑦𝑦2 − 8𝑦𝑦 = 0
⇒ 𝑦𝑦(𝑦𝑦 − 4) = 0
𝑦𝑦 = 0 or 4
We know that in a square diagonals are of equal length and bisect each other at 90o. Let O be the midpoint of AC so it will also be the midpoint of BD.
Coordinate of point O = �−1+32
, 2+22�
�1 + 𝑥𝑥1
2,𝑦𝑦 + 𝑦𝑦1
2� = (1, 2)
1 + 𝑥𝑥12
= 1
1 + 𝑥𝑥1 = 2
⇒ 𝑥𝑥1 = 1 𝑦𝑦 + 𝑦𝑦1
2= 2
⇒ 𝑦𝑦 + 𝑦𝑦1 = 4
If 𝑦𝑦 = 0
𝑦𝑦1 = 4
If 𝑦𝑦 = 4
𝑦𝑦1 = 0
So, Coordinates of other vertices are (1, 0)(1,4).
5. The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are
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planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure the students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆ PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?
Solution:
(i) Taking A as origin, we will take AD as 𝑥𝑥 axis and AB as 𝑦𝑦 aixs. Now we may observes that coordinates of points P, Q and R are (4, 6), (3, 2), (6, 5)
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6. The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD
AB= AE
AC= 1
4.
Calculate the area of the ∆ADE and compare it with the area of ∆ ABC.
Solution:
Given that ADAB
= AEAC
= 14
ADAD + BD
=AE
AE + EC=
14
ADDB
=AEEC
=13
So, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1 ∶ 3.
Coordinates of the point P(𝑥𝑥,𝑦𝑦) which divides the line segment joining the points points A(𝑥𝑥1,𝑦𝑦1) and B(𝑥𝑥2𝑦𝑦2) internally in the ratio 𝑚𝑚1 ∶ 𝑚𝑚2are
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Point R divides the side CF in a ratio 2 ∶ 1.
Coordinates of R = �2×5+1×12+1
,2×7
2+1×4
2+1� = �11
3, 113�
(iv) Now we may observe that coordinates of point P, Q, R are same. So, all these are representing same point on the plane i. e. centroid of the triangle.
(v) Now consider a ∆ABC having its vertices as A(𝑥𝑥1,𝑦𝑦1), B(𝑥𝑥2, 𝑦𝑦2), and C(𝑥𝑥3,𝑦𝑦3). Median AD of the triangle will divide the side BC in two equal parts. So D is the midpoint of side BC.
Coordinates of D = �𝑥𝑥2+𝑥𝑥32
, 𝑦𝑦2+𝑦𝑦32
�
Let centroid of this triangle is O.
Point O divides the side AD in a ratio 2 ∶ 1.
Coordinate of O = �2×𝑥𝑥2+𝑥𝑥3
2 +1×𝑥𝑥12+1
,2×𝑦𝑦2+𝑦𝑦3
2 +1×𝑦𝑦12+1
� = �𝑥𝑥1+𝑥𝑥2+𝑥𝑥33
, 𝑦𝑦1+𝑦𝑦2+𝑦𝑦33
�
8. ABCD is a rectangle formed by the points A(−1,−1), B(−1, 4), C(5, 4) and D(5,−1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.