NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.2 Coordinate Geometry

Download NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.2 – Coordinate Geometry. This Exercise contains 10 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.

NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.2 Coordinate Geometry

NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.2 Coordinate Geometry

1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.

Using section formula
x = $\cfrac { 3\times (-1)+2\times (4) }{ 2+3 }$ = $\cfrac { -3+8 }{ 5 }$ = $\cfrac { 5 }{ 5 }$ = 1
y = $\cfrac { 3\times (7)+2\times (-3) }{ 2+3 }$ = $\cfrac { 21-6 }{ 5 }$ = $\cfrac { 15 }{ 5 }$ = 3
Therefore, the coordinates of the point are (1, 3).

2. Find the coordinate of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Let A and B denote the points (4, -1) and (-2, -3) respectively and P and Q be the points of trisection of the line segment AB such that P is closer to A and Q is closer to B
AP = PQ = QB     …(i)

Ratio in which P divides AB = $\cfrac { AP }{ PB }$ = $\cfrac { AP }{ PQ+QB }$
= $\cfrac { AP }{ AP+AP }$ (Substituting values from equation (i))
= $\cfrac { 1 }{ 2 }$ = 1:2
Similarly, the ratio in which Q divides AB = 1:2
∴ Coordinates of P = $\left( \cfrac { 2\times (4)+1\times (-2) }{ 2+1 } ,\cfrac { 2\times (-1)+1\times (-3) }{ 2+1 } \right)$
= $\left( \cfrac { 8-2 }{ 3 } ,\cfrac { -2-3 }{ 3 } \right)$
= $\left( \cfrac { 6 }{ 3 } ,\cfrac { -5 }{ 3 } \right)$
= $\left( 2 ,\cfrac { -5 }{ 3 } \right)$
Coordinates of Q = $\left( \cfrac { 1\times (4)+2\times (-2) }{ 1+2 } ,\cfrac { 1\times (-1)+2\times (-3) }{ 1+2 } \right)$
= $\left( \cfrac { 4-4 }{ 3 } ,\cfrac { -1-6 }{ 3 } \right)$
= $\left( \cfrac { 0 }{ 3 } ,\cfrac { -7 }{ 3 } \right)$
=$\left( 0,\cfrac { -7 }{ 3 } \right)$

3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the figure. Niharika runs $\cfrac { 1 }{ 4 }$th the distance AD on the 2nd line and posts a green flag. Preet runs $\cfrac { 1 }{ 5 }$ 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 segments joining the two flags, where should she post her flag?

Assume a coordinate system fixed in the ground ABCD with A as origin, AB as x-axis and AD as y-axis

∵ Niharika runs on 2nd line
∴ x-coordinate of Niharika = 2 m

∵ Preet runs on 8nd line
∴ x-coordinate of Preet = 8 m

∵ Niharika posts the green flag after running $\cfrac { 1 }{ 4 }$ of the AD
∴ y-coordinate of green flag = $\cfrac { AD }{ 4 }$ = $\cfrac { 100 }{ 4 }$ = 25 m

∵ Preet posts the red flag after running $\cfrac { 1 }{ 5 }$ of the AD
∴ y-coordinate of red flag = $\cfrac { AD }{ 5 }$ = $\cfrac { 100 }{ 5 }$ = 20 m

Coordinates of green flag = (2, 25)
Coordinates of red flag = (8, 20)

Using distance formula
Distance between the two flags
=$\sqrt { (2-8)^{ 2 }+(25-20)^{ 2 } }$
= $\sqrt { (-6)^{ 2 }+5^{ 2 } }$
=$\sqrt { 36+25 }$
=$\sqrt { 61 }$m

∵ Blue flag has to be posted mid-way between the red and green flag
∴ The coordinates of blue flag will be the coordinates of mid-point of the line segment joining the green and red flag

Using section formula
Coordinates of blue flag
= $\left( \cfrac { 2+8 }{ 2 } ,\cfrac { 25+20 }{ 2 } \right)$
= $\left( \cfrac { 10 }{ 2 } ,\cfrac { 45 }{ 2 } \right)$
= (5, 22.5)
Therefore, the blue flag has to be posted on the 5th line at a distance of 22.5 m from the line AB.

4. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).

Let the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6) in the ratio m:1
∴ (-1, 6) = $\left( \cfrac { -3+6m }{ m+1 } ,\cfrac { 10-8m }{ m+1 } \right)$
From x-coordinate
-1 = $\cfrac { -3+6m }{ m+1 }$
-m – 1 = -3 + 6m
3 – 1 = 7m
m = $\cfrac { 2 }{ 7 }$
From y-coordinate
6 = $\cfrac { 10-8m }{ m+1 }$
6m + 6 = 10 – 8m
14m = 4
m = $\cfrac { 2 }{ 7 }$
Therefore, line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6) in the ratio 2:7

5. Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Let the line segment joining A(1, -5) and B(-4, 5) is divided by the point P(x, y) in the ratio m:1
∴(x,y)=$\left( \cfrac { 1\times (1)+m\times (5) }{ m+1 } ,\cfrac { 1\times (-5)+m\times (5) }{ m+1 } \right)$…(i)
∵ P lies on x-axis
∴ y-coordinate = 0
$\cfrac { -5+5m }{ m+1 }$ = 0
-5 + 5m = 0
5m = 5
m=1
Putting value in equation (i)
(x, y) = $\left( \cfrac { 1-4 }{ 1+1 } ,\cfrac { -5+5 }{ 1+1 } \right)$
(x, y) = $\left( \cfrac { -3 }{ 2 } ,\cfrac { 0 }{ 2 } \right)$
(x, y) = $\left( \cfrac { -3 }{ 2 } ,0 \right)$
Therefore, the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis at the point ($\cfrac { -3 }{ 2 }$ , 0) in 1:1

6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Let A, B, C and D denote the points (1, 2), (4, y), (x, 6) and (3, 5) respectively.
∵ Diagonals of a parallelogram bisect each other
∴ Coordinates mid-point of AC = Coordinates of mid-point of BD
$\left( \cfrac { 1+x }{ 2 } ,\cfrac { 2+6 }{ 2 } \right)$ = $\left( \cfrac { 3+4 }{ 2 } ,\cfrac { 5+y }{ 2 } \right)$
$\left( \cfrac { 1+x }{ 2 } ,\cfrac { 8 }{ 2 } \right)$ = $\left( \cfrac { 7 }{ 2 } ,\cfrac { 5+y }{ 2 } \right)$
From the x-coordinate
$\cfrac { 1+x }{ 2 }$ = $\cfrac { 7 }{ 2 }$
1 + x = 7
x = 6
From y-coordinate
$\cfrac { 8 }{ 2 }$ = $\cfrac { 5+y }{ 2 }$
8 = 5 + y
y = 3
Therefore, the values of x is 3 and y is 6.

7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

Let the coordinates of A be (x, y)
∵ AB is the diameter
∴ The centre of the circle is the mid-point of the line segment joining A and B
Using section formula
(2, -3) = $\left( \cfrac { x+1 }{ 2 } ,\cfrac { y+4 }{ 2 } \right)$
2 = $\cfrac { x+1 }{ 2 }$
4 = x + 1
x = 3
-3 = $\cfrac { y+4 }{ 2 }$
-6 = y + 4
y = -10
Therefore, the coordinates of A are (3, -10)

8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = $\cfrac { 3 }{ 7 }$ AB and P lies on the line segment AB.

Given, AP = $\cfrac { 3 }{ 7 }$AB  …(i)
AP + PB = AB
PB = AB – AP
PB = AB – $\cfrac { 3 }{ 7 }$ AB   (From equation (i))
PB = $\cfrac { 4 }{ 7 }$ AB
The ratio in which P divides the line segment AB = $\cfrac { AP }{ PB }$ = $\cfrac { \cfrac { 3 }{ 7 } AB }{ \cfrac { 4 }{ 7 } AB }$ = $\cfrac { 3 }{ 4 }$
Using section formula
P = $\left( \cfrac { 4\times (-2)+3\times (2) }{ 3+4 } ,\cfrac { 4\times (-2)+3\times (-4) }{ 3+4 } \right)$
= $\left( \cfrac { -8+6 }{ 7 } ,\cfrac { -8-12 }{ 7 } \right)$
= $\left( \cfrac { -2 }{ 7 } ,\cfrac { -20 }{ 7 } \right)$
Therefore, the coordinates of P are $\left( \cfrac { -2 }{ 7 } ,\cfrac { -20 }{ 7 } \right)$

9. Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Let P, Q and R be the points which divide the line segment AB into four equal parts
∴ AP = PQ = QR = RB     …(i)

Ratio in which Q divides AB = $\cfrac { AQ }{ QB }$ = $\cfrac { AP+PQ }{ QR+RB }$
=$\cfrac { AP+AP }{ AP+AP }$   (From equation (i))
= 1
∴ Q is the mid-point of line segment AB
Q = $\left( \cfrac { -2+2 }{ 2 } ,\cfrac { 2+8 }{ 2 } \right)$  (Using section formula)
= $\left( \cfrac { 0 }{ 2 } ,\cfrac { 10 }{ 2 } \right)$
= (0, 5)
Ratio in which P divides AQ = $\cfrac { AP }{ PQ }$
= $\cfrac { AP }{ AP }$  (From equation (i))
= 1
∴ P is the mid-point of line segment AQ
Similarly, R is the mid-point of QB
∴ P = $\left( \cfrac { -2+0 }{ 2 } ,\cfrac { 2+5 }{ 2 } \right)$ = $\left(-1 ,\cfrac { 7 }{ 2 } \right)$       (Using section formula)
R = $\left( \cfrac { 0+2 }{ 2 } ,\cfrac { 5+8 }{ 2 } \right)$ = $\left(1 ,\cfrac { 13 }{ 2 } \right)$      (Using section formula)
The coordinates of the points that divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts are (-1, $\cfrac { 7 }{ 2 }$), (0, 5) and (1, $\cfrac { 13 }{ 2 }$).

10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

Let A, B, C and D denote the points (3, 0), (4, 5), (-1, 4) and (-2, -1) respectively
Length of diagonals
AC = $\sqrt { (3-(-1))^{ 2 }+(0-4)^{ 2 } }$ = $\sqrt { 4^{ 2 }+(-4)^{ 2 } }$ = $\sqrt { 2\times 4^{ 2 } }$ = $4\sqrt { 2 }$
BD = $\sqrt { (4-(-2))^{ 2 }+(5-(-1))^{ 2 } }$ = $\sqrt { 6^{ 2 }+6^{ 2 } }$ = $\sqrt { 2\times 6^{ 2 } }$ = $6\sqrt { 2 }$
Area of rhombus = $\cfrac { 1 }{ 2 }$×(Product of length of diagonals)
= $\cfrac { 1 }{ 2 }$×AC×BD
= $\cfrac { 1 }{ 2 } \times 4\sqrt { 2 } \times 6\sqrt { 2 }$
= $\cfrac { 1 }{ 2 } \times 4\times 6\times 2$
= 24 sq. units

NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.2 – Coordinate Geometry, has been designed by the NCERT to test the knowledge of the student on the topic – Section Formula

Download NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.2 – Coordinate Geometry