# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Written by Team Trustudies
Updated at 2021-02-11

## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1

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)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Distance formula =$\sqrt{\left({x}_{2}?{x}_{1}{\right)}^{2}+\left({y}_{2}?{y}_{1}{\right)}^{2}}$
(i) Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
d = $\sqrt{\left(4?2{\right)}^{2}+\left(1?3{\right)}^{2}}=\sqrt{{2}^{2}+\left(?2{\right)}^{2}}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$

(ii) Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get
d = $\sqrt{\left(?5+1{\right)}^{2}+\left(7?3{\right)}^{2}}=\sqrt{\left(\left(?4{\right)}^{2}+\left(4{\right)}^{2}}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$

(iii) Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get
d = $\sqrt{\left(?a?a{\right)}^{2}+\left(?b?b{\right)}^{2}}=\sqrt{\left(\left(?2a{\right)}^{2}+\left(?2b{\right)}^{2}}=\sqrt{4{a}^{2}+4{b}^{2}}=2\sqrt{{a}^{2}+{b}^{2}}$

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.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

The coordinates of point A are (0, 0) and coordinates of point B are (36, 15).
To find the distance between them, we use Distance formula:
d = $\sqrt{\left(36?0{\right)}^{2}+\left(15?0{\right)}^{2}}=\sqrt{{36}^{2}?{15}^{2}}=\sqrt{1296+225}=\sqrt{1521}=39$km

3. Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let A = (1, 5), B = (2, 3) and C = (–2, –11)
Using Distance Formula to find distance AB, BC and CA.
AB = $\sqrt{\left(2?1{\right)}^{2}+\left(3?5{\right)}^{2}}=\sqrt{{1}^{2}?\left(?2{\right)}^{2}}=\sqrt{1+5}=\sqrt{5}$
BC = $\sqrt{\left(?2?2{\right)}^{2}+\left(?11?3{\right)}^{2}}=\sqrt{\left(?4{\right)}^{2}?\left(?14{\right)}^{2}}=\sqrt{16+196}=\sqrt{212}$
CA = $\sqrt{\left(?2?1{\right)}^{2}+\left(?11?5{\right)}^{2}}=\sqrt{\left(?3{\right)}^{2}?\left(?15{\right)}^{2}}=\sqrt{9+256}=\sqrt{265}$
Since AB + AC $?$ BC, BC + AC $?$ AB and AC $?$ BC.
Therefore, the points A, B and C are not collinear.

4. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let A = (5, –2), B = (6, 4) and C = (7, –2)
Using Distance Formula to find distances AB, BC and CA.
AB = $\sqrt{\left(6?5{\right)}^{2}+\left(4+2{\right)}^{2}}=\sqrt{{1}^{2}?{6}^{2}}=\sqrt{1+36}=\sqrt{37}$
BC = $\sqrt{\left(7?6{\right)}^{2}+\left(?2?4{\right)}^{2}}=\sqrt{{1}^{2}?\left(?6{\right)}^{2}}=\sqrt{1+36}=\sqrt{37}$
CA = $\sqrt{\left(7?5{\right)}^{2}+\left(?2+2{\right)}^{2}}=\sqrt{{2}^{2}?0}=\sqrt{4}=2$
Since AB = BC.
Therefore, A, B and C are vertices of an isosceles triangle.

5. In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). 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.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $\sqrt{\left(6?3{\right)}^{2}+\left(7?4{\right)}^{2}}=\sqrt{{3}^{2}+{3}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
BC = $\sqrt{\left(9?6{\right)}^{2}+\left(4?7{\right)}^{2}}=\sqrt{{3}^{2}+{3}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
CD = $\sqrt{\left(6?9{\right)}^{2}+\left(1?4{\right)}^{2}}=\sqrt{{3}^{2}+\left(?3{\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
DA = $\sqrt{\left(6?3{\right)}^{2}+\left(1?4{\right)}^{2}}=\sqrt{{3}^{2}+\left(?3{\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
Therefore, All the sides of ABCD are equal here. … (1)
Now, we will check the length of its diagonals.
AC = $\sqrt{\left(9?3{\right)}^{2}+\left(4?4{\right)}^{2}}=\sqrt{{6}^{2}?{0}^{2}}=\sqrt{36}=6$
BD = $\sqrt{\left(6?6{\right)}^{2}+\left(1?7{\right)}^{2}}=\sqrt{{0}^{2}?{6}^{2}}=\sqrt{36}=6$
So, Diagonals of ABCD are also equal. … (2)
From (1) and (2), we can definitely say that ABCD is a square.
Therefore, Champa is correct.

6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $\sqrt{\left(1+1{\right)}^{2}+\left(0+2{\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}$
BC = $\sqrt{\left(?1?1{\right)}^{2}+\left(2?0{\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}$
CD = $\sqrt{\left(?3+1{\right)}^{2}+\left(0?2{\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}$
DA = $\sqrt{\left(?3+1{\right)}^{2}+\left(0?2{\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}$
Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.
AC = $\sqrt{\left(?1+1{\right)}^{2}+\left(2+2{\right)}^{2}}=\sqrt{0+16}=4$
BD = $\sqrt{\left(?3?1{\right)}^{2}+\left(0+0{\right)}^{2}}=\sqrt{16+0}=4$
Therefore, diagonals of quadrilateral ABCD are also equal. … (2)
From (1) and (2), we can say that ABCD is a square.

(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $\sqrt{\left(?3?3{\right)}^{2}+\left(1?5{\right)}^{2}}=\sqrt{36+16}=2\sqrt{13}$
BC = $\sqrt{\left(0?3{\right)}^{2}+\left(3?1{\right)}^{2}}=\sqrt{9+4}=\sqrt{13}$
CD = $\sqrt{\left(?1?0{\right)}^{2}+\left(?4?3{\right)}^{2}}=\sqrt{1+49}=5\sqrt{2}$
DA = $\sqrt{\left(?1+3{\right)}^{2}+\left(?4?5{\right)}^{2}}=\sqrt{4+81}=\sqrt{85}$
We cannot find any relation between the lengths of different sides.
Therefore, we cannot give any name to the figure ABCD.

(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $\sqrt{\left(7?4{\right)}^{2}+\left(6?5{\right)}^{2}}=\sqrt{9+1}=\sqrt{10}$
BC = $\sqrt{\left(4?7{\right)}^{2}+\left(3?6{\right)}^{2}}=\sqrt{9+9}=\sqrt{18}$
CD = $\sqrt{\left(1?4{\right)}^{2}+\left(2?3{\right)}^{2}}=\sqrt{9+1}=\sqrt{10}$
DA = $\sqrt{\left(1?4{\right)}^{2}+\left(2?5{\right)}^{2}}=\sqrt{9+9}=\sqrt{18}$
Here opposite sides of quadrilateral ABCD are equal. … (1)
We can now find out the lengths of diagonals.
AC = $\sqrt{\left(4?4{\right)}^{2}+\left(3?5{\right)}^{2}}=\sqrt{0+4}=2$
BD = $\sqrt{\left(1?7{\right)}^{2}+\left(2?6{\right)}^{2}}=\sqrt{36+16}=2\sqrt{13}$
Here diagonals of ABCD are not equal. … (2)
From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.

7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).
Using Distance Formula and according to given conditions we have:
$\sqrt{\left(x?2{\right)}^{2}+\left(0?\left(?5\right){\right)}^{2}}=\sqrt{\left(x?\left(?2\right){\right)}^{2}+\left(0?9{\right)}^{2}}$
=>$\sqrt{{x}^{2}+4?4x+25}=\sqrt{{x}^{2}+4+4x+81}$
Squaring both sides, we get
= > ${x}^{2}+4?4x+25={x}^{2}+4+4x+81$
=> $?4x+29=4x+85$
=>$8x=?56$
=>$x=?7$
Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)

8. Find the values of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Using Distance formula, we have
$10=\sqrt{\left(2?10{\right)}^{2}+\left(?3?y{\right)}^{2}}$
=>$10=\sqrt{\left(?8{\right)}^{2}+9+{y}^{2}+6y}$
=>$10=\sqrt{64+9+{y}^{2}+6y}$
Squaring both sides, we get
100 =$73+{y}^{2}+6y$
=> ${y}^{2}+6y+27=0$

Solving this Quadratic equation by factorization, we can write
=>${y}^{2}+9y?3y?27=0$
=>$y\left(y+9\right)–3\left(y+9\right)=0$
=>$\left(y+9\right)\left(y?3\right)=0$
=>$y=3,?9$

9. If, Q (0, 1) is equidistant from P (5, –3) and R (x, 6), find the values of x. Also, find the distances QR and PR.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

It is given that Q is equidistant from P and R. Using Distance Formula, we get
PQ = RQ
PQ=$\sqrt{\left(5?0{\right)}^{2}+\left(?3?1{\right)}^{2}}=\sqrt{\left(?5{\right)}^{2}+\left(?4{\right)}^{2}}=\sqrt{25+16}=\sqrt{41}$
QR= $\sqrt{\left(0?x{\right)}^{2}+\left(1?6{\right)}^{2}}=\sqrt{\left(?x{\right)}^{2}+\left(?5{\right)}^{2}}=\sqrt{{x}^{2}+25}$
So, $\sqrt{41}=\sqrt{{x}^{2}+25}$
On squaring both sides,
$41={x}^{2}+25$
${x}^{2}=16$
$x=±4$
Coordinates of point R will be (4,6) or (-4,6),
If R is (4,6),
QR = $\sqrt{\left(0?4{\right)}^{2}+\left(1?6{\right)}^{2}}=\sqrt{{4}^{2}+{5}^{2}}=\sqrt{16+25}=\sqrt{41}$
PR = $\sqrt{\left(5?4{\right)}^{2}+\left(?3?6{\right)}^{2}}=\sqrt{{1}^{2}+{9}^{2}}=\sqrt{1+81}=\sqrt{82}$
If R is (-4,6),
QR = $\sqrt{\left(0+4{\right)}^{2}+\left(1?6{\right)}^{2}}=\sqrt{{4}^{2}+{5}^{2}}=\sqrt{16+25}=\sqrt{41}$
PR = $\sqrt{\left(5+4{\right)}^{2}+\left(?3?6{\right)}^{2}}=\sqrt{{9}^{2}+{9}^{2}}=\sqrt{81+81}=9\sqrt{2}$

10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

It is given that (x, y) is equidistant from (3, 6) and (–3, 4).
Using Distance formula, we can write
$\sqrt{\left(x?3{\right)}^{2}+\left(y?6{\right)}^{2}}=\sqrt{\left(x?\left(?3\right){\right)}^{2}+\left(y?4{\right)}^{2}}$
=>$\sqrt{{x}^{2}+9?6x+{y}^{2}+36?12y}=\sqrt{{x}^{2}+9+6x+{y}^{2}+16?8y}$
Squaring both sides, we get
=> ${x}^{2}+9?6x+{y}^{2}+36?12y={x}^{2}+9+6x+{y}^{2}+16?8y$
=> $?6x?12y+45=6x?8y+25$
=>$12x+4y=20$
=>$3x+y=5$

## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2

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

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the points be P(x,y), Using the section formula

$P\left(x,y\right)=\left[\frac{m{x}_{2}+n{x}_{1}}{m+n},\frac{m{y}_{2}+n{y}_{1}}{m+n}\right]$

$x=\frac{2\left(4\right)+3\left(?1\right)}{2+3}=\frac{8?3}{5}=1$
$y=\frac{2\left(?3\right)+3\left(7\right)}{2+3}=\frac{?6+21}{5}=3$
Therefore the point is(1,3)

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

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let P (x1, y1) and Q (x2, y2) 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 the ratio 1:2.
x1 = $\frac{\left(1×\left(?2\right)+2×4\right)}{1+2}=\frac{\left(?2+8\right)}{3}=\frac{6}{3}=2$
y1 = $\frac{\left(1×\left(?3\right)+2×?1\right)}{1+2}=\frac{\left(?3?2\right)}{3}=\frac{?5}{3}$
$?$P = (x1, y1) = P(2, $\frac{?5}{3}$)
Point Q divides AB internally in the ratio 2:1.
x2 = $\frac{\left(2×\left(?2\right)+1×4\right)}{2+1}=\frac{\left(?4+4\right)}{3}=0$
y2 = $\frac{\left(2×\left(?3\right)+1×?1\right)}{2+1}=\frac{\left(?6?1\right)}{3}=\frac{?7}{3}$
The coordinates of the point Q(x2 ,y2 ) = (0, $\frac{?7}{3}$ )

Q.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 following figure. Niharika runs $\frac{1}{4}$ th the distance AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$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 segment joining the two flags, where should she post her flag?

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

The green flag is $\frac{1}{4}th$ of total distance =
$\frac{1}{4}×100=25$ m in second line
$?$ the coordinate of green flag = (2 , 25 )
Similarly coordinate of red flag = (8 , 20)
Distance between red and green flag
= $\sqrt{\left(8?2{\right)}^{2}+\left(20?25{\right)}^{2}}=\sqrt{{6}^{2}+\left(?5{\right)}^{2}}=\sqrt{36+25}=\sqrt{61}m$
Now blue flag is posted at the mid point of two flag
Then let, the coordinate of blue flag = (x , y )
$?\left(x,y\right)=\left(\frac{\left(8+2\right)}{2},\frac{20+25}{2}\right)=\left(5,\frac{45}{2}\right)=\left(5,22.5\right)$
$?$ the blue flag is posted in fifth line at a distance of 22.5 m

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

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the required ratio be k : 1
By section formula ,
x = $\frac{m1x2+m2x1}{m1+m2}$
==> -1 = $\frac{k×6+1×?3\right)}{k+1}$
==> -k-1 = 6k -3
==> 7k = 2
==> k = $\frac{2}{7}$
The required ratio = k : 1 $\frac{2}{7}$ : 1 = $\frac{2}{7}×7:1×7$ = 2 : 7

Q.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.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let p(x , 0) be the point which devides the line segment joining A(1, -5 ) and B(-4,5 ) in the ratio m : 1
Then the using section formula
(x , 0 ) = $\left(\frac{m×\left(?4\right)+1×1}{m+1},\frac{m×5+1×\left(?5\right)}{m+1}\right)$
==> 0 = $\frac{m×5+1×\left(?5\right)}{m+1}$
==> 0 = 5m - 5
==> m = 5 /5 = 1
Hence the required ratio is 1 : 1
Since the ratio is 1 : 1 , so P is the mid point
$?$ x = $\frac{m×\left(?4\right)+1×1}{m+1}$
x = $\frac{1×\left(?4\right)+1×1}{1+1}=\frac{?4+1}{2}=\frac{?3}{2}$
$?\left(\frac{?3}{2},0\right)$ is the required point

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

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Mid point of AC = Mid point BD
==> $\frac{x+1}{2},\frac{6+2}{2}=\frac{4+3}{2},\frac{y+5}{2}$
==> $\frac{x+1}{2}=\frac{7}{2}$
==> x +1 = 7
==> x = 7-1= 6
==> $\frac{y+5}{2}=\frac{8}{2}$
==> y+5 = 8
==> y = 8 -5 = 3

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

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the coordinates of point A be (x, y).
Mid-point of AB is (2, – 3), which is the centre of the circle.
Coordinate of B = (1, 4)
(2, -3) =$\left(\frac{x+1}{2},\frac{y+4}{2}\right)$
$\frac{x+1}{2}=2==>x+1=4==>x=4?1=3$
and$\frac{y+4}{2}=?3==>y+4=?6==>y=?6?4=?10$
$?$ x = 3 and y = -10
The coordinates of A(3,-10).

Q.10 Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order.
[Hint: Area of a rhombus = 1/2(product of its diagonals)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let A(3, 0), B (4, 5), C( – 1, 4) and D ( – 2, – 1) are the vertices of a rhombus ABCD.
Length of diagonal AC
$=\sqrt{\left[3?\left(?1\right){\right]}^{2}+\left(0?4{\right)}^{2}}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$
Length of diagonal BD
$=\sqrt{\left[\left(?2\right)?4{\right]}^{2}+\left(?1?5{\right)}^{2}}=\sqrt{\left(?6{\right)}^{2}+\left(?6{\right)}^{2}}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}$
Area of rhombus = $\frac{1}{2}$ × product of diagonals
= $\frac{1}{2}×4\sqrt{2}×6\sqrt{2}=24$sq unit

## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3

1. Find the area of the triangle whose vertices are:

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

We know that formula for area of a triangle whose vertices are $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right),\left({x}_{3},{y}_{3}\right)$ is,
=

(i) So, here

So, area of triangle =

? Area of triangle is $\frac{21}{2}$ sq. units.

(ii) Similarly, here
So, area of triangle =

? Area of triangle is $32$ sq. units.

2. In each of the following find the value of ‘k’, for which the points are collinear.

(i) (7, -2), (5, 1), (3, -k)

(ii) (8, 1), (k, -4), (2, -5)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

We know that for collinear points area of triangle = 0 ,i.e.,

(i)
=>
=>
=>

(ii)
=>
=>
=>

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.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

The vertices of the triangle ABC are A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides AB,BC,CA of this triangle ABC respectively.
Coordinates of D, E, and F are given by :

D = = (1,0)
E = ( ) = (1,2)
F = ( ) = (0,1)

So, Area of

? Area of ? DEF = 1 sq. units

Area of ? ABC =

? Area of ? ABC = 4 sq. units

So, .

4. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Draw a line from B to D

Now we have,
Area of quad. ABCD = | Area of ? DAB | + | Area of ? BCD |

So,
Area of ? DAB = = = $\frac{23}{2}$ sq. units

Similarly, Area of ? BCD = = $\frac{33}{2}$ sq. units

? Area of quad. ABCD = = sq. units

5. You have studied in Class IX 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).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

The vertices of triangle ABC are A(4,-6), B (3, – 2) and C (5, 2).

Let D be the mid-point of side BC of ? ABC. Therefore, AD is the median in ? ABC.

So, coordinates of D are ( )
Therefore, coordinates of D = (4,0)

Now,
Area of ? ABD =
=
= sq. units

Similarly,
Area of ? ACD =
=
= sq. units

? Clearly, Area of ? ABD = Area of ? ACD = 3 sq. units.

Hence, median of a triangle divides it into two triangles of equal areas.

## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4

1. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry