NCERT solution for class 10 maths coordinate geometry ( Chapter 7)

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Solution for 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)

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Answer :

Distance formula =\(\sqrt { (x_2 - x_1)^2 + (y_2 - y_1)^2}\)
(i) Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
d = \(\sqrt { (4 - 2)^2 + (1 - 3)^2} = \sqrt{ 2^2 + (-2) ^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\)

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

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

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.

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Answer :

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 = \(\sqrt{(36-0)^2 + (15 - 0)^2} = \sqrt{36^2 - 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39\)km

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

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Answer :

Let A = (1, 5), B = (2, 3) and C = (–2, –11)
Using Distance Formula to find distance AB, BC and CA.
AB = \(\sqrt{(2-1)^2 + (3 - 5)^2} = \sqrt{1^2 - (-2)^2} = \sqrt{1 + 5} = \sqrt{5} \)
BC = \(\sqrt{(-2-2)^2 + (-11 - 3)^2} = \sqrt{(-4)^2 - (-14)^2} = \sqrt{16 + 196} = \sqrt{212}\)
CA = \(\sqrt{(-2-1)^2 + (-11 - 5)^2} = \sqrt{(-3)^2 - (-15)^2} = \sqrt{9 + 256} = \sqrt{265}\)
Since AB + AC \(\ne\) BC, BC + AC \(\ne\) AB and AC \(\ne\) 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.

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Answer :

Let A = (5, –2), B = (6, 4) and C = (7, –2)
Using Distance Formula to find distances AB, BC and CA.
AB = \(\sqrt{(6-5)^2 + (4 + 2)^2} = \sqrt{1^2 - 6^2} = \sqrt{1 + 36} = \sqrt{37}\)
BC = \(\sqrt{(7-6)^2 + (-2 - 4)^2} = \sqrt{1^2 - (-6)^2} = \sqrt{1 + 36} = \sqrt{37}\)
CA = \(\sqrt{(7-5)^2 + (-2 + 2)^2} = \sqrt{2^2 - 0} = \sqrt{4} = 2\)
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.

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Answer :

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 = \(\sqrt{(6-3)^2 + (7 - 4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
BC = \(\sqrt{(9- 6)^2 + (4 - 7)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
CD = \(\sqrt{(6-9)^2 + (1 - 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
DA = \(\sqrt{(6-3)^2 + (1 - 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
Therefore, All the sides of ABCD are equal here. … (1)
Now, we will check the length of its diagonals.
AC = \(\sqrt{(9-3)^2 + (4 - 4)^2} = \sqrt{6^2 - 0^2} = \sqrt{36} = 6\)
BD = \(\sqrt{(6-6)^2 + (1 - 7)^2} = \sqrt{0^2 - 6^2} = \sqrt{36} = 6\)
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)

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Answer :

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 = \(\sqrt{(1 + 1)^2 + (0 + 2)^2} = \sqrt{4 + 4} = 2 \sqrt{2}\)
BC = \(\sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{4 + 4} = 2 \sqrt{2}\)
CD = \(\sqrt{(-3 + 1)^2 + (0 - 2)^2} = \sqrt{4 + 4} = 2 \sqrt{2}\)
DA = \(\sqrt{(-3 + 1)^2 + (0 - 2)^2} = \sqrt{4 + 4} = 2 \sqrt{2}\)
Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.
AC = \(\sqrt{(-1 + 1)^2 + (2 + 2)^2} = \sqrt{0 + 16} = 4\)
BD = \(\sqrt{(-3 - 1)^2 + (0 + 0)^2} = \sqrt{16 + 0} = 4\)
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 = \(\sqrt{(-3 - 3)^2 + (1 - 5)^2} = \sqrt{36 + 16} = 2 \sqrt{13}\)
BC = \(\sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{9 + 4} = \sqrt{13}\)
CD = \(\sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{1 + 49} = 5\sqrt{2}\)
DA = \(\sqrt{(-1 + 3)^2 + (-4 - 5)^2} = \sqrt{4 + 81} = \sqrt{85}\)
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 = \(\sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{9 + 1} = \sqrt{10}\)
BC = \(\sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{9 + 9} = \sqrt{18}\)
CD = \(\sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{9 + 1} = \sqrt{10}\)
DA = \(\sqrt{(1 - 4)^2 + (2 - 5)^2} = \sqrt{9 + 9} = \sqrt{18}\)
Here opposite sides of quadrilateral ABCD are equal. … (1)
We can now find out the lengths of diagonals.
AC = \(\sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0 + 4} = 2 \)
BD = \(\sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{36 + 16} = 2 \sqrt{13}\)
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).

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Answer :

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:
\(\sqrt{(x - 2)^2 + ( 0 - (-5))^2} = \sqrt{ (x - (-2))^2 + ( 0 - 9)^2}\)
=>\(\sqrt {x^2 + 4 - 4x + 25} = \sqrt {x^2 + 4 + 4x + 81}\)
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.

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Answer :

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

Solving this Quadratic equation by factorization, we can write
=>\(y^2 + 9y - 3y - 27 = 0\)
=>\( y (y + 9) – 3 (y + 9) = 0\)
=>\( (y + 9) (y - 3) = 0\)
=>\( y = 3, -9\)

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.

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Answer :

It is given that Q is equidistant from P and R. Using Distance Formula, we get
PQ = RQ
PQ=\(\sqrt{(5-0)^2 + (-3 - 1)^2} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \)
QR= \(\sqrt{(0 - x)^2 + (1 - 6)^2} = \sqrt{(-x)^2 + (-5)^2} = \sqrt{x^2 + 25} \)
So, \(\sqrt{41} = \sqrt{x^2 + 25} \)
On squaring both sides,
\(41 = x^2 + 25\)
\(x^2 = 16\)
\(x = \pm4\)
Coordinates of point R will be (4,6) or (-4,6),
If R is (4,6),
QR = \(\sqrt{(0 - 4)^2 + (1 - 6)^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \)
PR = \(\sqrt{(5 - 4)^2 + (-3 - 6)^2} = \sqrt{1^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82} \)
If R is (-4,6),
QR = \(\sqrt{(0 + 4)^2 + (1 - 6)^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \)
PR = \(\sqrt{(5 + 4)^2 + (-3 - 6)^2} = \sqrt{9^2 + 9^2} = \sqrt{81 + 81} = 9\sqrt{2} \)

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

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Answer :

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

Solution for 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.

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Answer :

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

\(P(x, y) = [\frac{mx_2 + nx_1}{m + n} , \frac{my_2 + ny_1}{m + n}]\)

\(x = {{2(4) + 3(-1)} \over {2 + 3}} = {{8 - 3} \over {5}} = 1\)
\(y = {{2(-3) + 3(7)} \over {2 + 3}} = {{-6 + 21} \over {5}} = 3\)
Therefore the point is(1,3)

Solution for 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)


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Answer :

We know that formula for area of a triangle whose vertices are \( (x_1,y_1) , (x_2,y_2) , (x_3,y_3) \) is,
= \( \frac{1}{2} \ | \ [x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \ | \)

(i) So, here \(x_1 \ = \ 2 \ , \ y_1 \ = \ 3 \ , \ x_2 \ = \ -1 \ , \ y_2 \ = \ 0 \ x_3 \ = \ 2 \ , \ y_3 \ = \ -4 \)

So, area of triangle = \( \frac{1}{2} \ | \ [ 2( 0 + 4 ) \ - \ 1(-4 - 3) \ + \ 2(3 - 0)] \ | \ = \ \frac{1}{2} \ | \ [ 8 + 7 + 6 ] \ | \ = \ \frac{21}{2} \)

∴ Area of triangle is \( \frac{21}{2} \) sq. units.

(ii) Similarly, here \(x_1 \ = \ -5 \ , \ y_1 \ = \ -1 \ , \ x_2 \ = \ 3 \ , \ y_2 \ = \ -5 \ x_3 \ = \ 5 \ , \ y_3 \ = \ 2 \)
So, area of triangle = \( \frac{1}{2} \ | \ [ (-5)( -5 - 2 ) \ - \ 3(2 + 1) \ + \ 5(-1 + 5)] \ | \ = \ \frac{1}{2} \ | \ [ 35 + 9 + 20 ] \ | \ = \ 32 \)

∴ 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)


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Answer :


We know that for collinear points area of triangle = 0 ,i.e.,
\( 0 \ = \frac{1}{2}[x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \)

(i) \( 7(1-k) + 5(k+2) + 3(-2-1) \ = \ 0 \)
=> \( 7 - 7k + 5k + 10 - 9 \ = \ 0 \)
=> \( 2k \ = \ 8 \)
=> \( k \ = \ 4 \)

(ii) \( 8(-4+5) + k(-5-1) + 2(1+4) \ = \ 0 \)
=> \( 8 - 6k + 10 \ = \ 0 \)
=> \( 6k \ = \ 18 \)
=> \( k \ = \ 3 \)

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.

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Answer :



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 = \( (\frac{0+2}{2} \ , \ \frac{-1+1}{2}) \) = (1,0)
E = ( \( \frac{0+2}{2} \ , \ \frac{1+3}{2} \) ) = (1,2)
F = ( \( \frac{0+0}{2} \ , \ \frac{3-1}{2} \) ) = (0,1)

So, Area of \( ∆ \ DEF \ = \ \frac{1}{2} \ | \ [ 1(2-1) + 1(1-0) + 0(0-2)] \ | \ = \ 1 \)

∴ Area of ∆ DEF = 1 sq. units

Area of ∆ ABC = \( \frac{1}{2} \ | \ [0(1-3) \ + \ 2(3+1) \ + \ 0(-1-1) ] \ | \ = \ 4 \)

∴ Area of ∆ ABC = 4 sq. units

So, \( ∆ \ DEF \ : \ ∆ \ ABC \ = \ 1 \ : \ 4 \).

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

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Answer :


Draw a line from B to D



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

So,
Area of ∆ DAB = \( | \ \frac{1}{2} [2(-2+5) \ - \ 4(-5-3) \ - \ 3(3+2)] \ | \) = \( \ | \frac{1}{2} [6 + 32 - 15 ] \ | \) = \( \frac{23}{2} \) sq. units

Similarly, Area of ∆ BCD = \( | \ \frac{1}{2} [-3(-2-3) \ + \ 3(3+5) \ + \ 2(-5+2)] \ | \ = \ | \ \frac{1}{2} [15 + 24 - 6 ] \ | \) = \( \frac{33}{2} \) sq. units

∴ Area of quad. ABCD = \( \frac{23}{2} \ + \ \frac{33}{2} \) = \( \frac{56}{2} \ = \ 28 \) 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).

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Answer :

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 ( \( \frac{3+5}{2} \ , \ \frac{-2+2}{2} \) )
Therefore, coordinates of D = (4,0)

Now,
Area of ∆ ABD = \( \frac{1}{2} \ | \ [4(-2-0) \ + \ 3(0+6) \ + \ 4(-6+2)] | \)
= \( \frac{1}{2} \ | \ [ -8 + 18 - 16] \ | \)
= \( |-3| \ = \ 3 \) sq. units

Similarly,
Area of ∆ ACD = \( \frac{1}{2} \ | \ [4(0-2) \ + \ 4(2+6) \ + \ 5(-6-0)] \ | \)
= \( \frac{1}{2} \ | \ [ -8 + 32 - 30] \ | \)
= \( |-3| \ = \ 3 \) 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.

Solution for 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).

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Answer :

Let the line 2x + y - 4 = 0 divides AB in 1 : k ratio, then the coordinates of the point of division is


\( x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1} \)

The point of intersection lie on both lines

∴ \( x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1} \) should satisfy 2x + y - 4 = 0
=> \( 2( \frac{2+3k}{k+1}) \ + \ (\frac{-2+7k}{k+1}) \ - \ 4 \ = \ 0 \)  
=> \( 4 \ + \ 6k \ - \ 2 \ + \ 7k \ = \ 4k \ + \ 4 \)
=> \( 9k - 2 = 0\)
=> \( k \ = \ \frac{2}{9} \)

∴ The line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7) in ratio 2 : 9 internally.

2. Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

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Answer :

If given points are collinear then area of triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,

Area of triangle = \( \frac{1}{2} \ | \ [ (x_1(y_2 – y_3 ) + x_2(y_3 – y_1) + x_3(y_1 – y_2) ) ] \ | \ = \ 0 \)

=>   \( \frac{1}{2} \ | \ [x(2 – 0) + 1 (0 – y) + 7( y – 2)] \ | \ = \ 0 \)

=>   \( 2x \ – \ y \ + \ 7y \ – \ 14 \ = \ 0 \)

=>   \(2x \ + \ 6y \ – \ 14 \ = \ 0 \)

=>   \(x \ + \ 3y \ – \ 7 \ = \ 0 \) .

Hence this is the required relation between x and y.

3. Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).

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Answer :


Let the coordinate of centre O be (x,y)

Then, OA = OB = OC {∵ radius is equal}

Now,
OA = \( \sqrt{ (x-6)^2+(y+6)^2 } \)
OB = \( \sqrt{(x-3)^2 + (y+7)^2 } \)
OC = \( \sqrt{(x-3)^2 + (y-3)^2 } \)

Now,   OA = OB
=> \( \sqrt{(x-6)^2 + (y+6)^2 } \) = \( \sqrt{(x-3)^2 + (y+7)^2 } \)
=> \( x^2 \ + \ 36 \ - \ 12x \ + \ y^2 \ + \ 36 \ - \ 12 \ = \ x^2 \ + \ 9 \ - \ 6x \ + \ y^2 \ + \ 9 \ - \ 6y \)
=> \( 6x \ + 2y \ - \ 14 \ = \ 0 \) ..............(i)

Similarly, OB = OC
=> \( \sqrt{(x-3)^2 + (y+7)^2 } \) = \( \sqrt{(x-3)^2 + (y-3)^2 } \)
=> \( y^2 \ + \ 14y \ + \ 49 \ = \ y^2 \ - \ 6y \ + \ 9 \)
=> \( y \ = \ -2 \)

Putting this value of y in eq. (i) , we get

\( 6x \ + \ 2(-2) \ - \ 14 \ = \ 0 \)
=> \( x \ = \ 3 \)

∴ The centre of circle is at (3,-2)

4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

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Answer :




Let ABCD be a square with vertices A( \( x_1, y_1) , B(3 , 2) , C(x_2 , y_2) , D(-1 , 2)\)

Let each side of square be 'a'.
So applying Pythagoras theorem in ∆ABD, we get

\( BD^2 = AB^2 + DA^2 \)
=> \( \sqrt{(3+1)^2 \ + \ (2-2)^2} \ = \ a^2 \ + \ a^2 \)
=> \( 4 \ = \ 2 a^2 \)
=> \( a \ = \ 2 \sqrt{2} \)

Now, DA = AB         {∵ ABCD is a square }

=> \( \sqrt{(x_1 +1)^2 + (y_1 - 2)^2 } \ = \ \sqrt{(3-x_1)^2 + (2 - y_1)^2} \)     {by distance formula}
=> \( (x_1 +1)^2 \ = \ (3-x_1)^2 \)
=> \( x_1^2 \ + \ 1 \ + \ 2x_1 \ = \ x_1^2 \ + \ 9 \ - \ 6x_1 \)
=> \( x_1 \ = \ 1 \)

Now since each side is \( 2 \sqrt{2} \)

∴ AB \( = \ 2 \sqrt{2} \)   =>   \( \sqrt{(3-x_1)^2 + (2 - y_1)^2} \ = \ 2 \sqrt{2} \)   =>   \( (2-y_1)^2 = \ 4 \)   =>   \( y \ = \ 4 \)

Now, point O is id point of BD . SO coordinates of O are

x = \( \frac{3-1}{2} \) , y = \( \frac{2+2}{2} \)   =>   x = 1 , y = 2

Now, since ABCD is a square ∴ O is mid point of AC also

∴ \( \frac{1+x_2}{2} \ = \ 1 \) , \( \frac{4+ y_2}{2} \ = \ 2 \)   =>   \( x_2 \ = \ 1 \) , \( y_2 \ = \ 0 \)

∴ Coordinates of vertices square ABCD are A(1, 4) , B(3,2) , C(1,0) , D(-1,2)

5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot as shown in the fig. 7.14. The students are to sow the 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 triangle PQR if C is the origin?

Also calculate the areas of the triangles in these cases. What do you observe?


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Answer :

(i) Taking A as origin, coordinates of the vertices P, Q and R are,

From figure:
P = (4, 6)
Q = (3, 2)
R = (6, 5)

Here AD is the x-axis and AB is the y-axis.

Area of triangle PQR in case of origin A:

Area of a triangle = \( \frac{1}{2} \ | \ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)] \ | \ = \ \frac{1}{2} \ | \ [– 12 – 3 + 24] \ | \ = \ \frac{9}{2} \) sq. units


(ii) Taking C as origin, coordinates of vertices P, Q and R are,

From figure:
P = (12, 2)
Q = (13, 6)
R = (10, 3)

Here CB is the x-axis and CD is the y-axis.

Area of triangle PQR in case of origin C:

Area of a triangle = \( \frac{1}{2} \ | \ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)] \ | \ = \frac{1}{2} \ | \ [36 + 13 – 40] \ | \ = \ \frac{9}{2} \) sq unit

This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C

Area is same in both case because triangle remains the same no matter which point is considered as origin.

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 \( \frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4} \) . Calculate the area of the \( ∆ \ ADE \) and compare it with area of \( ∆ \ ABC \) . (Recall Theorem 6.2 and Theorem 6.6)

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Answer :




Given, the vertices of a \( ∆ \ ABC \) are A (4, 6), B (1, 5) and C (7, 2) and,
\( \frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4} \)
=> \( \frac{AB}{AD} \ = \ \frac{AC}{AE} \ = \ 4 \)
=> \( \frac{AD + DB}{AD} \ = \ \frac{AE + EC}{AE} \ = \ 4 \)
=> \( 1 + \frac{DB}{AD} \ = \ 1 + \frac{EC}{AE} \ = \ 4 \)
=> \( \frac{DB}{AD} \ = \ \frac{EC}{AE} \ = \ 3 \)

∴ Point D and E divides AB and AC in ratio 1 : 3

Now coordinates of D calculated by section formula are,
\( x \ = \ \frac{3(4) + 1(1)}{4} \ = \ \frac{13}{4} \) and,
\( y \ = \ \frac{3(6) + 1(5)}{4} \ = \ \frac{23}{4} \)

Similarly, coordinates of E are,
\( x \ = \ \frac{1(7) + 3(4)}{4} \ = \ \frac{19}{4} \) and,
\( y \ = \ \frac{1(2) + 3(6)}{4} \ = \ \frac{20}{4} \ = \ 5 \)

Now area of triangle ADE is,

\( \frac{1}{2} \ | \ [4(5-2) \ + \ 1(2-6) \ + \ 7(6-5)] \ | \)
= \( \frac{1}{2} \ | \ [(12-4+7)] \ | \)
= \( \frac{15}{2} \) sq. units

Similarly, area of triangle ABC is,

\( \frac{1}{2} \ | \ [4( \frac{23}{4} -5) \ + \ \frac{13}{4}(5-6) \ + \ \frac{19}{4}(6- \frac{23}{4})] \ | \)
= \( \frac{1}{2} \ | \ [(3- \frac{13}{4} + \frac{19}{4})] \ | \)
= \( \frac{15}{32} \) sq. units

∴ ratio of \(area \ of \ ∆ \ ADE \ to \ area \ of \ ∆ \ ABC \ = \ 1:16 \)

7. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of \( ∆ \ ABC \) .

(i) The median from A meets BC at D. Find the coordinates of point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.

(iv) What do you observe?

[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]

(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.


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Answer :


(i) Given, D is the median to BC. So, coordinates of D using distance formula are

\( x \ = \ \frac{6+1}{2} \ = \ \frac{7}{2} \) and \( y \ = \ \frac{5+4}{2} \ = \ \frac{9}{2} \)

(ii) Coordinates of P can be calculated using section formula

\( x \ = \ \frac{2( \frac{7}{2}) + 1(4)}{3} \ = \ \frac{11}{3} \) and \( y \ = \ \frac{2( \frac{9}{2}) + 1(2)}{3} \ = \ \frac{11}{3} \)

(iii) Coordinates of E = ( \( \frac{4+1}{2} \ , \ \frac{2+4}{2} \) ) = ( \( \frac{5}{2} , 3 \) )

So, coordinates of Q are, ( \( \frac{2( \frac{5}{2}) + 1(6)}{3} \ , \ \frac{2(3) + 1(5)}{3} \) ) = ( \( \frac{11}{3} , \frac{11}{3} \) )

Coordinates of F = ( \( \frac{4+6}{2} , \frac{2+5}{2} \) ) = ( \( 5, \frac{7}{2} \) )

So, coordinates of R are, ( \( \frac{2(5) + 1(1)}{3} , \frac{2 ( \frac{7}{2}) + 1(4)}{3} \) ) = ( \( \frac{11}{3} , \frac{11}{3} \) )


(iv) We observed that coordinates of P , Q, R are same i.e., ( \( \frac{11}{3} , \frac{11}{3} \) ) which shows that medians intersect each other at a common point which is called centroid of the triangle.

(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, then the coordinates of the centroid of the triangle is

\( x \ = \ \frac{x_1 + x_2 + x_3}{3} \ and \ \frac{y_1 + y_2 + y_3}{3} \)

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 midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

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Answer :


P, Q, R and S are the midpoints of AB, BC, CD and DA respectively

So, coordinate of P = ( \( \frac{-1-1}{2} , \frac{-1+4}{2} \) ) = ( \( -1 , \frac{3}{2} \) )

Similarly, coordinate of Q = ( \( \frac{5-1}{2} , \frac{4+4}{2} \) ) = ( 2,4)

R = ( \( \frac{5+5}{2} , \frac{4-1}{2} \) ) = ( \( 5 , \frac{3}{2} \) )

S = ( \( \frac{5-1}{2} , \frac{-1-1}{2} \) ) = ( 2,-1)

Now, PQ = \( \sqrt{(-1-2)^2 + ( \frac{3}{2} - 4)^2} \ = \ \sqrt{ \frac{61}{4} } \ = \ \frac{ \sqrt{61}}{2} \)

SP = \( \sqrt{(2+1)^2 + (-1- \frac{3}{2})^2} \ = \ \sqrt{ \frac{61}{4} } \ = \ \frac{ \sqrt{61}}{2} \)

QR = \( \sqrt{(2-5)^2 + (4- \frac{3}{2})^2} \ = \ \sqrt{ \frac{61}{4} } \ = \ \frac{ \sqrt{61}}{2} \)

RS = \( \sqrt{(5-2)^2 + ( \frac{3}{2} +1)^2} \ = \ \sqrt{ \frac{61}{4} } \ = \ \frac{ \sqrt{61}}{2} \)

PR = \( \sqrt{(-1-5)^2 + ( \frac{3}{2} - \frac{3}{2})^2 } \ = \ 6 \)

QS = \( \sqrt{(2-2)^2 + (4+1)^2} \ = \ 5 \)

Clearly, PQ = QR = RS = SP

But, \( PR \ne QS \)

Hence the lengths of all the sides are equal, while the lengths of the diagonals are different.

∴ The PQRS is a rhombus.