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

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)

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.

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.

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.

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.

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)

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).

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.

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.

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).

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.

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)

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)

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.

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).

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).

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).

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.

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).

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.

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?

(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)

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.

(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.

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.