Class 10 Maths

Coordinate Geometry

NCERT Exercise

7.4 Part 1

Question 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).

Solution: Let us assume the line divides AB in k : 1 ratio.

Coordinates of point of division can be given as follows:

x=(2+3k)/(k+1)

y=(-2+7k)/(k+1)

Substituting the values of x and y in following equation;

2x+y-4-0

Or, 2((2+3k)/(k+1))+((-2+7k)/(k+1))-4=0

Or, (4+6k)/(k+1)+(-2+7k)/(k+1)\-4=0

Or, 4+6k-2+7k-4(k+1)=0
Or, 4+6k-2+7k-4k-4=0
Or, -2+9k=0
Or, 9k=0
Or, k=2/9

Hence, the ratio is 2 : 9.

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

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

Or, ½ (x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3 (y_1 – y_2)) = 0

Or, ½ (x(2 – 0) + 1 (0 – y) + 7( y – 2)) = 0

Or, 2x – y + 7y – 14 = 0

Or, 2x + 6y – 14 = 0

Or, x + 3y – 7 = 0

The last equation gives the relation between x and y.

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

Solution: A = (6, -6), B = (3, -7), C = (3, 3)

If O is the centre, then OA = OB = OC (radii are equal)

If O = (x, y) then

OA=sqrt((x-6)^2+(y+6)^2)

OB=sqrt((x-3)^2+(y+7)^2)

OC=sqrt((x-3)^2+(y-3)^2)

Or, (x-6)^2+(y+6)^2=(x-3)^2+(y+7)^2
Or, x^2-12x+36+y^2+12y+36
=x^2-6x+9+y^2+14y+49
Or, x^2-12x+y^2+12y+72
=x^2-6x+y^2+14y+58
Or, x^2-12x-(x^2-6x)
=y^2+14y-(y^2+12y)+58-72
Or, -12x+6x=14y-12y-14
Or, -6x=2y-14 --------(1)

Similarly,

(x-3)^2+(y+7)^2
=(x-3)^2+(y-3)^2
Or, (y+7)^2=(y-3)^2
Or, y^2+14y+49=y^2-6y+9
Or, 14y+49=-6y+9
Or, 14y+6y=9-49
Or, 20y=-40
Or, y=-2

Substituting the value of y in equation (1), we get;

-6x=2y-14
Or, -6x=-4-14=-18
Or, x=3

Hence, x = 3, y = -2 are the coordinates of centre

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

Solution: Answer: O is the point of intersection of AC and BD. Coordinates of O can be calculated as follows:

x=(3-1)/(2)=1

y=(2+2)/(2)=2

Now, AC can be calculated as follows: (diagonals of a square are equal and bisect each other.

AC=sqrt((3+1)^2+(2-2)^2)=sqrt16=4

Hence, sides of square = 2√2 (Using Pythagoras theorem)

AD=2sqrt2=sqrt((x_1+1)^2+(y_1-2)^2)

Or, 8=(x_1+1)^2+(y_1-2)^2
CD=2sqrt2=sqrt((x_1-3)^2+(y_1-2)^2)

Or, 8=(x_1-3)^2+(y_1-2)^2

From these equations, it is clear that;

(x_1+1)^2+(y_1-2)^2
=(x_1-3)^2+(y_1-2)^2
Or, (x_1+1)^2=(x_1-3)^2
Or, x_1^2+2x+1=x_1^2-6x+9
Or, 2x+1=-6x+9
Or, 8x=8
Or, x=1

Value of y1 can be calculated as follows by using the value of x.

CD=2sqrt2=sqrt((x_1-3)^2+(y_1-2)^2)

Or, 8=(1-3)^2+(y_1-2)^2
Or, 4+(y_1-2)^2=8
Or, (y_1-2)^2=4
Or, y_1-2=2
Or, y_1=4

Hence, D = (1, 4)

Coordinates of B can be calculated using coordinates of O; as follows:

Earlier, we had calculated O = (1, 2)

For point BD;

1=(x+1)/(2)
Or, x+1=2
Or, x=1
Or, 2=(y+4)/(2)
Or, y+4=4
Or, y=0

Hence, B = (1, 0) and D = (1, 4)