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).

Answer: 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.

Answer: 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).

Answer: 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.

Answer: 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)