Coordinate Geometry

NCERT Exercise

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

Solution: Let given coordinates are A(2, –5), BC (–2, 9) and P(x, 0) is on the x-axis.

Since, point P is equidistanct from the given coordinates,

thus AP = BP

`AP=sqrt((x_2-x_2)^2+(y_2-y_1)^2)`

Here, x₁ = 2, x₂ = x, y₁ = -5 and y₂ = 0

So, `AP=sqrt((x-2)^2+(0+5)^2)`

`=sqrt(x^2+4-4x+25)`

`=sqrt(x^2-4x+29)`

`BP= sqrt((x_2-x_2)^2+(y_2-y_1)^2)`

Here, x₁ = -2, x₂ = x, y₁ = 9 and y₂ = 0

So, `BP=sqrt((x+2)^2+(0-9)^2)`

`=sqrt(x^2+4+4x+81)`

`=sqrt(x^2+4x+85)`

Now, since `AP=BP`

So, `sqrt(x^2-4x+29)=sqrt(x^2+4x+85)`

After squaring both sides we get:

`x^2-4x+29=x^2+4x+85`

Or, `-4x+29=4x+85`

Or, `-8x+29=85`

Or, `-8x=85-29=56`

Or, `-x=56/8=7`

Or, `x=-7`

Thus, required point is (-7, 0)

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

Solution: Given, P(2, –3) and Q(10, y) and QP = 10 unit

`PQ=sqrt((x_2-x_2)^2+(y_2-y_1)^2)`

Here, x₁ = 2, x₂ = 10, y₁ = -3 and y₂ = y

So, `PQ=sqrt((10-2)^2+(y+3)^2)`
Or `10=sqrt(8^2+(y+3)^2)`
Or `10=sqrt(64+y^2+9+6y)`

After squaring both sides we get;

`10^2=64+9+y^2+6y`
Or, `100=73+y^2+6y`
Or, `y^2+6y=100-73=27`
Or, `y^2+6y-27=0`
Or, `y^2+9y-3y-27=0`
Or, `y(y+9)-3(y+9)=0`
Or, `(y-3)(y+9)=0`

Thus, if `y-3=0` Then `y=3`
And if `y+9=0`
Then `y=-9`

Thus, the required value = 3 or -9

Thus, required value = 3 or – 9