Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is

10 units

#### Solution

The distance *d* between two points `(x_1,y_1)` and `(x_2,y_2)` is given by the formula

`d = sqrt((x_1-x_2)^2 + (y_1 - y_2)^2)`

The distance between two points *P*(2*,**−*3) and *Q*(10*,y*) is given as 10 units. Substituting these values in the formula for distance between two points we have,

`10 = sqrt((2 - 10)^2 + (-3 - y)^2)`

Now, squaring the above equation on both sides of the equals sign

`100 = (-8)^2 + (-3 - y)^2`

`100 = 64 + 9 + y^2 + 6y`

`27 = y^2 + 6y`

Thus we arrive at a quadratic equation. Let us solve this now,

`y^2 + 6y - 27 = 0`

`y^2 + 9y - 3y - 27 = 0`

`y(y + 9) - 3(y + 9) = 0`

(y - 3)(y + 9) = 0

The roots of the above quadratic equation are thus 3 and −9.

Thus the value of ‘*y*’ could either be 3 or -9