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
