Question

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

Answer 1

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

⇒ `sqrt((10 - 2)^2 + (y + 3)^2) = 10`

⇒ (8)² + (y + 3)² = 100

⇒ 64 + y² + 6y + 9 = 100

⇒ y² + 6y + 73 - 100 = 0

⇒ y² + 6y - 27 = 0

⇒ y² + 9y - 3y - 27 = 0

⇒ y(y + 9) - 3(y + 9) = 0

⇒ (y + 9) (y - 3) = 0

⇒ y + 9 = 0

⇒ y = -9

and y - 3 = 0

⇒ y = 3

Answer 2

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 + y² + 6y

27 = y² + 6y

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

y² + 6y - 27 = 0

y² + 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.