Question

If the distance between the points (3, 0) and (0, y) is 5 units and y is positive, then what is the value of y?

Answer 1

It is given that distance between P (3, 0) and Q (0, y) is 5.

In general, the distance between `A(x_1, y_1)` and `B(x_2, y_2)` is given by,

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

So,

`5^2 = (0 - 3)^2 + (y - 0)^2`

On further simplification,

`y^2 = 16`

`y = +-4`

We will neglect the negative value. So,

y = 4