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_12)`  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