Question

What could be the coordinates of the third vertex, z, of triangle xyz so that it would have a hypotenuse with a length of square root 45 units

Answer 1

`z = (6,-1)`

Explanation:

The missing parameters are:

`x = (3,-1)`

`y = (3,5)`

`yz= \sqrt{45` --- Hypotenuse

Required

The coordinate of Z

First, calculate the distance xy

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

`xy = √((3 - 3)^2 + (-1 - 5)^2)`

`xy = √(0 + 36)`

`xy = √(36)`

`xy = 6`

By Pythagoras theorem, distance xz is:

`yz^2 = xz^2 + xy^2`

`(\sqrt 45)^2 = xz^2 + 6^2`

`45 = xz^2 + 36`

Collect like terms

`xz^2 =45-36`

`xz^2 =9`

`xz = √(9)`

`xz = 3`

This means that z is 3 units to the right of x

We have:

`x = (3,-1)`

The rule to determine z is:

`(x,y) \to (x + 3, y)`

So, we have:

`z = (3 + 3,-1)`

`z = (6,-1)`