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

Answer:

$z = (6,-1)$

Step-by-step 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

$xy = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$xy = \sqrt{(3 - 3)^2 + (-1 - 5)^2}$

$xy = \sqrt{0 + 36}$

$xy = \sqrt{36}$

$xy = 6$

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

$xz^2 =45-36$

$xz^2 =9$

$xz = \sqrt{9}$

$xz = 3$

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:

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

$z = (6,-1)$