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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

Sagot :

MrRoyalidn

Answer:

[tex]z = (6,-1)[/tex]

Step-by-step explanation:

The missing parameters are:

[tex]x = (3,-1)[/tex]

[tex]y = (3,5)[/tex]

[tex]yz= \sqrt{45[/tex] --- Hypotenuse

Required

The coordinate of Z

First, calculate the distance xy

[tex]xy = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]

[tex]xy = \sqrt{(3 - 3)^2 + (-1 - 5)^2}[/tex]

[tex]xy = \sqrt{0 + 36}[/tex]

[tex]xy = \sqrt{36}[/tex]

[tex]xy = 6[/tex]

By Pythagoras theorem, distance xz is:

[tex]yz^2 = xz^2 + xy^2[/tex]

[tex](\sqrt 45)^2 = xz^2 + 6^2[/tex]

[tex]45 = xz^2 + 36[/tex]

Collect like terms

[tex]xz^2 =45-36[/tex]

[tex]xz^2 =9[/tex]

[tex]xz = \sqrt{9}[/tex]

[tex]xz = 3[/tex]

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

We have:

[tex]x = (3,-1)[/tex]

The rule to determine z is:

[tex](x,y) \to (x + 3, y)[/tex]

So, we have:

[tex]z = (3 + 3,-1)[/tex]

[tex]z = (6,-1)[/tex]

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