Question

In the standard (x,y) coordinate plane a right triangle has vertices at (-3,4)(3,4)(3,-4) what is the length in coordinates units of the hypotenuse of this triangle

Answer 1

Given:

The vertices of a right triangle are:

$(-3,4),(3,4),(3,-4)$

To find:

The length of the hypotenuse of the given right triangle.

Solution:

Let the vertices of the right triangle are $A(-3,4),B(3,4),C(3,-4)$.

The distance formula is:

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

Using distance formula, we get

$AB=\sqrt{(3-(-3))^2+(4-4)^2}$

$AB=\sqrt{(3+3)^2+(0)^2}$

$AB=\sqrt{(6)^2+0}$

$AB=\sqrt{36}$

$AB=6$

Similarly,

$BC=\sqrt{(3-3)^2+(-4-4)^2}$

$BC=\sqrt{(0)^2+(-8)^2}$

$BC=\sqrt{64}$

$BC=8$

And,

$AC=\sqrt{(3-(-3))^2+(-4-4)^2}$

$AC=\sqrt{(6)^2+(-8)^2}$

$AC=\sqrt{36+64}$

$AC=\sqrt{100}$

$AC=10$

Now, taking sum of squares of two smaller sides, we get

$AB^2+BC^2=6^2+8^2$

$AB^2+BC^2=36+64$

$AB^2+BC^2=100$

$AB^2+BC^2=AC^2$

By the definition of the Pythagoras theorem, AC is the hypotenuse of the given triangle.

Therefore, the length of the hypotenuse is 10 units.