Given:
The vertices of a right triangle are:
[tex](-3,4),(3,4),(3,-4)[/tex]
To find:
The length of the hypotenuse of the given right triangle.
Solution:
Let the vertices of the right triangle are [tex]A(-3,4),B(3,4),C(3,-4)[/tex].
The distance formula is:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula, we get
[tex]AB=\sqrt{(3-(-3))^2+(4-4)^2}[/tex]
[tex]AB=\sqrt{(3+3)^2+(0)^2}[/tex]
[tex]AB=\sqrt{(6)^2+0}[/tex]
[tex]AB=\sqrt{36}[/tex]
[tex]AB=6[/tex]
Similarly,
[tex]BC=\sqrt{(3-3)^2+(-4-4)^2}[/tex]
[tex]BC=\sqrt{(0)^2+(-8)^2}[/tex]
[tex]BC=\sqrt{64}[/tex]
[tex]BC=8[/tex]
And,
[tex]AC=\sqrt{(3-(-3))^2+(-4-4)^2}[/tex]
[tex]AC=\sqrt{(6)^2+(-8)^2}[/tex]
[tex]AC=\sqrt{36+64}[/tex]
[tex]AC=\sqrt{100}[/tex]
[tex]AC=10[/tex]
Now, taking sum of squares of two smaller sides, we get
[tex]AB^2+BC^2=6^2+8^2[/tex]
[tex]AB^2+BC^2=36+64[/tex]
[tex]AB^2+BC^2=100[/tex]
[tex]AB^2+BC^2=AC^2[/tex]
By the definition of the Pythagoras theorem, AC is the hypotenuse of the given triangle.
Therefore, the length of the hypotenuse is 10 units.