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=√((x_2-x_1)^2+(y_2-y_1)^2)`

Using distance formula, we get

`AB=√((3-(-3))^2+(4-4)^2)`

`AB=√((3+3)^2+(0)^2)`

`AB=√((6)^2+0)`

`AB=√(36)`

`AB=6`

Similarly,

`BC=√((3-3)^2+(-4-4)^2)`

`BC=√((0)^2+(-8)^2)`

`BC=√(64)`

`BC=8`

And,

`AC=√((3-(-3))^2+(-4-4)^2)`

`AC=√((6)^2+(-8)^2)`

`AC=√(36+64)`

`AC=√(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.