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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

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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.

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