Show that the points form the vertices of the indicated polygon. ?
Right triangle: (6,2), (3,4), (-3,-5)
?
A) $( \sqrt { 25 } ) ^ { 2 } + ( \sqrt { 117 } ) ^ { 2 } = ( \sqrt { 130 } ) ^ { 2 }$
B) $( \sqrt { 7 } ) ^ { 2 } + ( \sqrt { 117 } ) ^ { 2 } = ( \sqrt { 130 } ) ^ { 2 }$
C) $( \sqrt { 11 } ) ^ { 2 } + ( \sqrt { 117 } ) ^ { 2 } = ( \sqrt { 130 } ) ^ { 2 }$
D) $( \sqrt { 13 } ) ^ { 2 } + ( \sqrt { 117 } ) ^ { 2 } = ( \sqrt { 130 } ) ^ { 2 }$
E) $( \sqrt { 40 } ) ^ { 2 } + ( \sqrt { 117 } ) ^ { 2 } = ( \sqrt { 130 } ) ^ { 2 }$

Question

Quizplus · Accepted Answer

To show that the points form the vertices of a right triangle, we can use the Pythagorean theorem. First, we calculate the distances between the points, which represent the sides of the triangle. The distances (squared) between the points are:1. Between (6,2) and (3,4): $(6-3)^2 + (2-4)^2 = 3^2 + (-2)^2 = 9 + 4 = 13$2. Between (3,4) and (-3,-5): $(3-(-3))^2 + (4-(-5))^2 = 6^2 + 9^2 = 36 + 81 = 117$3. Between (6,2) and (-3,-5): $(6-(-3))^2 + (2-(-5))^2 = 9^2 + 7^2 = 81 + 49 = 130$The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Here, $13 + 117 = 130$, which matches option D: $( \sqrt { 13 } ) ^ { 2 } + ( \sqrt { 117 } ) ^ { 2 } = ( \sqrt { 130 } ) ^ { 2 }$. This confirms that the points form a right triangle.

Show That the Points Form the Vertices of the Indicated (25)2+(117)2=(130)2( \sqrt { 25 } ) ^ { 2 } + ( \sqrt { 117 } ) ^ { 2 } = ( \sqrt { 130 } ) ^ { 2 }(25​)2+(117​)2=(130​)2

Show That the Points Form the Vertices of the Indicated $( \sqrt { 25 } ) ^ { 2 } + ( \sqrt { 117 } ) ^ { 2 } = ( \sqrt { 130 } ) ^ { 2 }$