The points (5,-4,5), (-3,-3,2) and (-1,-6,8) | vedclass

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(A) Let the points be $A(5, -4, 5)$,$B(-3, -3, 2)$,and $C(-1, -6, 8)$.We calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2

Vedclass · Accepted Answer

an isosceles triangle

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(A) Let the points be $A(5, -4, 5)$,$B(-3, -3, 2)$,and $C(-1, -6, 8)$.We calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.$AB = \sqrt{(-3-5)^2 + (-3-(-4))^2 + (2-5)^2} = \sqrt{(-8)^2 + (1)^2 + (-3)^2} = \sqrt{64 + 1 + 9} = \sqrt{74}$.$BC = \sqrt{(-1-(-3))^2 + (-6-(-3))^2 + (8-2)^2} = \sqrt{(2)^2 + (-3)^2 + (6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.$CA = \sqrt{(5-(-1))^2 + (-4-(-6))^2 + (5-8)^2} = \sqrt{(6)^2 + (2)^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.Since $BC = CA = 7$,two sides of the triangle are equal.Therefore,the points form an isosceles triangle.

The points $(5,-4,5), (-3,-3,2)$ and $(-1,-6,8)$ form ......

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