Show that the points form the vertices of the indicated polygon. ?
Isosceles triangle: (7,-6), (9,8), (-5,6)
?
A)Distances between the points: $$\sqrt { 37 } , \sqrt { 200 } , \sqrt { 288 }$$
B)Distances between the points: $\sqrt { 200 } , \sqrt { 200 } , \sqrt { 288 }$
C)Distances between the points: $\sqrt { 26 } , \sqrt { 200 } , \sqrt { 288 }$
D)Distances between the points: $$\sqrt { 43 } , \sqrt { 200 } , \sqrt { 288 }$$
E)Distances between the points: $\sqrt { 19 } , \sqrt { 200 } , \sqrt { 288 }$

Question

Quizplus · Accepted Answer

To show that the points form the vertices of an isosceles triangle, we calculate the distances between each pair of points using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. For an isosceles triangle, two sides must be of equal length. Calculating the distances:1. Between (7,-6) and (9,8): $d = \sqrt{(9-7)^2 + (8-(-6))^2} = \sqrt{2^2 + 14^2} = \sqrt{4 + 196} = \sqrt{200}$.2. Between (9,8) and (-5,6): $d = \sqrt{(-5-9)^2 + (6-8)^2} = \sqrt{-14^2 + (-2)^2} = \sqrt{196 + 4} = \sqrt{200}$.3. Between (-5,6) and (7,-6): $d = \sqrt{(7-(-5))^2 + (-6-6)^2} = \sqrt{12^2 + (-12)^2} = \sqrt{144 + 144} = \sqrt{288}$.Thus, two sides are of equal length ($\sqrt{200}$), and the third side is different ($\sqrt{288}$), indicating an isosceles triangle.

Show That the Points Form the Vertices of the Indicated 37,200,288\sqrt { 37 } , \sqrt { 200 } , \sqrt { 288 }37​,200​,288​

Show That the Points Form the Vertices of the Indicated $\sqrt { 37 } , \sqrt { 200 } , \sqrt { 288 }$