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Mathematics

Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

Coordinate Geometry

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Answer

Let vertices be A(5, -2), B(6, 4) and C(7, -2).

By formula,

Distance between two points (D) = (y2−y1)2+(x2−x1)2\sqrt{(y2 - y1)^2 + (x2 - x1)^2}

Substituting values we get :

AB=[4−(−2)]2+(6−5)2=[4+2]2+(1)2=62+1=36+1=37.BC=(−2−4)2+(7−6)2=(−6)2+(1)2=36+1=37.AC=[−2−(−2)]2+(7−5)2=[−2+2]2+(2)2=0+4=4=2.AB = \sqrt{[4 - (-2)]^2 + (6 - 5)^2} \\[1em] = \sqrt{[4 + 2]^2 + (1)^2} \\[1em] = \sqrt{6^2 + 1} \\[1em] = \sqrt{36 + 1} \\[1em] = \sqrt{37}. \\[1em] BC = \sqrt{(-2 - 4)^2 + (7 - 6)^2} \\[1em] = \sqrt{(-6)^2 + (1)^2} \\[1em] = \sqrt{36 + 1} \\[1em] = \sqrt{37}. \\[1em] AC = \sqrt{[-2 - (-2)]^2 + (7 - 5)^2} \\[1em] = \sqrt{[-2 + 2]^2 + (2)^2} \\[1em] = \sqrt{0 + 4} \\[1em] = \sqrt{4} \\[1em] = 2.

Since, AB = BC.

Hence, points (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

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