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

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

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The distance between two points A(x_{1},y_{1})\ and\ B(x_{2},y_{2}) is given by:

D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangle A, B, and C respectively.

AB = \sqrt{(5-6)^2+(-2-4)^2} = \sqrt{1+36} = \sqrt{37}

BC = \sqrt{(6-7)^2+(4+2)^2} = \sqrt{1+36} = \sqrt{37}

CA = \sqrt{(5-7)^2+(-2+2)^2} = \sqrt{4+0} = 2

Therefore, AB = BC

Here two sides are equal in length.

Therefore, ABC is an isosceles triangle.

Posted by

Divya Prakash Singh

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