  Question

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

Solution

The vertices of a triangle ABC is given as $$A\left( {5, - 2} \right)$$, $$B\left( {6,4} \right)$$ and $$C\left( {7, - 2} \right)$$. The distance AB is, $$AB = \sqrt {{{\left( {6 - 5} \right)}^2} + {{\left( {4 - \left( { - 2} \right)} \right)}^2}}$$ $$= \sqrt {{{\left( 1 \right)}^2} + {{\left( 6 \right)}^2}}$$ $$= \sqrt {1 + 36}$$ $$= \sqrt {37}$$ The distance BC is, $$BC = \sqrt {{{\left( {7 - 6} \right)}^2} + {{\left( { - 2 - 4} \right)}^2}}$$ $$= \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 6} \right)}^2}}$$ $$= \sqrt {1 + 36}$$ $$= \sqrt {37}$$ The distance CA is, $$CA = \sqrt {{{\left( {5 - 7} \right)}^2} + {{\left( { - 2 - \left( { - 2} \right)} \right)}^2}}$$ $$= \sqrt {{{\left( { - 2} \right)}^2} + 0}$$ $$= \sqrt 4$$ $$= 2$$ Since, $$AB = BC$$, that is the two sides of a triangle are equal, therefore, the triangle is an isosceles triangle.Mathematics

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