Question

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

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Answer 1

${ \bold{ \underline{Given \: that \: vertices \: of \: a \: triangle \: are-}}}$

$\rightarrow \red{A=(5,2)}$

$\rightarrow \blue {B=(6,4)}$

$\rightarrow \pink {C=(7,2)}$

$\bold {So,lets \: prove \: by \: using \: distance \: formula}$

$\:AB=√(6−5)^2+(4−2)^2$

$\:AB=√(1)^2+(2)^2$

$\: AB=√1+4$

$\: AB=√5units$

$\bold{and}$

$\: BC=√(7−6)^2+(2−4)^2$

$\: BC=√(1)^2+(−2)^2$

$\: BC=√1+4$

$\: BC=√5units$

$\bold{and}$

$\: AC=(7−5)^2+(2−2)^2$

$\ AC=(2)^2+(0)^2$

$\: AC=4+0$

$\: AC=2 units$

$\bold{Here}$

$\boxed{AB=BC}$

so the given vertices are of an isosceles triangle

Answer 2

Since two sides of any isosceles triangle are equal. To check whether given points are vertices of an isosceles triangle, we will find the distance between all the points.

Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C respectively.

This implies, whether given points are vertices of an isosceles triangle.