Question

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

Answer 1

Let $\mathrm{A}\left(5,-2\right),\mathrm{B}\left(6,4\right)$ and $\mathrm{C}\left(7,-2\right)$ are the vertices of a triangle.

To verify whether the given points are vertices of an isosceles triangle, we will determine the distance between all the points.

Distance formula$=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}$

To find the distance between points $\mathrm{A}(5,-2)$ and$\mathrm{B}(6,4)$, let${\mathrm{x}}_1=5,{\mathrm{y}}_1=-2$and ${\mathrm{x}}_2=6,{\mathrm{y}}_2=4$

$\begin{array}{rcl}\mathrm{AB}& =& \sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ & =& \sqrt{{\left(6-5\right)}^2+{\left(4-(-2)\right)}^2}\\ & =& \sqrt{1^2+6^2}\\ & =& \sqrt{37}\mathrm{unit}\end{array}$

To find the distance between points $\mathrm{B}(6,4)$ and$\mathrm{C}(7,-2)$, let${\mathrm{x}}_1=6,{\mathrm{y}}_1=4$ and${\mathrm{x}}_2=7,{\mathrm{y}}_2=-2$

$\begin{array}{rcl}\mathrm{BC}& =& \sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ & =& \sqrt{{\left(7-6\right)}^2+{\left(-2-\left(4\right)\right)}^2}\\ & =& \sqrt{1^2+{(-6)}^2}\\ & =& \sqrt{37}\mathrm{unit}\end{array}$

To find the distance between points $\mathrm{C}(7,-2)$ and$\mathrm{A}(5,-2)$, let${\mathrm{x}}_1=7,{\mathrm{y}}_1=-2,$ and ${\mathrm{x}}_2=5,{\mathrm{y}}_2=-2$

$\begin{array}{rcl}\mathrm{CA}& =& \sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ & =& \sqrt{{\left(5-7\right)}^2+{\left(-2-(-2)\right)}^2}\\ & =& \sqrt{{(-2)}^2+{\left(0\right)}^2}\\ & =& \sqrt{4}\\ & =& 2\mathrm{unit}\end{array}$

$\because$ The sides $\mathrm{AB}$ and $\mathrm{BC}$ are equal, this indicates the given points are vertices of an isosceles triangle.

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