Question
Show that the triangle whose vertices are (8, 2), (5, -3) and (0, 0) is an isosceles.
✓
Answer
Let, the given vertices be $A\left(x_1, y_1\right)=(8,2), B\left(x_2, y_2\right)$ $=(5,-3)$ and $C\left(x_3, y_3\right)=(0,0)$
$\therefore$ $A B=\left|\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\right|$
$=\left|\sqrt{(8-5)^2+(2+3)^2}\right|$
$=\sqrt{34}$ units.
$B C=\left|\sqrt{\left(x_2-x_3\right)^2+\left(y_2-y_3\right)^3}\right|$
$=\left|\sqrt{(5-0)^2+(-3-0)^2}\right|$
$=\sqrt{34}$ units.
$\therefore A B=B C$ i.e; $\angle A C B=\angle B A C$
Hence, $\triangle A B C$ is isosceles.
$\therefore$ $A B=\left|\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\right|$
$=\left|\sqrt{(8-5)^2+(2+3)^2}\right|$
$=\sqrt{34}$ units.
$B C=\left|\sqrt{\left(x_2-x_3\right)^2+\left(y_2-y_3\right)^3}\right|$
$=\left|\sqrt{(5-0)^2+(-3-0)^2}\right|$
$=\sqrt{34}$ units.
$\therefore A B=B C$ i.e; $\angle A C B=\angle B A C$
Hence, $\triangle A B C$ is isosceles.
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