Question

The points (5, - 2), (6, 4) and (7, - 2) are the vertices of an _________ triangle.

Answer 1

Let the points (5, - 2), (6, 4) and (7, - 2) represent the vertices A, B, and C of the given triangle respectively.
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$AB=\sqrt{{(5-6)}^2+{(-2-4)}^2}$
$=\sqrt{{(-1)}^2+{\left(6\right)}^2}$
$=\sqrt{1+36}=\sqrt{37}$
$BC=\sqrt{{(6-7)}^2+{(4-(-2))}^2}$
$=\sqrt{{(-1)}^2+{\left(6\right)}^2}$
$=\sqrt{1+36}=\sqrt{37}$
$CA=\sqrt{{(5-7)}^2+{(-2(-2))}^2}$
$=\sqrt{{(-2)}^2+0^2}=2$
$\Rightarrow$ AB = BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.