Proving Relationships of Parallel and Perpendicular Lines

Let's draw the triangle.

We see that $\angle N$ and $\angle M$ are acute angles. Therefore, their are not right angles. To determine whether $\angle L$ is a right angle, we will calculate the slopes of $\overline{LN}$ and $\overline{LM}.$ If these segments are perpendicular to each other, we know that the triangle is a right triangle.

Side	Points	$\frac{y_2-y_1}{x_2-x_1}$	Slope
$\overline{LN}$	$(0,6),(4,\text{-}1)$	$\frac{\text{-}1-6}{4-0}$	$\text{-}\frac{7}{4}$
$\overline{LM}$	$(0,6),(5,8)$	$\frac{8-6}{5-0}$	$\frac{2}{5}$

Let's multiply the obtained slopes to see if their product is $\text{-}1.$ If this happens, then they are opposite reciprocals and therefore the segments are perpendicular. Since the product of the slopes is not $\text{-}1,$ the segments are not perpendicular. Therefore, the angle they form is not a right angle. This means that the triangle is not a right triangle.