Question

The vertices of a $△ABC$ are $A(1,2),B(-4,5)$ and $C(0,1)$. Find the slopes of the altitudes of the triangle.

Answer 1

Let $AD,BE$ and $CF$ be the altitudes of a $△ABC$. Slope of $BC=\frac{1-5}{0+4}=-1$ Since the altitude $AD$ is a perpendicular to $BC$, Slope of $AD=1$ $\because$ $m_1{m}_2=-1$ Slope of $AC=\frac{1-2}{0-1}=1$ Thus, Slope of $BE=-1$ $\because$ $BE\perp AC$ Also, Slope of $AB=\frac{5-2}{-4-1}=-\frac{3}{5}$ $\therefore$ Slope of $CF=\frac{5}{3}$ $\because$ $CF\perp AB$