Question

The vertices of a $△ABC$ are $A(-5,7),B(-4,-5)$ and $C(4,5)$. Find the slopes of the altitudes of the triangle.

• A. $\frac{1}{12},-\frac{4}{5},\frac{9}{2}$
• B. $\frac{1}{12},\frac{4}{5},\frac{9}{2}$
• C. $\frac{1}{12},-\frac{4}{5},\frac{-9}{2}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{12},-\frac{4}{5},\frac{9}{2}$

Consider the given vertices.

$A(-5,7),B(-4,-5)$ and $C(4,5)$

We know that

$slope\left(m\right)=\frac{y_2-y_1}{x_2-x_1}$

So,

Slope of $AB=\frac{-5-7}{-4+5}$

$=\frac{-12}{1}=-12$

Slope of the altitude of the triangle $=-\frac{1}{AB}$

$[\because \text{the altitude drawn to the side is perpendicular}\Rightarrow productofslopes=-1]$

$=-\frac{1}{-12}=\frac{1}{12}$

Similarly,

Slope of $BC=\frac{5+5}{4+4}$

$=\frac{10}{8}=\frac{5}{4}$

Slope of the altitude of the triangle $=-\frac{1}{BC}$

$=-\frac{1}{\frac{5}{4}}=-\frac{4}{5}$

Similarly,

Slope of $CA=\frac{7-5}{-5-4}$

$=\frac{2}{-9}=-\frac{2}{9}$

Slope of the altitude of the triangle $=-\frac{1}{BC}$

$=-\frac{1}{-\frac{2}{9}}=\frac{9}{2}$

Hence, this is the answer.