Question

If A$(-1,8)$, B$(4,-2)$ and C$(-5,-3)$ are the vertices of a triangle, then the equation of the altitude through $(-1,8)$ is given as $ax+b=y$. Find the value of $a+b$.

Answer 1

In the given figure, AD is perpendicular to BC.

Slope of a line passing through (a, b) and (c, d) $=\frac{d-b}{c-a}$
Slope of BC $=\frac{-3-(-2)}{-5-4}=\frac{1}{9}$
$\left[0.5mark\right]$

Product of slopes of perpendicular lines is -1.
Let the slope of AD be m.
AD $\perp$ BC
So, $\frac{m}{9}=-1\to m=-9$
$\left[0.5mark\right]$

Equation of a line passing through (a, b) and slope 'm' = $\frac{y-b}{x-a}=m$

Equation of AD $=\frac{y-8}{x-(-1)=9\to y-8=9x+9\to y=9x+17}$

Comparing $y=9x+17$ with $y=ax+b$, we get $a=9$ and $b=17a+b=9+17=26$
$\left[1mark\right]$