Q51. - Algebra 2

Question

51. CRITICAL THINKING Given $\Delta \mathbf{ABC}$ with vertices $\mathbf{A}\left(-\mathbf{6},-\mathbf{8}\right)$,$\mathbf{B}\left(\mathbf{6},\mathbf{4}\right)$ and $\mathbf{C}\left(-\mathbf{6},\mathbf{10}\right)$, write an equation of the line containing the altitude from $\mathbf{A}$.

(Hint: The altitude from $\mathbf{A}$ is a segment that is perpendicular to $\overline{\mathbf{BC}}$)

Step-by-Step Solution

Answer

The equation of the required straight line is $y=2x+4$.

Step 1 – State the concept

The slope intercept form of a straight-line equation is $y=mx+c$ where $m$ is the slope and $c$ is the y-intercept.

The product of slopes of perpendicular lines is $-1$.

The slope of a line passing through $\left(a,b\right)$ and $\left(c,d\right)$ is $m=\frac{d-b}{c-a}$.

The equation of a straight-line having slope $m$ and passing through the point $\left(h,k\right)$ is given as$y-k=m\left(x-h\right)$.

Step 2 – List the given data

It is given that the three vertices of $\Delta ABC$ are $A\left(-6,-8\right)$, $B\left(6,4\right)$ and $C\left(-6,10\right)$.

The objective is to find the line containing the altitude from $A$.

The altitude from $A$ is a line segment that passes through $A$ and is perpendicular to $\overline{BC}$.

Then, $\left(h,k\right)=\left(-6,-8\right)$, $\left(a,b\right)=\left(6,4\right)$ and $\left(c,d\right)=\left(-6,10\right)$.

Step 3 – Determine the slope

Put $\left(a,b\right)=\left(6,4\right)$ and $\left(c,d\right)=\left(-6,10\right)$ in $m=\frac{d-b}{c-a}$ to get,

$\begin{array}{l}m=\frac{10-4}{-6-6}\\ =\frac{6}{-12}\\ =-\frac{1}{2}\end{array}$

So, the slope of $\overline{BC}$ is $-\frac{1}{2}$.

Then, the slope of the altitude from $A$, which is perpendicular to $\overline{BC}$, is $2$.

Then, $m=2$.

Step 4 – Write the equation

Put $m=2$ and $\left(h,k\right)=\left(-6,-8\right)$ in $y-k=m\left(x-h\right)$ to get,

$y-\left(-8\right)=2\left(x-\left(-6\right)\right)$ 

$y+8=2\left(x+6\right)$ (Simplify) 

$y+8=2x+12$  (Distributive property)

$y+8-8=2x+12-8$  (Subtract 8 from both sides)

$y=2x+4$   (Simplify)

So, the required equation of the straight line is $y=2x+4$.