Q9. - Algebra 2

Question

Write an equation in slope-intercept form for the line that satisfies each set of conditions.

passes through$\left(-\mathbf{3},\mathbf{5}\right)$ and $\left(\mathbf{2},\mathbf{2}\right)$

Step-by-Step Solution

Answer

The equation of the required straight line in slope-intercept form is $y=-\frac{3}{5}x+\frac{16}{5}$.

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 equation of a straight-line passing through the points $\left(h,k\right)$ and $\left(a,b\right)$ is given as $\frac{y-k}{x-h}=\frac{b-k}{a-h}$.

Step 2 – List the given data

It is given that the line passes through $\left(-3,5\right)$ and $\left(2,2\right)$.

Then, $\left(h,k\right)=\left(-3,5\right)$ and $\left(a,b\right)=\left(2,2\right)$.

Step 3 – Write the equation

Put $\left(h,k\right)=\left(-3,5\right)$ and $\left(a,b\right)=\left(2,2\right)$ in $\frac{y-k}{x-h}=\frac{b-k}{a-h}$ to get,

$\frac{y-5}{x-\left(-3\right)}=\frac{2-5}{2-\left(-3\right)}$ 

$\frac{y-5}{x+3}=\frac{-3}{2+3}$  (Simplify) 

$\frac{y-5}{x+3}=-\frac{3}{5}$  (Simplify)

$\frac{y-5}{x+3}\left(x+3\right)=-\frac{3}{5}\left(x+3\right)$  (Multiply both sides by $x+3$)

$y-5=-\frac{3}{5}\left(x+3\right)$ (Simplify) 

$y-5=-\frac{3}{5}x-\frac{9}{5}$  (Distributive property) 

$y-5+5=-\frac{3}{5}x-\frac{9}{5}+5$  (Add 5 to both sides) 

$y=-\frac{3}{5}x+\frac{16}{5}$  (Simplify) 

So, the required equation of the straight line in slope-intercept form is $y=-\frac{3}{5}x+\frac{16}{5}$.