Q20. - Algebra 2

Question

Write an equation in slope-intercept form for each graph.

Step-by-Step Solution

Answer

The equation of the required straight line in slope-intercept form is $y=-\frac{5}{3}x+\frac{29}{3}$.

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 clear, from the graph, that the line passes through $\left(4,3\right)$ and $\left(7,-2\right)$.

Then, $\left(h,k\right)=\left(4,3\right)$ and $\left(a,b\right)=\left(7,-2\right)$.

Step 3 – Write the equation

Put $\left(h,k\right)=\left(4,3\right)$ and $\left(a,b\right)=\left(7,-2\right)$ in $\frac{y-k}{x-h}=\frac{b-k}{a-h}$ to get,

 

$\frac{y-3}{x-4}=\frac{-2-3}{7-4}$

 

$\frac{y-3}{x-4}=-\frac{5}{3}$  (Simplify)

 

$\frac{y-3}{x-4}\left(x-4\right)=-\frac{5}{3}\left(x-4\right)$  (Multiply both sides by $x-4$)

 

$y-3=-\frac{5}{3}\left(x-4\right)$  (Simplify)

 

$y-3=-\frac{5}{3}x+\frac{20}{3}$  (Distributive property)

 

$y-3+3=-\frac{5}{3}x+\frac{20}{3}+3$  (Add 3 to both sides)

 

$y=-\frac{5}{3}x+\frac{29}{3}$

 

So, the required equation of the straight line in slope-intercept form is $y=-\frac{5}{3}x+\frac{29}{3}$.