Q. 32 - Algebra 2

Question

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

passes through $\left(-\mathbf{2},-\mathbf{3}\right)$ and $\left(\mathbf{0},\mathbf{0}\right)$

Step-by-Step Solution

Answer

The equation of the required straight line in slope-intercept form is $y=\frac{3}{2}x$.

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 $\left(-2,-3\right)$ and $\left(0,0\right)$.

Then, $\left(h,k\right)=\left(-2,-3\right)$ and $\left(a,b\right)=\left(0,0\right)$.

Step 3. Write the equation

Put $\left(h,k\right)=\left(-2,-3\right)$ and $\left(a,b\right)=\left(0,0\right)$ in $\frac{y-k}{x-h}=\frac{b-k}{a-h}$ to get,

 

$\frac{y-\left(-3\right)}{x-\left(-2\right)}=\frac{0-\left(-3\right)}{0-\left(-2\right)}$

 

$\frac{y+3}{x+2}=\frac{0+3}{0+2}$  (Simplify)

 

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

 

$\frac{y+3}{x+2}\left(x+2\right)=\frac{3}{2}\left(x+2\right)$  (Multiply both sides by $x+2$)

 

$y+3=\frac{3}{2}\left(x+2\right)$  (Simplify)

 

$y+3=\frac{3}{2}x+3$  (Distributive property)

 

$y+3-3=\frac{3}{2}x+3-3$  (Subtract 3 from both sides)

 

$y=\frac{3}{2}x$  (Simplify)

 

So, the required equation of the straight line in slope-intercept form is $y=\frac{3}{2}x$.