Q. 38 - Algebra 2

Question

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

passes through, parallel to the line that passes through $\left(\mathbf{3},\mathbf{3}\right)$ and $\left(\mathbf{0},\mathbf{6}\right)$

Step-by-Step Solution

Answer

The equation of the required straight line in slope-intercept form is $y=-x-4$.

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 slopes of parallel lines are equal.

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 line passes through $\left(-3,-1\right)$ and is parallel to the line passing through $\left(3,3\right)$ and $\left(0,6\right)$.

Then, $\left(h,k\right)=\left(-3,-1\right)$, $\left(a,b\right)=\left(3,3\right)$ and $\left(c,d\right)=\left(0,6\right)$.

Step 3. Determine the slope

Put $\left(a,b\right)=\left(3,3\right)$ and $\left(c,d\right)=\left(0,6\right)$ in $m=\frac{d-b}{c-a}$ to get,

$\begin{array}{c}m=\frac{6-3}{0-3}\\ =\frac{3}{-3}\\ =-1\end{array}$

So, $m=-1$

So, the slope of the line passing through $\left(3,3\right)$ and $\left(0,6\right)$, and thus of the required line, as they are parallel, is $-1$.

Then, $m=-1$.

Step 4. Write the equation

Put $m=-1$ and $\left(h,k\right)=\left(-3,-1\right)$ in $y-k=m\left(x-h\right)$ to get,

 

$y-\left(-1\right)=-1\left(x-\left(-3\right)\right)$

 

$y+1=-1\left(x+3\right)$  (Simplify)

 

$y+1=-x-3$  (Distributive property)

 

$y+1-1=-x-3-1$  (Subtract 1 from both sides)

 

$y=-x-4$   (Simplify)

 

So, the required equation of the straight line in slope-intercept form is $y=-x-4$.