Q39. - Algebra 2

Question

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

 

Passes through $\left(3,2\right)$, perpendicular to the graph of $4x-3y=12$ 

Step-by-Step Solution

Answer

The equation of the line is $y=-\frac{3}{4}x+\frac{17}{4}$.

Step-1 – Apply the concept of slope-intercept form

Equation of line in slope intercept form is expressed below.

$y=mx+c$

Where m is the slope and c is the intercept of y-axis.

Step-2 –Compute the intercept

The y-intercept of a graph is when the x-coordinate is 0. The point where the graph of the equation cuts the y-axis. 

It is provided that line passes through the point $\left(3,2\right)$ and perpendicular to the graph of $4x-3y=12$.

Product of slopes of perpendicular lines is $-1$. This means slope of line is $-1$ divided by slope of the line $4x-3y=12$.

Rewrite the equation in slope-intercept form to find the slope of the line.

Subtract both sides by 4x.

$\begin{array}{c}4x-3y-4x=12-4x\\ -3y=12-4x\end{array}$

Divide both sides by –3.

$\begin{array}{c}y=-\frac{12}{3}+\frac{4x}{3}\\ y=\frac{4}{3}x-4\end{array}$

Therefore, the slope of the line is $m=\frac{4}{3}$.

Now that line passes through the point $\left(3,2\right)$ with slope $m=-\frac{3}{4}$.

So, to find the y-intercept that is value of c, follow the step provided below.

Substitute x as 3, y as 2 and m as $-\frac{3}{4}$ in the equation $y=mx+c$.

 $\begin{array}{c}2=-\frac{3}{4}\left(3\right)+c\\ 2=-\frac{9}{4}+c\\ 2+\frac{9}{4}=c\\ c=\frac{17}{4}\end{array}$

Therefore, y-intercept is $\frac{17}{4}$.

Step-3 – Construct the equation

Recall the equation of line in slope-intercept form that is $y=mx+c$.

Where m is the slope and c is the intercept of y-axis.

Replace m by $-\frac{3}{4}$ and c by $\frac{17}{4}$ to obtain equation of line in slope-intercept form.

$y=-\frac{3}{4}x+\frac{17}{4}$

Hence, equation of line passing through the point $\left(3,2\right)$ and perpendicular to the graph of $4x-3y=12$ is $y=-\frac{3}{4}x+\frac{17}{4}$.