Chapter 4.3, Problem 10CYU

To determine

To Write: An equation of line in slope intercept form that passes through point $\left(3,6\right)$ and is perpendicular to line $3x-4y=-2$

Answer to Problem 10CYU

An equation of line in slope- intercept form is $y=-\frac{4}{3}+10$

Explanation of Solution

Given information:

Line passes through point $\left(3,6\right)$ and is perpendicular to line $3x-4y=-2$

Concept and Formula Used:

Slope intercept form $y=mx+c$ where $m$ is slope and $c$ is intercept.

Point slope form of a line: $y-y_1=m\left(x-x_1\right)$

Perpendicular lines: If lines are perpendicular then the product of their slope is $-1$

Calculation:

The required line is perpendicular to

  $\begin{array}{l}3x-4y=-2\\ \text{Add }4y\text{ on both side }\\ 3x-4y+4y=-2+4y\\ 3x=-2+4y\\ \text{Add 2 to both sides}\\ 3x+2=-2+4y+2\\ 3x+2=4y\\ \text{Divide both sides by 4}\\ \frac{4y}{4}=\frac{3}{4}x+\frac{2}{4}\\ y=\frac{3}{4}x+\frac{1}{2}\end{array}$

Now, find the slope of given line by compairing with $y=mx+c$ .

The slope of line $y=\frac{3}{4}x+\frac{1}{2}$ is $\frac{3}{4}$ .

Let the slope of required line be ‘m’. As the required line is perpendicular to $y=\frac{3}{4}x+\frac{1}{2}$

Therefore, product of slopes is $-1$

  $\begin{array}{l}m\times \frac{3}{4}=-1\\ m=-\frac{4}{3}\end{array}$

The required line passes through $\left(3,6\right)$ and having slope $-\frac{4}{3}$ . Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(3,6\right)\&m=-\frac{4}{3}\\ y-\left(6\right)=-\frac{4}{3}\left(x-\left(3\right)\right)\\ y-6=-\frac{4}{3}x+4\left(\text{Distributive property}\right)\\ \text{Add 6 to both sides}\\ y-6+6=-\frac{4}{3}x+4+6\\ y=-\frac{4}{3}+10\end{array}$

Conclusion:

An equation of line in slope- intercept form is $y=-\frac{4}{3}+10$