Chapter 4, Problem 13PT

To determine

To Write: An equation of line in slope intercept form that passes through point $\left(1,4\right)$ and is perpendicular to line $y=-2x+5$

Answer to Problem 13PT

An equation of line in slope- intercept form is $y=\frac{1}{2}x+\frac{7}{2}$

Explanation of Solution

Given information:

Line passes through point $\left(1,4\right)$ and is perpendicular to line $y=-2x+5$

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 $y=-2x+5$ .First find the slope of given line by comparing with $y=mx+c$ .

The slope of line $y=-2x+5$ is $-2$ .

Let the slope of required line be ‘m’. As the required line is perpendicular to $y=-2x+5$

Therefore, product of slopes is $-1$

  $\begin{array}{l}m\times -2=-1\\ m=\frac{1}{2}\end{array}$

The required line passes through $\left(1,4\right)$ and having slope $\frac{1}{2}$ . Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(1,4\right)\&m=\frac{1}{2}\\ y-\left(4\right)=\frac{1}{2}\left(x-\left(1\right)\right)\\ y-\left(4\right)=\frac{1}{2}\left(x-1\right)\\ y-4=\frac{1}{2}x-\frac{1}{2}\left(\text{Distributive property}\right)\\ \text{Add 4 to both sides}\\ y-4+4=\frac{1}{2}x-\frac{1}{2}+4\\ y=\frac{1}{2}x-\frac{1-8}{2}\\ y=\frac{1}{2}x+\frac{7}{2}\end{array}$

Conclusion:

An equation of line in slope- intercept form is $y=\frac{1}{2}x+\frac{7}{2}$