Q. 5 - Precalculus Enhanced with Graphing Utilities

Question

Find the equation of the line perpendicular to the line y = 2x + 1 and containing the point (3, 5). Express your answer in slope–intercept form and graph the line. 

Step-by-Step Solution

Answer

The equation of line is $y=-\frac{1}{2}x+\frac{13}{2}$.

Step 1. Given information.

Find the equation of the line perpendicular to the line y = 2x + 1 and containing the point (3, 5). Express your answer in slope–intercept form and graph the line. 

Step 2. Step 2. Definition.

The equation of line ( in slope - intercept form) is given by:

$y=mx+b; m\ne 0$

If two directions are vertical, their direction coefficients are inversely proportional:

$m_2=-\frac{1}{m_1}$ 

Step 3. Equation of line.

Slope:

There is given the equation of line $y=2x+1$, so the $m_1=2$. The line we are searching for is perpendicular to the given line.

Then $m_2=-\frac{1}{2}$.

The y - intercept:

The point $\left(3,5\right)$is on the, because we want to find b , put the coordinates of the points inside the equation: 

$$\begin{gathered} y=-\frac{1}{2}x+b \\ 5=-\frac{1}{2}·3+b \\ 5=-\frac{3}{2}+b \\ b=5+\frac{3}{2} \\ b=\frac{13}{2} \end{gathered}$$

So, the equation of line is :

$y=-\frac{1}{2}x+\frac{13}{2}$y

Step 4. Conclusion.

The equation of line is $y=-\frac{1}{2}x+\frac{13}{2}y$.