Chapter 4.3, Problem 26PPS

To determine

To find the equation of line in slope-intercept form.

Answer to Problem 26PPS

  $y=4x-2$

Explanation of Solution

Given:

Line passing through the point $\left(2,6\right)$ and is perpendicular to the line $y=-\frac{1}{4}x+3$ .

Concept used:

Two lines are perpendicular if and only if the product of their slopes is $-1.$

Equation of a line passing through the point $\left(x_1,y_1\right)$ with slope m is $y-y_1=m\left(x-x_1\right)$ .

Calculation:

Comparing $y=-\frac{1}{4}x+3$ with $y=mx+c$ , we get

  $m=\text{slope}=-\frac{1}{4}$

Now let m₁ be the slope of the required line.

Since required line is perpendicular to the line $y=-\frac{1}{4}x+3$ , therefore product of their slopes must be $-1.$

  $\begin{array}{l}\therefore m_1\cdot m=-1\\ \Rightarrow m_1\cdot \left(-\frac{1}{4}\right)=-1\\ \Rightarrow m_1=4\end{array}$

Hence equation of the line passing through the point $\left(2,6\right)$ with slope 4 is,

  $\begin{array}{l}y-6=4\left(x-2\right)\\ \Rightarrow y-6=4x-8\\ \Rightarrow y=4x-8+6\\ \Rightarrow y=4x-2\end{array}$