Chapter 4, Problem 22MCQ

To determine

To Write: An equation of line in slope intercept form that passes through point $\left(-4,-5\right)$ and is perpendicular to line $-4x+5y=-6$

Answer to Problem 22MCQ

An equation of line in slope- intercept form is $y=-\frac{5}{4}x-10$

Explanation of Solution

Given information:

Line passes through point $\left(-4,-5\right)$ and is perpendicular to line $-4x+5y=-6$ .

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 $-4x+5y=-6$ .First find the slope of given line by comparing with $y=mx+c$ .

The slope of line $-4x+5y=-6\Rightarrow 5y=4x-6\Rightarrow y=\frac{4}{5}x-\frac{6}{5}$ is $m=\frac{4}{5}$ .

Let the slope of required line be ‘m’. As the required line is perpendicular to $-4x+5y=-6$

Therefore, product of slopes is $-1$

  $\begin{array}{l}m\times \frac{4}{5}=-1\\ m=-\frac{5}{4}\end{array}$

The required line passes through $\left(-4,-5\right)$ and having slope $-\frac{5}{4}$ . Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(-4,-5\right)\&m=-\frac{5}{4}\\ y-\left(-5\right)=-\frac{5}{4}\left(x-\left(-4\right)\right)\\ y+5=-\frac{5}{4}x-5\left(\text{Distributive property}\right)\\ \text{Subtract 5 from both sides}\\ y+5-5=-\frac{5}{4}x-5-5\\ y=-\frac{5}{4}x-10\end{array}$

Conclusion:

An equation of line in slope- intercept form is $y=-\frac{5}{4}x-10$