Question

Find the equation of the line passing the poin (-7,2) that is perpendicular to line 4x-3y=10

Answer 1

$y=-\frac{3}{4}x-\frac{13}{4}$

Explanation

Step 1

find the slope of the given equation:

two lines are perpendicular it the product of the slopes equals -1, so

$\begin{gathered} \text{ line 1 is perpendicular to line 2} \\ L1\parallel L2 \\ \text{if} \\ m_1\cdot m_2=-1 \end{gathered}$

then , let

$\begin{gathered} \text{ Line 1} \\ 4x-3y=10 \end{gathered}$

to know the slope, we need to convert the equation into the form:

$\begin{gathered} y=mx+b \\ \text{where m is the slope} \end{gathered}$

to do that, let's isolate y

so

$\begin{gathered} 4x-3y=10 \\ \text{subtract 4x in both sides} \\ 4x-3y-4x=10-4x \\ -3y=10-4x \\ \text{divide both sides by -3} \\ \frac{-3y}{-3}=\frac{10-4x}{-3} \\ y=-\frac{10}{3}+\frac{4}{3}x \\ y=\frac{4}{3}x-\frac{10}{3} \\ y=\frac{4}{3}x-\frac{10}{3}\rightarrow y=\text{ mx+b} \end{gathered}$

hence,

$\text{slope 1=m}_1=\frac{4}{3}$

Step 2

now, let's find the slope of the line 2

$\begin{gathered} \text{let} \\ m_1=\frac{4}{3} \\ \end{gathered}$

replacing

$\begin{gathered} m_1\cdot m_2=-1 \\ \frac{4}{3}\cdot m_2=-1 \\ \text{ to isolate, multiply both sides by 3/4} \\ \frac{4}{3}\cdot m_2\cdot\frac{3}{4}=-1\cdot\frac{3}{4} \\ m_2=-\frac{3}{4} \end{gathered}$

so, the slope 2 is -3/4

Step 3

finally, let's find the equation of the line

use the expression

$\begin{gathered} y-y_0=m(x-x_0) \\ \text{where} \\ \text{ m is the slope} \\ \text{and (x}_0,y_0)\text{ are the coordinates of a known point} \end{gathered}$

let

$\begin{gathered} m=m_2=-\frac{3}{4} \\ (x_0,y_0)=(-7,2) \end{gathered}$

replace and isolate y

$\begin{gathered} y-y_0=m(x-x_0) \\ y-2=-\frac{3}{4}(x-(-7)) \\ y-2=-\frac{3}{4}(x+7) \\ y-2=-\frac{3}{4}x-\frac{21}{4} \\ \text{add 2 in both sides} \\ y-2+2=-\frac{3}{4}x-\frac{21}{4}+2 \\ y=-\frac{3}{4}+\frac{-21+8}{4} \\ y=-\frac{3}{4}x-\frac{13}{4} \\ y=-\frac{3}{4}x-\frac{13}{4} \end{gathered}$

therefore, the answer is

$y=-\frac{3}{4}x-\frac{13}{4}$

I hope this helps you