Question

Write the equation of a line perpendicular to the line:
y=43x−1 that goes through the point (-6, -6).
Write your answer in slope- intercept form, using reduced fractions for the slope and intercept.
y-__=__(x-__)
y=__x+__

Answer 1

Given:

The equation of perpendicular line is:

$y=\dfrac{4}{3}x-1$

The required line passes through the point is (-6,-6).

To find:

The equation of the line.

Solution:

The slope intercept form of a line is:

$y=mx+b$               ...(i)

Where, m is the slope and b is the y-intercept.

We have,

$y=\dfrac{4}{3}x-1$           ...(ii)

On comparing (i) and (ii), we get

$m=\dfrac{4}{3}$

Slope of given line is $\dfrac{4}{3}$.

The product of slopes of two perpendicular line is -1.

$m_1\times \dfrac{4}{3}=-1$

$m_1=-\dfrac{3}{4}$

So, the slope of the required line is $m_1=-\dfrac{3}{4}$. It passes through the point is (-6,-6). So, the equation of the line is:

$y-y_1=m_1(x-x_1)$

$y-(-6)=-\dfrac{3}{4}(x-(-6))$

$y+6=-\dfrac{3}{4}(x+6)$

On further simplification, we get

$y+6=-\dfrac{3}{4}(x)-\dfrac{3}{4}(6)$

$y+6=-\dfrac{3}{4}(x)-4.5$

$y=-\dfrac{3}{4}(x)-4.5-6$

$y=-\dfrac{3}{4}(x)-10.5$

Therefore, equations of the required line are $y+6=-\dfrac{3}{4}(x+6)$ and $y=-\dfrac{3}{4}(x)-10.5$.