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=(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=(4)/(3)x-1` ...(ii)

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

`m=(4)/(3)`

Slope of given line is
`(4)/(3)`.

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

`m_1* (4)/(3)=-1`

`m_1=-(3)/(4)`

So, the slope of the required line is
`m_1=-(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)=-(3)/(4)(x-(-6))`

`y+6=-(3)/(4)(x+6)`

On further simplification, we get

`y+6=-(3)/(4)(x)-(3)/(4)(6)`

`y+6=-(3)/(4)(x)-4.5`

`y=-(3)/(4)(x)-4.5-6`

`y=-(3)/(4)(x)-10.5`

Therefore, equations of the required line are
`y+6=-(3)/(4)(x+6)` and
`y=-(3)/(4)(x)-10.5`.