Question

Find the standard form for the equation of the line that contains the point (-5, -12) and that is perpendicular to 2x + 3y = 12.

Answer 1

The equation of the line perpendicular to the given line is

`y=(3)/(2)x-(9)/(2)`

Step-by-step explanation:

Given the equation:

2x + 3y = 12

Let us rewrite it in standard form to have the slope and y-intercept.

Subtract 2x from both sides

3y = 12 - 2x

Divide both sides by 3

`y=(12)/(3)-(2)/(3)x`

or

`y=-(2)/(3)x+4`

Here, the slope is -2/3, and the y-intercept is 4

An line perpendicular to this line has it's slope as the negative reciprocal of the slope -2/3

The negative reciprocal of -2/3 is 3/2

The perpendicular line is in the form:

`y=(3)/(2)x+b`

Where b is the y-intercept.

Since the line contains the point (-5, -12), -5 and -12 are the coordinates of the x and y axes respectively, using them, we can obtain a value for b, the y-intercept.

`\begin{gathered} -12=(3)/(2)(-5)+b \\ \\ -12=-(15)/(2)+b \\ \\ \text{Add 15/2 to both sides} \\ -12+(15)/(2)=b \\ \\ b=-(9)/(2) \end{gathered}`

Therefore, the equation of the line perpendicular to the given line is

`y=(3)/(2)x-(9)/(2)`