Question

How do you write an equation of the line in standard form that is Perpendicular to 3y-2x=6 and through (-1,2)?

Answer 1

`y=-3/2x+1/2`

Explanation:

The first step to solve this is to transform the original equation into `y=mx+b` form.

`3y-2x=6`
→Add `2x` to both sides

→`3y=2x+6`
→Divide `3` on both sides

→`y=2/3x+2`

A perpendicular line to this equation means that the slope of the new line will be the opposite of the original equation's slope.

For example if the original equation was `y=2x+3`, the slope of the perpendicular line would be `-1/2`.

So for this problem, the slope of perpendicular line will be `-3/2`.

Now that you have the slope, you have to solve for the `b`-value:

`y=-3/2x+b`
→ `2=-3/2(-1)+b`
→ `2=3/2+b`
→Subtract `3/2` to both sides
→`1/2=b`

Now that you have both the `b` and the `m`-value, you plug them into the `y=mx+b` form to get your answer.

`y=-3/2x+1/2`