Question

What is the equation of the line that passes through the point (-2,3) and that is perpendicular to the line represented by 3x-2y= -2?

Answer 1

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

Or

`y = -3/2x`

Explanation:

First, we need to convert the line into slope-intercept form to find the slope.

The slope-intercept form of a linear equation is:

`y = color(red)(m)x + color(blue)(b)`

Where `color(red)(m)` is the slope and `color(blue)(b` is the y-intercept value.

We can solve the equation in the problem for `y`:

`3x - 2y = -2`

`3x - color(red)(3x) - 2y = -2 - color(red)(3x)`

`0 - 2y = -3x - 2`

`-2y = -3x - 2`

`(-2y)/color(red)(-2) = (-3x - 2)/color(red)(-2)`

`(color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2)) = (-3x)/color(red)(-2) - 2/color(red)(-2)`

`y = 3/2x + 1`

So for this equation the slope is `3/2`

A line perpendicular to this line will have a slope which is the negative inverse of our line or `-3/2`

We can now use the point-slope formula to write the equation for the perpendicular line:

The point-slope formula states: `(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))`

Where `color(blue)(m)` is the slope and `color(red)(((x_1, y_1)))` is a point the line passes through.

Substituting the point from problem and the slope we calculated gives:

`(y - color(red)(3)) = color(blue)(-3/2)(x - color(red)(-2))`

`(y - color(red)(3)) = color(blue)(-3/2)(x + color(red)(2))`

Or, we can put the equation in the more familiar slope-intercept form by solving for `y`:

`y - color(red)(3) = color(blue)(-3/2)x + (color(blue)(-3/2) xx color(red)(2))`

`y - color(red)(3) = -3/2x - 3`

`y - color(red)(3) + 3 = -3/2x - 3 + 3`

`y = -3/2x + 0`

`y = -3/2x`