Question

What is the equation of the line perpendicular to y=3x-6 that runs through the point (-1,-3) ?

Answer 1

`y=-1/3x-10/3`

Explanation:

The given line, `y=3x-6` is in slope-intercept form:

`y=mx+b,`

where `m` is the slope `(3)`, and `b` is the y-intercept, `(-6)`.

The slope of a perpendicular line is the negative reciprocal of the slope of the original line. The formula for finding the slope of a perpendicular line is:

`m_1m_2=-1,`

where:

`m_1` is the slope of the first line, and `m_2` is the slope of the perpendicular line.

`m_2=-1/(m_1)`

`m_2=-1/3`

Now that we have the slope of the perpendicular line, we can use the slope and the given point to determine the point-slope form for a linear equation.

`y-y_1=m(x-x_2),`

where:

`m=-1/3`, and `(x_1,y_1)=(-1,-3)`

Plug in the known values.

`y-(-3)=-1/3(x-(-1))`

Simplify.

`y+3=-1/3(x+1)` `larr` point-slope form.

We can solve the point-slope form for `y` to get the slope-intercept form.

`y+3=-1/3(x+1)`

Simplify.

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

Subtract `3` from both sides.

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

`1/3` and `3` must have the same denominator in order to subtract. A whole number has a denominator of `1`, so `3=3/1`.

Multiply `3/1` by `3/3` to convert it to the equivalent fraction `9/3`.

`y=-1/3x-1/3-9/3`

`y=-1/3x-10/3` `larr` perpendicular line

graph{(y-3x+6)(y+1/3x+10/3)=0 [-14.24, 14.23, -7.12, 7.12]}