Question

Line L has equation `2x-3y= 5` and Line M passes through the point (2, 10) and is perpendicular to line L. How do you determine the equation for line M?

Answer 1

In slope-point form, the equation of line M is `y-10=-3/2(x-2)`.
In slope-intercept form, it is `y=-3/2x+13`.

Explanation:

In order to find the slope of line M, we must first deduce the slope of line L.

The equation for line L is `2x-3y=5`. This is in standard form, which does not directly tell us the slope of L. We can rearrange this equation, however, into slope-intercept form by solving for `y`:

`2x-3y=5`

`color(white)(2x)-3y=5-2x"    "`(subtract `2x` from both sides)

`color(white)(2x-3)y=(5-2x)/(-3)"     "`(divide both sides by `-3`)

`color(white)(2x-3)y=2/3 x-5/3"     "`(rearrange into two terms)

This is now in slope-intercept form `y=mx+b`, where `m` is the slope and `b` is the `y`-intercept. So, the slope of line L is `2/3`.

(Incidentally, since the slope of `2x-3y=5` was found to be `2/3`, we can show that the slope of any line `Ax+By=C` will be `-A/B`. This may be useful to remember.)

Okay. Line M is said to be perpendicular to line L—that is, lines L and M create right angles where they cross.

The slopes of two perpendicular lines will be negative reciprocals of each other. What does this mean? It means that if the slope of a line is `a/b`, then the slope of a perpendicular line will be `-b/a`.

Since the slope of line L is `2/3`, the slope of line M will be `-3/2`.

Alright—now we know the slope of line M is `-3/2`, and we know a point that it passes through: `(2,10)`. We now simply pick an equation for a line that allows us to plug in this data. I will choose to insert the data into the slope-point equation for a line:

`y-y_1=m(x-x_1)`

`y-10=-3/2(x-2)`

Choosing slope-point form allows us to simply stop here. (You could choose to use `y=mx+b`, where `(x,y)=(2,10)` and `m=-3/2`, then solve for `b`, and finally use this `b` along with `m` in slope-intercept form again:

`y="        "mx"  "+b`

`10=-3/2 (2)+b`

`10="   "-3"    "+b`

`13=b`

`:.y=mx+b`
`=>y=-3/2 x + 13`

Same line, different form.)