Question

How do you write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (1.7) and perpendicular to 3x+7y= 1?

Answer 1

The slope-intercept form is `y=7/3x+14/3` and standard form is `-7x+3y=14`.

Explanation:

1. Find the slope for `3x+7y=1`.
The slope of a line in `Ax+By=C` form is always `-A/B`. (Rearranging `Ax+By=C` into slope-intercept form gives `y=-A/Bx+C/B.`) Thus, the slope of this line is `-3/7`.

2. Find the slope of a line perpendicular to this given line.
Lines that are perpendicular have slopes that are negative reciprocals of each other. (One line's `"rise"/"run"` will be the other's `(-"run")/"rise"`.) So the slope of a line perpendicular to this one will be `m=7/3`.

3. Plug in the known point `(x_1,y_1)` and `m` into the slope-point equation `y-y_1=m(x-x_1)`.
Assuming the point given is `(x_1,y_1)=(1,7)`, our slope-point equation is `y-7=7/3(x-1)`.

4. Rearrange this equation into slope-intercept form.
Solving for `y` yields:

`y=7/3x-7/3+7`
`=>y=7/3x+14/3`.

5. Rearrange (3) or (4) into standard form.
Starting from the equation in (3), we get

`3y-21=7x-7`
`=>-7x+3y=14`

(Alternatively, starting from (4), we get

`3y=7x+14`
`=>-7x+3y=14`

which is the same equation.)

A graph of `3x+7y=1`:
graph{3x+7y=1 [-12.74, 9.76, -2.16, 9.095]}
For every 3 it goes down, it goes 7 right.

A graph of `-7x+3y=14`:
graph{-7x+3y=14 [-12.74, 9.76, -2.16, 9.095]}
For every 7 it goes up, it goes 3 right.
And as you can see, this second line does indeed pass through `(1,7)`.