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?

1 Answer
Nov 25, 2016

Answer:

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)#.