Question

Indicate the equation of the given line in standard form.

The line that is the perpendicular bisector of the segment whose endpoints are R(-1, 6) and S(5, 5)

Answer 1

Do you know how to find the eq. of the line containing \(RS\)?

Answer 2

eq?

Answer 3

equation

Answer 4

i think but i just don't know what to do when they say there is a perpendicular bisector

Answer 5

The line you're looking for is \(perpendicular\) to \(RS\), and it also \(bisects\) it, which means the line you're looking for intersects \(RS\) at its midpoint.

Answer 6

Perpendicular lines have slopes that are opposite reciprocals of each other, so the first thing you have to do is find the slope of the line containing \(RS\).

Then you find its midpoint.

Answer 7

so -6 is the slope

Answer 8

Yep, so that means the slope of the perpendicular line is...

Answer 9

well it was 1/6 so i did what you said to make it -6

Answer 10

i got 1/6 for RS

Answer 11

Sorry, you're right. I misread the points.

Answer 12

all good :)

Answer 13

so i end up with y=-6x what do i do from there? or is that even right?

Answer 14

That's not the right equation, no. You have to use the point-slope formula:
\[y-y_0=m(x-x_0)\]
\((x_0,y_0)\) is either given point, and \(m\) is the slope. Plug in those values and solve for y.

Answer 15

can you show me how to do that?

Answer 16

Just a moment, I'm getting ahead of myself... Something doesn't look right to me.

Answer 17

The slope of \(RS\) should be \(-\dfrac{1}{6}\), for one thing:
\[\text{slope}=\frac{5-6}{5-(-1)}=\frac{-1}{6}\]
which means the slope of the perp. bisector is actually \(6\).

Answer 18

And my first instinct to use the point-slope formula was right, but I used the wrong info. The point I mention in the comment with the formula should be the midpoint of \(RS\).
\[\text{midpoint}=\left(\frac{-1+5}{2},\frac{6+5}{2}\right)=\left(2,\frac{11}{2}\right)\]
Now plug all this information into the point-slope formula. This is the equation of the line you're looking for:
\[y-\frac{11}{2}=6(x-2)\]
Do you know how to write the eq. of a line in standard form?

Answer 19

no i'm not quite sure

Answer 20

First thing you would do is distribute the 6, then move all the x's and y's to one side, and the constant to the other. Doing this, you have
\[y-6x=-12+\frac{11}{2}\\
-6x+y=-\frac{13}{2}\]
Standard form is
\[Ax+By=C\]