# Thread: Calculating the equation and slope of a Right Bisector

1. ## Calculating the equation and slope of a Right Bisector

I'm doing a math course through correspondence and I'm having some trouble with a couple of things that haven't been properly explained in the work book.

There are two things I need help with:

1) Calculating the slope of a Right Bisector;

The work book says that the Right Bisector must be perpendicular to the original line segment, and that the product of their slopes must equal -1.

They give two examples,

2 x (right bisector) = -1
2 x -½ = -1

That one is pretty obvious, but the next example they give has me completely stumped. I can't figure out the method they used to get the answer. I'll write what they have in the book below.

-7/4 x (right bisector) = -1
-7/4 x 4/7 = -1

The math adds up, but how did they figure out that the slope of the Right Bisector was 4/7? They don't explain it! Anyways, could someone help me figure out their method or give me one of their own?

2) Calculating the equation of a Right Bisector;

The second problem I need help with is one where I'm pretty sure I have the right answer. What's wrong is that the answer in the back of the book is different from mine. It involves finding the equation of a Right Bisector. I'll write my answer and then the answer that is in the back of the book.

y - 1/2 = 4/7 (x - 1)
y - 1/2 = 4/7x - 4/7
y = 4/7x - 4/7 + 1/2
y = 4/7x - 8/14 + 7/14
y = 4/7x - 1 1/14

y - 1/2 = 4/7 (x - 1)
y - 1/2 = 4/7x - 4/7
y = 4/7x - 4/7 + 1/2
y = 4/7x - 8/14 + 7/14
y = 4/7x - 1/14

Did they mess up or what? They didn't say anywhere in the lesson that you have to throw away whole numbers, so I'm assuming they didn't take the time to proofread their answers. I might be wrong though, and if I am could you please explain why?

Thanks.

2. Originally Posted by mathdonkey
1) Calculating the slope of a Right Bisector;

The work book says that the Right Bisector must be perpendicular to the original line segment, and that the product of their slopes must equal -1.

They give two examples,

2 x (right bisector) = -1
2 x -½ = -1

That one is pretty obvious, but the next example they give has me completely stumped. I can't figure out the method they used to get the answer. I'll write what they have in the book below.

-7/4 x (right bisector) = -1
-7/4 x 4/7 = -1

The math adds up, but how did they figure out that the slope of the Right Bisector was 4/7? They don't explain it! Anyways, could someone help me figure out their method or give me one of their own?
Say that we have two lines with slopes $m_1$ and $m_2$ respectively and that the two lines are perpendicular. Then we have that
$m_1m_2 = -1$

Thus
$m_2 = -\frac{1}{m_1}$

So if we have our first line with a slope $m_1 = 2$, then
$m_2 = -\frac{1}{2}$

If we have our first line with a slope $m_1 =-\frac{7}{4}$, then
$m_2 = - \frac{1}{-\frac{7}{4}} = -\frac{1}{-\frac{7}{4}} \cdot \frac{4}{4} = \frac{4}{7}$

-Dan

If (-7/4)*(Right Bisector) = -1, then (Right Bisector) = (-1)/(-7/4) = (-1)*(-4/7) = 4/7

Your second answer is, sorry, the book got it right.

-8 + 7 = -1, not -15

Good work. You do seem to be on the right track.

4. Originally Posted by TKHunny

If (-7/4)*(Right Bisector) = -1, then (Right Bisector) = (-1)/(-7/4) = (-1)*(-4/7) = 4/7

Your second answer is, sorry, the book got it right.

-8 + 7 = -1, not -15

Good work. You do seem to be on the right track.
Oh, I understand my mistake now. The negative has to be attributed to the 8.

Thanks topsquark & TKHunny, this all seems much simpler now!

5. Yes, you have it. It is a common point of confusion how that "-" means subtraction in one moment and negative in another.