# finding slope

• Mar 29th 2010, 01:45 AM
dannyc
finding slope
Given that triangle $\displaystyle ABC$ has points, $\displaystyle A(8,2), B(0,6),$ and $\displaystyle C(-3,2)$, point $\displaystyle C$ can be moved along a certain line with points $\displaystyle A$ and $\displaystyle B$ remaining stationary, and the area of Δ$\displaystyle ABC$ will not change. What is the slope of that line?

(It says the answer in my book which is $\displaystyle -\frac {1}{2}$)
• Mar 29th 2010, 01:51 AM
pickslides
Quote:

Originally Posted by dannyc
Given that triangle $\displaystyle ABC$ has points, $\displaystyle A(8,2), B(0,6),$ and $\displaystyle C(-3,2)$, point C can be moved along a certain line with points A and B remaining stationary, and the area of Δ$\displaystyle ABC$ will not change. What is the slope of that line?

(It says the answer in my book which is $\displaystyle -\frac {1}{2}$)

The line you are after is parallel to AB so all you need to do is find the gradient of the line between A and B.

Follow?
• Mar 29th 2010, 02:00 AM
dannyc
Yep--it comes out to the right slope. But what's the logic behind it? (Why is the correct line parallel, is there a mini proof or is it more common knowledge that I somehow missed?)
• Mar 29th 2010, 02:05 AM
pickslides
Area of a rectangle is always $\displaystyle \frac{1}{2}\times \text{base}\times \text{height}$

The base and height will only always be the same (so the area is constant) for C, if C itself is only moving parallel to the line AB.

Draw a picture of ABC and move C around parallel to AB, you will see the base and height are always the same.
• Mar 29th 2010, 02:48 AM
HallsofIvy
Quote:

Originally Posted by pickslides
Area of a rectangle is always $\displaystyle \frac{1}{2}\times \text{base}\times \text{height}$

The area of a triangle, of course.

Quote:

The base and height will only always be the same (so the area is constant) for C, if C itself is only moving parallel to the line AB.

Draw a picture of ABC and move C around parallel to AB, you will see the base and height are always the same.