# midpoint on line

• May 24th 2011, 02:40 AM
anthonye
midpoint on line
Hi;
With the pythagorean distance formula can I find the midpoint on a line
by dividing the result by 2.
• May 24th 2011, 02:54 AM
HallsofIvy
Quote:

Originally Posted by anthonye
Hi;
With the pythagorean distance formula can I find the midpoint on a line
by dividing the result by 2.

It's hard to understand what you are trying to say here. "by dividing the result by 2". Dividing what result by 2?

I guess you mean to first use the Pythagorean formula to find the length of the line segment, then use the Pythagorean formula with an unknown point, equal to half the previous distance to get an equation for x and y. Yes, you can do that but it far harder than necessary.

Suppose $\displaystyle (x_0, y_0)$ and $\displaystyle (x_1, y_1)$ are the endpoints of the line segment. Then $\displaystyle \sqrt{(x_1- x_0)^2+ (y_1- y_0)^2}$ is the length of the line segment. Let (x, y) be the midpoint. Then we must have $\displaystyle \sqrt{(x_1- x)^2+ (y_1- y)^2}= \frac{1}{2}\sqrt{(x_1- x_0)^2+ (y_1- y_0)^2}$. In order to find the two coordinates, x and y, we need another equation. Okay, use the other endpoint: \sqrt{(x- x_0)^2+ (y- y_0)^2= \frac{1}{2}\sqrt{(x_1- x_0)^2+ (y_1- y_0)^2}[/tex].

If you square those two equations you get $\displaystyle (x_1- x)^2+ (y_1- y)^2= \frac{1}{4}((x_1- x_0)^2+ (y_1- y_0)^2)$ and $\displaystyle (x- x_0)^2+ (y- y_0)^2= \frac{1}{4}((x_1- x_0)^2+ (y_1- y_0)^2)$

But it is far simpler to use "similar triangles" and get that the midpoint between $\displaystyle (x_0, y_0)$ and $\displaystyle (x_1, y_1)$ is $\displaystyle \left(\frac{x_1+ x_0}{2}, \frac{y_1+ y_0}{2}\right)$.
• May 24th 2011, 05:57 AM
anthonye
Yeah answered my own question earlier but internet was down.

Thanks