Determine the equation of the median from vertex

**Scott9909** · January 16th 2006, 04:28 PM

ABC has the following co-ordinates: A(3,7)B(-1,-6) and C(-5,3). Determine the equation of the median from vertex C.

This question is giving me a bit of trouble. If someone could please help me out i would greatly appreciate it.

**ThePerfectHacker** · January 16th 2006, 06:32 PM

Originally Posted by Scott9909

ABC has the following co-ordinates: A(3,7)B(-1,-6) and C(-5,3). Determine the equation of the median from vertex C.

This question is giving me a bit of trouble. If someone could please help me out i would greatly appreciate it.

Part the First, find the coordinates of midpoint: By definition, the median from vertex $C=(-5,3)$ is a line which joins the midpoint of side $AB$ . But by the midpoint formula the midpoint of $AB$ is $(\frac{3-1}{2},\frac{7-6}{2})=(1,1/2)$ . Thus, the median passes through points $C=(-5,3)$ and $(1,1/2)$ .

Part the Second, find the equation of median: Using the slope-point formula which states that the equation of a line passing through point $(x_0,y_0)$ having slope $m$ is $y-y_0=m(x-x_0)$ . Thus, the slope of $(1,1/2),(-5,3)$ is $m=-5/12$ . Thus, the equation of line is (use any point for $(x_0,y_0)$ )
$y+5=-5/12(x-3)$ Open and simplify,
$y=-\frac{5}{12}x-\frac{15}{4}$
Q.E.D.

**Scott9909** · January 16th 2006, 06:56 PM

Im a bit confused on part 2.
Do you have to find the slope of the line? Im not furmilur with the formula you put up. Ive been taught to do it Y=X2-X1/Y2-y1

and i dont seem to be getting the same slope.

**ThePerfectHacker** · January 16th 2006, 07:09 PM

That is exactly what I did $(y_2-y_1)/(x_2-x_1)$ . You mean the formula for the equation of the line?

**Scott9909** · January 16th 2006, 07:13 PM

Im not exactly sure. I really dont understand math that well.

are you supposed to do y2-y/x2-x1 with your midpoint and C(-5,3)?

And also if it is what is considered y2 and x2? C or the midpoint.

Sorry if these questions are stupid.

**earboth** · February 18th 2006, 01:43 AM

Originally Posted by Scott9909

Im not exactly sure. I really dont understand math that well.

are you supposed to do y2-y/x2-x1 with your midpoint and C(-5,3)?

And also if it is what is considered y2 and x2? C or the midpoint.

Hello,

you've got the answer to your problem already. I'll give you only a few additional informations:

1. If you have 2 points $P_1,\ P_2$ with the coordinates $P_1(x_1,\ y_1),\ P_2(x_2,\ y_2)$ then you'll get the midpoint $M \left( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$

2. A line through 2 points is described completely by the following equation: $\frac{y-y_1}{x-x_1} = \frac{y_2 - y_1}{x_2 - x_2}$

Solve this equation for y and you'll get: $y = \frac{y_2 - y_1}{x_2 - x_2}\cdot (x-x_1) + y_1$ where $\frac{y_2 - y_1}{x_2 - x_2}$ is the slope of the line.

I hope that these additional remarks helped a little bit.

Greetings

EB

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