# Determine the equation of the median from vertex

• Jan 16th 2006, 04:28 PM
Scott9909
Determine the equation of the median from vertex
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.
• Jan 16th 2006, 06:32 PM
ThePerfectHacker
Quote:

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 $\displaystyle C=(-5,3)$ is a line which joins the midpoint of side $\displaystyle AB$. But by the midpoint formula the midpoint of $\displaystyle AB$ is $\displaystyle (\frac{3-1}{2},\frac{7-6}{2})=(1,1/2)$. Thus, the median passes through points $\displaystyle C=(-5,3)$ and $\displaystyle (1,1/2)$.

Part the Second, find the equation of median: Using the slope-point formula which states that the equation of a line passing through point $\displaystyle (x_0,y_0)$ having slope $\displaystyle m$ is $\displaystyle y-y_0=m(x-x_0)$. Thus, the slope of $\displaystyle (1,1/2),(-5,3)$ is $\displaystyle m=-5/12$. Thus, the equation of line is (use any point for $\displaystyle (x_0,y_0)$)
$\displaystyle y+5=-5/12(x-3)$ Open and simplify,
$\displaystyle y=-\frac{5}{12}x-\frac{15}{4}$
Q.E.D.
• Jan 16th 2006, 06:56 PM
Scott9909
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.
• Jan 16th 2006, 07:09 PM
ThePerfectHacker
That is exactly what I did $\displaystyle (y_2-y_1)/(x_2-x_1)$. You mean the formula for the equation of the line?
• Jan 16th 2006, 07:13 PM
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.

Sorry if these questions are stupid.
• Feb 18th 2006, 01:43 AM
earboth
Quote:

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,

1. If you have 2 points $\displaystyle P_1,\ P_2$ with the coordinates $\displaystyle P_1(x_1,\ y_1),\ P_2(x_2,\ y_2)$ then you'll get the midpoint $\displaystyle M \left( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
2. A line through 2 points is described completely by the following equation: $\displaystyle \frac{y-y_1}{x-x_1} = \frac{y_2 - y_1}{x_2 - x_2}$
Solve this equation for y and you'll get: $\displaystyle y = \frac{y_2 - y_1}{x_2 - x_2}\cdot (x-x_1) + y_1$ where $\displaystyle \frac{y_2 - y_1}{x_2 - x_2}$ is the slope of the line.