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.

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- Jan 16th 2006, 04:28 PMScott9909Determine 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 PMThePerfectHackerQuote:

Originally Posted by**Scott9909**

**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 PMScott9909
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 PMThePerfectHacker
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 PMScott9909
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 AMearbothQuote:

Originally Posted by**Scott9909**

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

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.

I hope that these additional remarks helped a little bit.

Greetings

EB