Answer to Question #158898 in Analytic Geometry for Brayden

Question

Determine an equation AND length for the median from vertex C for the triangle with vertices:

C(5, 2) | A(‐3, 3) | B(2, ‐5)

Answer 1

Solution:

Median: A median is the line segment drawn from the vertex to the midpoint of the opposite side.

As per the image below we have M as the mid point of side AB.

Mid Point(M) = ( $\frac{x1 + x2}{2}$ , $\frac{y1 + y2}{2}$ )

M = ($\frac{-3 + 2}{2}$ ,$\frac{3 + (-5)}{2}$ )

M = ($\frac{-1}{2}$ ,$-1$ )

Now we need to find the equation of the line CM

We have the formula to find the equation of the line passing through two points.

$y-y1=m(x-x1)$ ------------------------ (1)

Where m = $\frac{y2 - y1}{x2 - x1}$

Calculate m

m = $\frac{-1-2}{-1/2 - 5}$

m = $\frac{6}{11}$

Plug the value of m and ($x1$ ,$y1$ ) into equation (1)

$y- y1 = m(x-x1)$

$y-2 =\frac{6}{11}(x-5)$

$y-2 = \frac{6x}{11} - \frac{30}{11}$

$y= 2+\frac{6x}{11} - \frac{30}{11}$

$y= \frac{6x}{11}- \frac{8}{11}$ --------------------------(2)

Distance:

Distance formula = $\sqrt{(x1-x2)^2 + (y1-y2)^2}$

Distance between Point C and Point M

D = $\sqrt{(5+\frac{1}{2})^2 +(2+1)^2}$

D = $\sqrt{(\frac{11}{2})^2 + 3^2}$

D = $\sqrt{\frac{121}{4} + 9}$

D = $\sqrt{\frac{157}{4}}$

D = 6.26