# Math Help - Finding Equations. 3 questions actually.

1. ## Finding Equations. 3 questions actually.

Determine the equation of the median BN, where N is the midpoint of AC.

The vertices of the triangle are A (13,13) B (9,3) and C (-1,-1)

And the equation I got fo the median AM is 4x-3y=13

I'm not even sure how i got the median AM equation.

I have no idea were to go with BN, i got 7x-7y = 13 for BN.

I really don't think thats right, someone help me please.

For the second question it asks this

Use the equations of AM and BN to determine the coordinates of P, the point of intersection of AM and BN.

Prove that the distance between P and M is one third and the length of the median, AM.

I'm sorry it's alot, but im having difficulity, alot of it.
Details will be EXTREMLY helpful thanks!

2. Hi

A (13,13) B (9,3) and C (-1,-1)
N is the midpoint of [AC] so

$x_N = \frac{x_A + x_C}{2} = 6$

$y_N = \frac{y_A + y_C}{2} = 6$

$\overrightarrow{BN}$ coordinates are
$x_{BN} = x_N - x_B = -3$
$y_{BN} = y_N - y_B = 3$

$\overrightarrow{BN}$ is a direction vector of the line (BN), so is $\frac{1}{3} \overrightarrow{BN}$ whose coordinates are (-1,1)

For a general Cartesian equation ax+by+c=0, we know that one direction vector has coordinates (-b,a)
Here -b=-1 and a=1
A Cartesian equation of (BN) is therefore x+y+c=0
The value of c is found by substituting B (or N) coordinates into the equation : c=-12
x+y-12=0

Your equation of (AM) is right : 4x-3y=13

P is the point of intersection of AM and BN therefore
$4 x_P - 3 y_P - 13 = 0$
$x_P + y_P - 12 = 0$

Which gives P(7,5)

M(4,1) and P(7,5) $\Rightarrow \overrightarrow{MP}$(3,4)

and $MP = \|\overrightarrow{MP}\| = \sqrt{3^2+4^2} = 5$

M(4,1) and A(13,13) $\Rightarrow \overrightarrow{MA}(9,12)$

and $MA = \|\overrightarrow{MA}\| = \sqrt{9^2+12^2} = 15$

Therefore MP = MA / 3