Two Problems

2.  M =  ( 1.5 sqrt (3), 3)      N  =  (0, 3)

The slope of AM  =   3 / (1.5sqrt (3) )   =  2/sqrt(3)

And the equation of  AM  =  

y  = [2/sqrt(3) ] x

The slope of CN  =   [ 3 / -3sqrt (3)]  = -1/sqrt(3)

And the equation of AN  =

y  =  [ -1/sqrt(3)] x  +  3

To find the x coordinate of P, we have

[2/sqrt(3) ] x  =   [ -1/sqrt(3)] x  +  3

3/sqrt (3)x  =  3

x =  3 *sqrt (3) / 3   =  sqrt (3)

And the y coordinate  of P

y  = [2/sqrt(3)] (sqrt (3)  =  2

So  CP  =  √ [ ( 3sqrt (3) - sqrt (3) )^2  + (2 - 0)^2 ]  = √ (2sqrt(3))^2 + 4 )  =

√  [ (sqrt (12) )^2  + 4 ]  =  √ [ 12 + 4]  =  √16   =   

4

Both questions are to do with medians of a triangle, and the solutions come easiest from the fact that the medians trisect each other. You should have been made aware of this prior to looking at these questions.

The second question is the easier to see, so look at your  diagram for that.

The medians AM and CN cross at the point P.

By trisect, I mean that the point P splits the line AM into two parts in the ratio 2 : 1, and similarly P splits the line CN into two parts in the ratio 2 : 1.

The 2 part is the bit from the vertex, the 1 part is the bit from the side.

So, AP = 2PM (sounds like some time in the afternoon !), and CP = 2PN.

In question 2 then, rather than the earlier ' round the houses ' approach, say that

\(\displaystyle BN=AB/2=3,\text{ so }CN^{2}=3^{2}+(3\sqrt{3})^{2}=36\text{ (by Pythagoras)}\),

\(\displaystyle \text{so }CN=6,\text{ so }CP=(2/3)6=4.\)

Question 2 also can be used to give an insight into question 1.

Look at the diagram and in particular the triangles APC and PMC.

Considering the line AM to be a base line for both triangles, both APC and PMC have the same height.

Therefore, since AP is twice the length of PM it follows that the area of the triangle APC will be twice the area of the triangle PMC.

That's a clue as to how to tackle question 1.

Try it, and post again if it's still a problem.