There is a difficulty here with conflicting notation because the 
formulas you must use, which are:

The midpoint between (x1, y1) and (x2, y2) has coordinates:

(, ) 

and

The distance d between (x1, y1) and (x2, y2) is given by:



already have x1, x2, y1 and y2 in them.  When such a situation of conflicting
notation arises, we must either change the notation in the problem, or else
change the notation in the formula. We will change the notation in the
problem.  So, let's rewrite the problem as this instead:

The vertices of a triangle are A(0,0) B(p,0) C(q,r). 
M, N are the midpoints of AB,AC.
a) write the coordinates of M,N.




Use the midpoint formula to find the coordinates of M:

M(, )

Substitute 

(x1, y1) = (0,0) and (x2, y2) = (p,0)  


M(, )

M(, )

M(, 0)



Use the midpoint formula to find the coordinates of N:

N(, )

Substitute 

(x1, y1) = (0,0) and (x2, y2) = (q,r)  

N(, )

N(, )

N(, )





b) show that MN = BC 

Let's find MN and BC, using the distance formula.  Let's find BC first:

(x1, y1) = (p,0) and (x2, y2) = (q,r)







Now let's find MN:

(x1, y1) = (,0) and (x2, y2) = (,)







Write  as ,  as , and  as ,



Factor out  inside the first parentheses:



Remove the outer parentheses in both expression under the radical
by squaring the two factors individually:



Write the  as 



Factor  out under the radical:



Take indivisual square roots:



Write  as 



Now we notice that the right side of that is  times the right side of



So we have proved that MN = BC

Edwin