Question 388362
<pre><font face = "BatangChe" size = 4><b>
There is a difficulty here with conflicting notation because the 
formulas you must use, which are:

The midpoint between (x<sub>1</sub>, y<sub>1</sub>) and (x<sub>2</sub>, y<sub>2</sub>) has coordinates:

({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}}) 

and

The distance d between (x<sub>1</sub>, y<sub>1</sub>) and (x<sub>2</sub>, y<sub>2</sub>) is given by:

{{{d}}}{{{""=""}}}{{{sqrt((x[2]-x[1]))^2+(y[2]-y[1])^2)}}}

already have x<sub>1</sub>, x<sub>2</sub>, y<sub>1</sub> and y<sub>2</sub> 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.

{{{drawing(400,400,-3,10,-3,10,

red(line(-11,0,11,0),line(0,-11,0,11)), locate(10,0,x),locate(0,10,y),
circle(7,0,.1),locate(6.5,-.2,"B(p,0)"),
circle(3.5,.1),locate(3,-.2,"M(?,?)"),
circle(5,8,.1),locate(4.5,9.2,"C(q,r)"),
circle(2.5,4,.1),locate(.8,4.5,"N(?,?)"),
circle(0,0,.1), locate(-2.5,1,"A(0,0)"),
green(triangle(0,0,5,8,7,0),triangle(2.5,4,3.5,0,2.5,4))
 )}}}


Use the midpoint formula to find the coordinates of M:

M({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

Substitute 

(x<sub>1</sub>, y<sub>1</sub>) = (0,0) and (x<sub>2</sub>, y<sub>2</sub>) = (p,0)  


M({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

M({{{(0+p)/2}}}, {{{(0+0)/2}}})

M({{{p/2}}}, 0)

{{{drawing(400,400,-3,10,-3,10,

red(line(-11,0,11,0),line(0,-11,0,11)), locate(10,0,x),locate(0,10,y),
circle(7,0,.1),locate(6.5,-.2,"B(p,0)"),
circle(3.5,.1),locate(3,-.2,M(p/2,0)),
circle(5,8,.1),locate(4.5,9.2,"C(q,r)"),
circle(2.5,4,.1),locate(.8,4.5,"N(?,?)"),
circle(0,0,.1), locate(-2.5,1,"A(0,0)"),
green(triangle(0,0,5,8,7,0),triangle(2.5,4,3.5,0,2.5,4))
 )}}}

Use the midpoint formula to find the coordinates of N:

N({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

Substitute 

(x<sub>1</sub>, y<sub>1</sub>) = (0,0) and (x<sub>2</sub>, y<sub>2</sub>) = (q,r)  

N({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

N({{{(0+q)/2}}}, {{{(0+r)/2}}})

N({{{q/2}}}, {{{r/2}}})

{{{drawing(400,400,-3,10,-3,10,

red(line(-11,0,11,0),line(0,-11,0,11)), locate(10,0,x),locate(0,10,y),
circle(7,0,.1),locate(6.5,-.2,"B(p,0)"),
circle(3.5,.1),locate(3,-.2,M(p/2,0)),
circle(5,8,.1),locate(4.5,9.2,"C(q,r)"),
circle(2.5,4,.1),locate(.6,4.7,N(q/2,r/2)),
circle(0,0,.1), locate(-2.5,1,"A(0,0)"),
green(triangle(0,0,5,8,7,0),triangle(2.5,4,3.5,0,2.5,4))
 )}}}



b) show that MN = {{{expr(1/2)}}}BC 

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

(x<sub>1</sub>, y<sub>1</sub>) = (p,0) and (x<sub>2</sub>, y<sub>2</sub>) = (q,r)

{{{d}}}{{{""=""}}}{{{sqrt((x[2]-x[1]))^2+(y[2]-y[1])^2)}}}

{{{BC}}}{{{""=""}}}{{{sqrt((q-p)^2+(r-0)^2)}}}

{{{BC}}}{{{""=""}}}{{{sqrt((q-p)^2+r^2)}}}

Now let's find MN:

(x<sub>1</sub>, y<sub>1</sub>) = ({{{p/2}}},0) and (x<sub>2</sub>, y<sub>2</sub>) = ({{{q/2}}},{{{r/2}}})

{{{d}}}{{{""=""}}}{{{sqrt((x[2]-x[1]))^2+(y[2]-y[1])^2)}}}

{{{MN}}}{{{""=""}}}{{{sqrt((q/2-p/2)^2+(r/2-0)^2)}}}

{{{MN}}}{{{""=""}}}{{{sqrt((q/2-p/2)^2+r^2)}}}

Write {{{q/2}}} as {{{expr(1/2)q}}}, {{{p/2}}} as {{{expr(1/2)p}}}, and {{{r/2}}} as {{{expr(1/2)r}}},

{{{MN}}}{{{""=""}}}{{{sqrt((expr(1/2)q-expr(1/2)p)^2+(expr(1/2)r)^2)}}}

Factor out {{{1/2}}} inside the first parentheses:

{{{MN}}}{{{""=""}}}{{{sqrt((expr(1/2)(q-p))^2+(expr(1/2)r)^2)}}}

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

{{{MN}}}{{{""=""}}}{{{sqrt((expr(1/2))^2(q-p)^2+(expr(1/2))^2r^2)}}}

Write the {{{(1/2)^2}}} as {{{1/4}}}

{{{MN}}}{{{""=""}}}{{{sqrt(expr(1/4)(q-p)^2+expr(1/4)r^2)}}}

Factor {{{1/4}}} out under the radical:

{{{MN}}}{{{""=""}}}{{{sqrt(expr(1/4)((q-p)^2+r^2))}}}

Take indivisual square roots:

{{{MN}}}{{{""=""}}}{{{sqrt(expr(1/4))sqrt(((q-p)^2+r^2))}}}

Write {{{sqrt(1/4)}}} as {{{1/2}}}

{{{MN}}}{{{""=""}}}{{{expr(1/2)sqrt(((q-p)^2+r^2))}}}

Now we notice that the right side of that is {{{1/2}}} times the right side of

{{{BC}}}{{{""=""}}}{{{sqrt((q-p)^2+r^2)}}}

So we have proved that MN = {{{expr(1/2)}}}BC

Edwin</pre>