Question 144172
Given: triangle ABC A(-1,2) B(7,0) C(1,-6) and a point D(4,-3) on segment BC
Prove: segment AD is the perpendicular bisector of segment BC
<pre><font size = 4 color = "indigo"><b>
{{{drawing(400,400,-10,10,-8,8,
graph(400,400,-10,10,-8,8), triangle(-1,2,7,0,1,-6),locate(-4,2.5,"A(-1,2)"), locate(7,1,"B(7,0)"),
locate(.5,-6,"C(1,-6)"), locate(4.4,-3+.3,"D(4,-3)"),
triangle(-1,2,7,0,4,-3) )}}} 

First we use the midpoint formula to show that D 
is the midpoint of BC.  That will show that AD is 
a bisector of BC.  Then we will use the slope 
formula to show that AD is perpendicular to BC.

The midpoint of the segment joining ({{{x1}}},{{{y1}}}) and ({{{x2}}},{{{y2}}}) is
given by the formula


{{{MIDPOINT}}} = ({{{(x1+x2)/2}}},{{{(y1-y2)/2}}})

We use B({{{7}}},{{{0}}}) as ({{{x1}}},{{{y1}}})
and C({{{1}}},{{{-6}}}) as ({{{x2}}},{{{y2}}}) 

{{{MIDPOINT}}} = ({{{(7+1)/2}}},{{{(0+(-6))/2}}})

{{{MIDPOINT}}} = ({{{8/2}}},{{{(0-6)/2}}})

{{{MIDPOINT}}} = ({{{4}}},{{{-6/2}}})  

{{{MIDPOINT}}} = ({{{4}}},{{{-3}}})

Since D has those coordinates, AD bisects BC.

Now we need to show AD and BC are perpendicular.

The slope of the segment joining ({{{x1}}},{{{y1}}}) and ({{{x2}}},{{{y2}}}) is
given by the formula

{{{m = (y2-y1)/(x2-x1)}}}

So we now find the slope of BC, again using
B({{{7}}},{{{0}}}) as ({{{x1}}},{{{y1}}})
and C({{{1}}},{{{-6}}}) as ({{{x2}}},{{{y2}}})

{{{m = (y2-y1)/(x2-x1)}}}

{{{m = ((-6)-(0))/((1)-(7))}}}

{{{ m = (-6)/(-6)}}}

{{{m = 1}}}

So the slope of BC is {{{1}}}

Now So we now find the slope of AD, using
A({{{-1}}},{{{2}}}) as ({{{x1}}},{{{y1}}})
and D({{{4}}},{{{-3}}}) as ({{{x2}}},{{{y2}}})

{{{m = ((-3)-(2))/((4)-(-1))}}}

{{{m = (-3-2)/((4)+(1))}}}

{{{ m = (-5)/5}}}

{{{m = -1}}}

So the slope of AD is {{{-1}}}

Since {{{1}}} and {{{-1}}} are reciprocals with 
opposite signs, this proves AD is perpendicular to
BC.

Therefore AD is the perpendicular bisector of BC.

Edwin</pre>