Question 721940
<pre>
You don't need the Pythagorean theorem.

Given &#2017;ABC with vertices A(-1,2), B(5,-4), C(9,6), prove that CD
is both an altitude (perpendicular to base AB) and a median
(bisects the base) where D is the point D(-4,3).

Draw &#2017;ABC and AD

{{{drawing(400,2800/9,-8,10,-4,10,

graph(400,2800/9,-8,10,-4,10), triangle(-2,-1,-6,7,6,8),
green(line(6,8,-4,3)), locate(6,8,"C(6,8)"),locate(-6.5,3,"D(-4,3)"), 
locate(-5,-1,"B(-2,-1)"), locate(-8,7.9,"A(-6,7)") )}}}

We need to show that 

(1) CD &#8869; AB  and  (2) D is the midpoint of AB.

For (1), we find the slopes of CD and AB and show that they are negative
reciprocals, that their product is -1.

Slope formula:

m = {{{(y[2]-y[1])/(x[2]-x[1])}}}

To find the slope of CD:
where (x<sub>1</sub>,y<sub>1</sub>) = C(6,8)
and where (x<sub>2</sub>,y<sub>2</sub>) = D(-4,3)

m = {{{((3)-(8))/((-4)-(6))}}} = {{{-5/-10}}} = {{{1/2}}}

To find the slope of AB:
where (x<sub>1</sub>,y<sub>1</sub>) = A(-6,7)
and where (x<sub>2</sub>,y<sub>2</sub>) = B(-2,-1)

m = {{{((-1)-(7))/((-2)-(-6))}}} = {{{-8/(-2+6)}}} = {{{(-8)/4}}} = -2

Since {{{1/2}}} and {{{-2}}} are negative reciprocals,
their product is -1, we have proved that CD &#8869; AB,
which proves that CD is an altitude to base AB.

For (2), we use the midpoint formula to show that D(-4,3) is the
midpoint of AB.

Midpoint formula:

Midpoint = {{{(matrix(1,3,(x[1]+x[2])/2, ",",(y[1]+y[2])/2))}}}

where (x<sub>1</sub>,y<sub>1</sub>) = A(-6,7)
and where (x<sub>2</sub>,y<sub>2</sub>) = B(-2,-1)

Midpoint = {{{(matrix(1,3,((-6)+(-2))/2, ",",((7)+(-1))/2))}}} = {{{(matrix(1,3,(-8)/2, ",",(6)/2))}}} = (-4,3) which is point D.

Thus CD is also a median to the base AB of &#2017;ABC.

Edwin</pre>