Question 1082805
<b><pre>

First you draw the triangle on graph paper:

{{{drawing(400,320,-6.5,3.5,-3.5,4.5,grid(1),

locate(-4+.05,-2,"C(-4,-2)"),
locate(1+.05,-2,"B(1,-2)"),
locate(1+.05,3.4,"A(1,3)"),

line(-4,-2,1,-2), line(1,-2,1,3), line(-4,-2,1,3) )}}}

Then we check out each answer:
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A. d(B;C) = {{{sqrt(41)}}}
<pre>
We can just count the blocks from B to C since BC is
horizontal and find that the distance from B to C is 5,
so if it's 5, it's NOT {{{sqrt(41)}}}
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B. &#8736;C = 90°
<pre>
We can certainly see from the graph that &#8736;C in NOT 90°,
since it is an acute angle much smaller than 90°.  It's
&#8736;B that's 90°, not &#8736;C.
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C. The midpoint of the hypotenuse is (-3/2, 1/2)
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That's the only one left, so it must be the one that's true, 
since the others are false.  But let's see why anyway.

We use the midpoint formula:

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

The hypotenuse is AC, so we substitute:

x<sub>1</sub> = x-coordinate of A = 1
x<sub>2</sub> = x-coordinate of C = -4
y<sub>1</sub> = y-coordinate of A = 3
y<sub>2</sub> = y-coordinate of C = -2

Midpoint of AC = {{{(matrix(1,3,(1+(-4))/2, ",",(3+(-2))/2))}}}
Midpoint of AC = {{{(matrix(1,3,(-3)/2, ",",1/2))}}}
Midpoint of AC = {{{(matrix(1,3,-3/2, ",",1/2))}}}

So we see this is the only one that is true.

Edwin</pre><b>