Question 934933
<pre>
I'll try to do a little more for you here than the other tutor,
although what he says is very correct.

A(3,-1), B(5,2), C(-2,0), P(-3,4), Q(-5,-3), R(-6,2)

To do such problems you must draw graphs, like this:

{{{drawing(400,880/3,-8,7,-5,6,grid(1),
circle(3,-1,0.15),circle(3,-1,0.13),circle(3,-1,0.11),circle(3,-1,0.09),circle(3,-1,0.07),circle(3,-1,0.05),circle(3,-1,0.03),circle(3,-1,0.01),

locate(3+.15,-1-.1,A),
locate(5+.15,2+.65,B),
locate(-2+.15,0+.65,C),
locate(-3+.15,4+.65,P),
locate(-5+.15,-3+.65,Q),
locate(-6+.15,2+.65,R),
triangle(3,-1,5,2,-2,0),
triangle(-3,4,-5,-3,-6,2),



circle(5,2,0.15),circle(5,2,0.13),circle(5,2,0.11),circle(5,2,0.09),circle(5,2,0.07),circle(5,2,0.05),circle(5,2,0.03),circle(5,2,0.01),

circle(-2,0,0.15),circle(-2,0,0.13),circle(-2,0,0.11),circle(-2,0,0.09),circle(-2,0,0.07),circle(-2,0,0.05),circle(-2,0,0.03),circle(-2,0,0.01),

circle(-3,4,0.15),circle(-3,4,0.13),circle(-3,4,0.11),circle(-3,4,0.09),circle(-3,4,0.07),circle(-3,4,0.05),circle(-3,4,0.03),circle(-3,4,0.01),

circle(-5,-3,0.15),circle(-5,-3,0.13),circle(-5,-3,0.11),circle(-5,-3,0.09),circle(-5,-3,0.07),circle(-5,-3,0.05),circle(-5,-3,0.03),circle(-5,-3,0.01),

circle(-6,2,0.15),circle(-6,2,0.13),circle(-6,2,0.11),circle(-6,2,0.09),circle(-6,2,0.07),circle(-6,2,0.05),circle(-6,2,0.03),circle(-6,2,0.01)

)}}}

Only by drawing the graph, can you determine what you need
to do.

You must use the distance formula:

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

You must use it 6 times to show that

AB = RP
AC = RQ
BC = PQ

Here is how you show AB = RP

{{{AB}}}{{{""=""}}}{{{sqrt((5^""-3)^2+(2^""-(-1))^2)}}}{{{""=""}}}{{{sqrt(2^2+(2^""+1)^2)}}}{{{""=""}}}{{{sqrt(4+3^2)}}}{{{""=""}}}{{{sqrt(4+9)}}}{{{""=""}}}{{{sqrt(13)}}} 

{{{RP}}}{{{""=""}}}{{{sqrt((4^""-2)^2+(-3^""-(-6))^2)}}}{{{""=""}}}{{{sqrt((2)^2+(-3+6)^2)}}}{{{""=""}}}{{{sqrt(4+3^2)}}}{{{""=""}}}{{{sqrt(4+9)}}}{{{""=""}}}{{{sqrt(13)}}} 

Now do the same to show AC = RQ and BC = PQ.

Then the two triangles are congruent by SSS.

Edwin</pre>