Question

In the coordinate plan the vertices of Triangle ABC are A(-4, -3), B(2, 5), and C(8, -3), and the coordinates of Triangle XYZ(-10,-7), Y(-7,-3) and Z(-4,-7)

Prove that Triangle ABC ~ Triangle XYZ

There is a graph including on the question as well.

Answer 1

you can find the sides of the tringles... and you will see that AB=10. BC=10. AC=12.
XY=5, YZ=5, ZX=6. so the sides are in the ratio 2:1
so ABC~XYZ

Answer 2

Oh I see, but how would I explain that on paper. And to find the sides I use distance formula correct

Answer 3

to find the side use the formula \[\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})^{2}}\]

Answer 4

Alright Thank You

Answer 5

But how do I explain the ratio part, I don't understand the ratio part at all.

Answer 6

see AB/XY =BC/YZ =CA/ZX=2 i.e. the sides are in the same ratio...

Answer 7

I guess I kinda do, but how do I say ABC ~ XYZ I use the Ratio 2:1 to Say that?