Question

: Coordinate Proof.Consider AABC with vertices A(2, 3) and B(6, 6) and C(8,-5).a) Draw AABC on graph paper. Is this a right triangle? JUSTIFY your answer.b) Reflect A ABC across AC to create AA'B'C' Find the location of B'. What kind oftriangle is ABB'C':? JUSTIFY your answer.

Answer 1

Yes, this is a right triangle. B'(-2,0)

1) Given those coordinates let's plot that triangle:

Let's find the legs:

`\begin{gathered} d_(AC)=\sqrt[]{(8-2)^2+(-5-3)^2}=10 \\ d_(BC)=\sqrt[]{(8-6)^2+(-5-6)^2}=5\sqrt[]{5} \\ d_(AB)=\sqrt[]{(6-2)^2+(6-3)^2}=5 \end{gathered}`

Let's test whether this is or not a Right Triangle, by using the Pythagorean Theorem:

`\begin{gathered} (5\sqrt[]{5})^2=10^2+5^2 \\ 125=100+25 \\ 125=125\text{ TRUE} \end{gathered}`

Hence, this is a right triangle.

b) Reflecting that triangle across leg AC, we have:

Counting to the left of Vertix A, 2 units to the left we have the Vertix B' and since its a reflection across AC, A = A' and C=C'

Therefore

B'(-2,0)