Question

Triangle ABC has vertices at A(-2, 3), B(-3,-6), and C(2,-

Answer 1

Answer:

Yes, the right angle is angle B

Step-by-step explanation:

we have

$A(-2, 3), B(-3,-6),C(2,-1)$

Plot the vertices

see the attached figure

we know that

If triangle ABC is a right triangle

then

Applying the Pythagoras Theorem

$AB^{2} =AC^{2}+BC^{2}$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(-2, 3), B(-3,-6)$

substitute in the formula

$d=\sqrt{(-6-3)^{2}+(-3+2)^{2}}$

$d=\sqrt{(-9)^{2}+(-1)^{2}}$

$AB=\sqrt{82}\ units$

Find the distance BC

$B(-3,-6),C(2,-1)$

substitute in the formula

$d=\sqrt{(-1+6)^{2}+(2+3)^{2}}$

$d=\sqrt{(5)^{2}+(5)^{2}}$

$BC=\sqrt{50}\ units$

Find the distance AC

$A(-2, 3),C(2,-1)$

substitute in the formula

$d=\sqrt{(-1-3)^{2}+(2+2)^{2}}$

$d=\sqrt{(-4)^{2}+(4)^{2}}$

$AC=\sqrt{32}\ units$

Verify the Pythagoras theorem

$(\sqrt{82})^{2} =(\sqrt{32})^{2}+(\sqrt{50})^{2}$

$82=82$ ---> is true

therefore

Is a right triangle and the right angle is B