Question

triangle abc has vertices a(1, 1), b(-2, 4), and c(1, 7). which of these options correctly justifies the claim that the triangle is an isosceles right triangle?

a) the claim is correct because the slope of $overline{ab}$ is 1, which is the reciprocal of the slope of $overline{bc}$, and $ab = bc=sqrt{6}$
b) the claim is correct because the slope of $overline{ab}$ is 1, which is the reciprocal of the slope of $overline{bc}$, and $ab = bc = 3sqrt{2}$
c) the claim is correct because the slope of $overline{ab}$ is -1, which is the negative reciprocal of the slope of $overline{bc}$, and $ab = bc=sqrt{6}$
d) the claim is correct because the slope of $overline{ab}$ is -1, which is the negative reciprocal of the slope of $overline{bc}$, and $ab = bc = 3sqrt{2}$

Explanation:

Step1: Calculate the slope of $\overline{AB}$

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For points $A(1,1)$ and $B(-2,4)$, $m_{AB}=\frac{4 - 1}{-2 - 1}=\frac{3}{-3}=- 1$.

Step2: Calculate the slope of $\overline{BC}$

For points $B(-2,4)$ and $C(1,7)$, $m_{BC}=\frac{7 - 4}{1-(-2)}=\frac{3}{3}=1$. The slope of $\overline{AB}$ is the negative - reciprocal of the slope of $\overline{BC}$, so $\angle ABC = 90^{\circ}$.

Step3: Calculate the length of $AB$

The distance formula is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For $A(1,1)$ and $B(-2,4)$, $AB=\sqrt{(-2 - 1)^2+(4 - 1)^2}=\sqrt{(-3)^2+3^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}$.

Step4: Calculate the length of $BC$

For $B(-2,4)$ and $C(1,7)$, $BC=\sqrt{(1+2)^2+(7 - 4)^2}=\sqrt{3^2+3^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}$.

Answer:

D. The claim is correct because the slope of $\overline{AB}$ is $-1$, which is the negative reciprocal of the slope of $\overline{BC}$, and $AB = BC = 3\sqrt{2}$