Question

a=(-3,6); b=(5,0); c=(6,3). choose the correct graph that shows points a, b, c, and triangle abc. to show that the triangle is a right triangle, show that the sum of the squares of the lengths of two of the sides (the legs) equals the square of the length of the third side (the hypotenuse). find the length of each line - segment. d(a,b)= d(a,c)= d(b,c)= (simplify your answers. type exact answers, using radicals as needed.)

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate $d(A,B)$

For $A=(-3,6)$ and $B=(5,0)$, we have $x_1=-3,y_1 = 6,x_2=5,y_2 = 0$. Then $d(A,B)=\sqrt{(5+3)^2+(0 - 6)^2}=\sqrt{64 + 36}=\sqrt{100}=10$.

Step3: Calculate $d(A,C)$

For $A=(-3,6)$ and $C=(6,3)$, we have $x_1=-3,y_1 = 6,x_2=6,y_2 = 3$. Then $d(A,C)=\sqrt{(6 + 3)^2+(3 - 6)^2}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10}$.

Step4: Calculate $d(B,C)$

For $B=(5,0)$ and $C=(6,3)$, we have $x_1=5,y_1 = 0,x_2=6,y_2 = 3$. Then $d(B,C)=\sqrt{(6 - 5)^2+(3 - 0)^2}=\sqrt{1 + 9}=\sqrt{10}$.

Step5: Check Pythagorean - theorem

We check if $d(B,C)^2+d(A,C)^2=d(A,B)^2$ or $d(B,C)^2+d(A,B)^2=d(A,C)^2$ or $d(A,B)^2+d(A,C)^2=d(B,C)^2$.
$d(B,C)^2=(\sqrt{10})^2 = 10$, $d(A,C)^2=(3\sqrt{10})^2=90$, $d(A,B)^2 = 100$. And $d(B,C)^2+d(A,C)^2=10 + 90=100=d(A,B)^2$. So it is a right - triangle.

Answer:

$d(A,B)=10$
$d(A,C)=3\sqrt{10}$
$d(B,C)=\sqrt{10}$