Question

Find the measure of the sides of angle ABC with vertices A(1,5), B(3,-2), and C(-3,0). Classify the triangle

Answer 1

Answer:

$AB=\sqrt{53},\ BC=\sqrt{40},\ CA=\sqrt{41}\\It\ is\ an\ acute\ triangle$

Step-by-step explanation:

The distance between two points $(x_1,y_1),\ (x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

Here $A(1,5),\ B(3,-2),\ C(-3,0)$ are the vertices of a triangle.

$AB=\sqrt{(-2-5)^2+(3-1)^2}=\sqrt{(-7)^2+(2)^2}=\sqrt{49+4}=\sqrt{53}=7.28\ unit\\\\BC=\sqrt{(0+2)^2+(-3-3)^2}=\sqrt{(2)^2+(-6)^2}=\sqrt{4+36}=\sqrt{40}=6.32\ unit\\\\CA=\sqrt{(5-0)^2+(1+3)^2}=\sqrt{(5)^2+(4)^2}=\sqrt{25+16}=\sqrt{41}=6.40\ unit$

Longest side$=AB$

$AB^2=53\\BC^2+CA^2=40+41=81\\\Rightarrow AB^2$

Hence this is an acute triangle.