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

`AB=√(53),\ BC=√(40),\ CA=√(41)\\It\ is\ an\ acute\ triangle`

Explanation:

The distance between two points
`(x_1,y_1),\ (x_2,y_2)` is given by
`√((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=√((-2-5)^2+(3-1)^2)=√((-7)^2+(2)^2)=√(49+4)=√(53)=7.28\ unit\\\\BC=√((0+2)^2+(-3-3)^2)=√((2)^2+(-6)^2)=√(4+36)=√(40)=6.32\ unit\\\\CA=√((5-0)^2+(1+3)^2)=√((5)^2+(4)^2)=√(25+16)=√(41)=6.40\ unit`

Longest side
`=AB`

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

Hence this is an acute triangle.