Question

Find the perimeter of a triangle with vertices A(3, 1), B(2, -1), and C(-3,2). Leave your answer in decimal form, rounded to the nearest hundredth.

Answer 1

Answer: P≈14,15.

Explanation:

Line length formula:

`\overline {L}=√((x_2-x_1)^2+(y_2-y_1)^2)`

`P=\overline {AB}+\overline {BC}+\overline {AC}\\A(3;1)\ \ \ \ \ B(2;-1)`

`x_1=3\ \ \ \ x_2=2\ \ \ \ y_1=1\ \ \ \ y_2=-1\\\\\overline {AB}=√((2-3)^2+(-1-1)^2)\\\\ \overline {AB}=√((-1)^2+(-2)^2)\\\\ \overline {AB}=√((1+4)\\\\ \overline {AB}=√(5).\\\\`

`B(2;-1)\ \ \ \ C(-3;2)\\\\x_1=2\ \ \ \ \ x_2=-3\ \ \ \ y_1=-1\ \ \ \ y_2=2\\\\\overline {BC}=√((-3-2)^2+(2-(-1))^2)\\\\\overline {BC}=√((-5)^2+(2+1)^2)\\\\\overline {BC}=√(25+3^2)\\\\\overline {BC}=√(25+9)\\\\\overline {BC}=√(34).\\\\`

`A(3;1)\ \ \ \ C(-3;2)\\\\x_1=3\ \ \ \ \ x_2=-3\ \ \ \ y_1=1\ \ \ \ \ \ y_2=2\\\\\overline {AC}=√((-3-3)^2+(2-1)^2) \\\\\overline {AC}=√((-6)^2+1^2) \\\\\overline {AC}=√(36+1) \\\\\overline {AC}=√(37). \\\\`

`P=√(5)+√(34) +√(37) \\\\P\approx14,15.`