Triangle: perimeter, line through two vertices and area

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Given the triangle with vertices $A(3;1)$, $B(-2;8)$, $C(0;-5)$:

• (a) compute the perimeter;
• (b) find the equation of line $AB$;
• (c) find the area of the triangle.

Solution

(a) Side lengths: $\overline{AB}=\sqrt{(-2-3)^2+(8-1)^2}=\sqrt{25+49}=\sqrt{74}\approx 8{,}60,$ $\overline{BC}=\sqrt{(0+2)^2+(-5-8)^2}=\sqrt{4+169}=\sqrt{173}\approx 13{,}15,$ $\overline{AC}=\sqrt{(0-3)^2+(-5-1)^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt5\approx 6{,}71.$ Perimeter: $2p=\sqrt{74}+\sqrt{173}+3\sqrt5\approx 28{,}46$.

(b) Slope $m_{AB}=\dfrac{8-1}{-2-3}=-\dfrac75$; through $A(3;1)$: $y-1=-\tfrac75(x-3)\Rightarrow 7x+5y-26=0$.

(c) Area by the determinant (shoelace) formula: $A=\tfrac12\left|3\bigl(8-(-5)\bigr)+(-2)\bigl(-5-1\bigr)+0\bigl(1-8\bigr)\right|=\tfrac12|39+12|=\tfrac{51}{2}=25{,}5.$

$\boxed{2p\approx 28{,}46,\quad A=\tfrac{51}{2}=25{,}5}$