∆ABC–oarecare
A(–1;2)
B(4;–2)
C(0;3)
P∆=?
AB=√(xB–xA)²+(yB–yA)²
AB=√(4+1)²+(–2–2)²
AB=√5²+(–4)²
AB=√25–16
AB=√9
AB=3
AC=√(xC–xA)²+(yC–yA)²
AC=√(0+1)²+(3–2)²
AC=√1²+1²
AC=√2
BC=√(xC–xB)²+(yC–yB)²
BC=√(0–4)²+(3+2)²
BC=√–4²+5²
BC=√–16+25
BC=√9
BC=3
P∆ABC=AB+AC+BC=3+√2+3=(6+√2)
