Find the area of ΔABC whose vertices are:
A(10, –6), B(2, 5) and C(–1, 3)
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(10(5–3) + 2(3 + 6)–1(–6–5))
= 1/2(20 + 18 + 11)
= 1/2(49)
= 24.5 sq units
1
Find the area of ΔABC whose vertices are:
A(10, –6), B(2, 5) and C(–1, 3)
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(10(5–3) + 2(3 + 6)–1(–6–5))
= 1/2(20 + 18 + 11)
= 1/2(49)
= 24.5 sq units