Question

Find the area of $△ABC$ whose vertices are :

(i) A(1, 2), B (-2, 3) and C(-3, -4)

(ii) A (-5,7),B(-4, -5) and C(4,5)

(iii) A(3, 8), B(-4, 2) and C(5, -1)

(iv) A (10, -6),B (2,5) and C(-1, 3)

Answer 1

(i) A (1, 2), B (-2, 3) and C (-3, -4) are the vertices of ΔABC. Then

$(x_1=1,y_1=2),(x_2=-2,y_2=3),(x_3=-3,y_3=-4)$

Area of triangle ABC

$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=\frac{1}{2}\left[1\right(3-(-4))+(-2\left)\right(-4-2)+(-3\left)\right(2-3\left)\right]=\frac{1}{2}\left[1\right(3+4)-2(-6)-3(-1\left)\right]=\frac{1}{2}[7+12+3]=\frac{1}{2}\left[22\right]=11sq.units$

(ii) A(-5, 7), B(-4, -5) and C(4, 5)

A (-5, 7), B (-4, -5) and C (4, 5) are the vertices of ΔABC. Then

$(x_1=-5,y_1=7),(x_2=-4,y_2=-5),(x_3=4,y_3=5)$

Area of triangle ABC

$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=\frac{1}{2}[-5(-5-5)+(-4\left)\right(5-7)+4(7-(-5)\left)\right]=\frac{1}{2}[-5(-10)-4(-2)+4(12\left)\right]=\frac{1}{2}[50+8+48]=\frac{1}{2}\left[106\right]=53sq.units$

(iii) A(3, 8), B(-4, 2) and C(5, -1)

A (3, 8), B (-4, 2) and C (5, -1) are the vertices of ΔABC. Then

$(x_1=3,y_1=8),(x_2=-4,y_2=2),(x_3=5,y_3=-1)$

Area of triangle ABC

$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=\frac{1}{2}\left[3\right(2–(-1))+(-4\left)\right(-1–8)+5(8–2\left)\right)]=\frac{1}{2}[3(2+1)–4(-9)+5\left(6\right)]=\frac{1}{2}[9+36+30]=\frac{1}{2}[75]=37.5sq.units$

(iv) A(10, -6), B(2, 5) and C(-1, 3)

A (10, -6), B (2, 5) and C (-1, -3) are the vertices of ΔABC. Then

$(x_1=10,y_1=-6),(x_2=2,y_2=5),(x_3=-1,y_3=3)$

Area of triangle ABC

$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=\frac{1}{2}\left[10\right(5-3)+(2\left)\right(3-(-6))+(-1\left)\right(-6–5\left)\right]=\frac{1}{2}\left[10\right(2)+2(9)–1(11\left)\right]=\frac{1}{2}[20+18+11]=\frac{1}{2}\left[49\right]=24.5sq.units$