# Find the area of ∆ABC whose vertices are:

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)

Solution:

(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of ∆ABC. Then,
(x1 = 1, y1 = 2), (x2 = −2, y2 = 3) and (x3 = −3, y3 = -4)

Area of triangle $A B C$

$=\frac{1}{2}\left\{x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right\}$

$=\frac{1}{2}\{1(3-(-4))+(-2)(-4-2)+(-3)(2-3)\}$

$=\frac{1}{2}\{1(3+4)-2(-6)-3(-1)\}$

$=\frac{1}{2}\{7+12+3\}$

$=\frac{1}{2}(22)$

$=11$ sq. units

(ii) A(−5, 7), B(−4, −5) and C(4, 5) are the vertices of ∆ABC. Then,
(x1 = −5, y1 = 7), (x2 = −4, y2 = −5) and (x3 = 4, y3 = 5)

Area of triangle $A B C$

$=\frac{1}{2}\left\{x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right\}$

$=\frac{1}{2}\{(-5)(-5-5)+(-4)(5-7)+4(7-(-5))\}$

$=\frac{1}{2}\{(-5)(-10)-4(-2)+4(12)\}$

$=\frac{1}{2}\{50+8+48\}$

$=\frac{1}{2}(106)$

$=53$ sq. units

(iii) A(3, 8), B(−4, 2) and C(5, −1) are the vertices of ∆ABC. Then,
(x1 = 3, y1 = 8), (x2 = −4, y2 = 2) and (x3 = 5, y3 = −1)

Area of triangle $A B C$

$=\frac{1}{2}\left\{x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right\}$

$=\frac{1}{2}\{3(2-(-1))+(-4)(-1-8)+5(8-2)\}$

$=\frac{1}{2}\{3(2+1)-4(-9)+5(6)\}$

$=\frac{1}{2}\{9+36+30\}$

$=\frac{1}{2}(75)$

$=37.5$ sq. units

(iv) A(10, −6), B(2, 5) and C(−1, −3) are the vertices of ∆ABC. Then,
(x1 = 10, y1 = −6), (x2 = 2, y2 = 5) and (x3 = −1, y3 = 3)

Area of triangle $A B C$

$=\frac{1}{2}\left\{x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right\}$

$=\frac{1}{2}\{10(5-3)+2(3-(-6))+(-1)(-6-5)\}$

$=\frac{1}{2}\{10(2)+2(9)-1(-11)\}$

$=\frac{1}{2}\{20+18+11\}$

$=\frac{1}{2}(49)$

$=24.5$ sq. units