Question

find the area of triangle whose vertices are : (2,3),(-1,0),(2,-4)

Answer 1

$area\:of\:traingle=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\\\=\frac{1}{2}[2(0-(-4))-1(-4-3)+2(3-0)]\\\\=\frac{1}{2}[8+7+6]\\\\=\frac{21}{2}\\\\10.5$

Answer 2

Here,
x1 = 2 , x2 = -1 , x3 = 2
y1 = 3 , y2 = 0 , y3 = -4

hence,
the area of angle :
1/2 [ x1 ( y2 -y3 ) ] + x2 ( y3 - y1 ) + x3 ( y1 - y2 ) ]

= 1/2 [ 2 ( 0 - ( - 4 ) ) ] - 1 ( - 4 - 3 ) + 2 ( 3 - 0 )

= 1/2 [ 2 × 4 - 1 × ( - 7 ) + 2 × 3 ]

= 1/2 [ 8 + 7 + 6 ]

= 1/2 × 21

= 21/2

= 10.5

so 10.5 is the answer

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