Find the area of a triangle with vertices $A (3,\ 0),\ B (7,\ 0)$ and $C (8,\ 4)$.
Given: Vertices of the triangle $A (3,\ 0),\ B (7,\ 0)$ and $C (8,\ 4)$.
To do: To find the area of the triangle.
Solution:
As given, $x_1=3,\ y_1=0,\ x_2=7,\ y_2=0,\ x_3=8,\ y_3=4$
$\therefore$ Area of the triangle$=\frac{1}{2}[x_1( y_2-y_3)+x_2( y_3-y_1)+x_3( y_1-y_2)]$
$=\frac{1}{2}[3( 0-4)+7( 4-0)+8( 0-7)]$
$=\frac{1}{2}[-12+28-56]$
$=\frac{1}{2}[-40]$
$=-20$
$\because$ Area can't be negative.
$\therefore$ Area of the triangle is $20$ sq. unit.
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