Find the area of triangle whose vertices are : A (5,8), B (5,0), … - Vidyadip

Question

Find the area of triangle whose vertices are : A (5,8), B (5,0), C (1,0)

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Answer

Here, $A\left(x_1, y_1\right) \equiv A(5,8), B\left(x_2, y_2\right)=B(5,0), C\left(x_3, y_3\right)=C(1,0)$

Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$

$\mathrm{A}(\triangle \mathrm{ABC})=\frac{1}{2}\left|\begin{array}{lll}5 & 8 & 1 \\ 5 & 0 & 1 \\ 1 & 0 & 1\end{array}\right|$

$\begin{aligned} & =\frac{1}{2}[5(0-0)-8(5-1)+1(0-0)] \\ & =\frac{1}{2}[0-8(4)+0] \\ & =\frac{1}{2}(-32) \\ & =-16\end{aligned}$

Since area cannot be negative, A(ΔABC) = 16 sq. units

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