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# Find area of the triangle with vertices at the point given in each of the following (2, 7), (1, 1), (10, 8)

$\begin{array}{1 1} 47 sq\; units \\47/2 sq\; units \\ 47 \;cubic\; units \\ 47/2 cubic\; units \end{array}$

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A)
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• The area of a triangle whose vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
• $\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
• $\bigtriangleup=\frac{1}{2}[x_1(y_2-y_3)-y_1(x_2-x_3)+1(x_2y_3-y_2x_3)]$
Let $(x_1,y_1)$ be=(2,7)

Let $(x_2,y_2)$ be=(1,1)

Let $(x_3,y_3)$ be=(10,8)

Now area of a triangle is given by

$\bigtriangleup=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}$

Substituting the respective values we get,

$\bigtriangleup=\frac{1}{2}\begin{vmatrix}2 & 7 & 1\\1 & 1 & 1\\10 & 8 & 1\end{vmatrix}$

Now expanding along the row $R_1$ we get

$\mid\bigtriangleup\mid=\frac{1}{2}\begin{bmatrix}2\begin{vmatrix}1 & 1\\8 & 1\end{vmatrix}-7\begin{vmatrix}1 & 1\\10 & 1\end{vmatrix}+1\begin{vmatrix}1 & 1\\10 & 8\end{vmatrix}\end{bmatrix}$

$=\frac{1}{2}[2(1-8)-7(1-10)+1(8-10)]$

$=\frac{1}{2}[2(-7)-7(-9)+1(-2)]$

$=\frac{1}{2}[-14+63-2]=\frac{1}{2}[47]=47/2.$

Hence the area is $\frac{47}{2}$sq.units.