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If the vertices of a triangle are $(-2,6),(3,-6)$ and $(1,5)$, then the area of the triangle is
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Verified Answer
The correct answer is:
15.5 sq. units
Given vertices of triangle are $(-2,6),(3,-6)$ and $(1,5)$.
$$
\begin{aligned}
& \text { Area of triangle }=\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right| \\
&= \frac{1}{2}\left|\begin{array}{ccc}
-2 & 6 & 1 \\
3 & -6 & 1 \\
1 & 5 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
-5 & 12 & 0 \\
2 & -11 & 0 \\
1 & 5 & 1
\end{array}\right| \\
& {\left[R_1 \leftrightarrow R_1-R_2, R_2 \leftrightarrow R_2-R_3\right] } \\
&= \frac{1}{2}[1(55-24)]=\frac{31}{2}=15.5 \text { sq. units }
\end{aligned}
$$
$$
\begin{aligned}
& \text { Area of triangle }=\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right| \\
&= \frac{1}{2}\left|\begin{array}{ccc}
-2 & 6 & 1 \\
3 & -6 & 1 \\
1 & 5 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
-5 & 12 & 0 \\
2 & -11 & 0 \\
1 & 5 & 1
\end{array}\right| \\
& {\left[R_1 \leftrightarrow R_1-R_2, R_2 \leftrightarrow R_2-R_3\right] } \\
&= \frac{1}{2}[1(55-24)]=\frac{31}{2}=15.5 \text { sq. units }
\end{aligned}
$$
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