Question

find the area of triangle whose vertices are: (-5,-1), (3,-5), (5,2) ​

Answer 1

we know that :-

$\sf \: area \: of \: triangle \: with \: vertices \: (x_1,x_2) , (y_1,y_2) and (x_3,y_3) : \\ \\ \rightarrow\frac{1}{2} \left|\begin{array}{ccc}\sf1&\sf x_1&\sf y_1\\\sf1&\sf x_2&\sf y_2\\ \sf1 & \sf x_3&\sf y_3\end{array}\right|$

Now , we have to find area of triangle whose vertices are: (-5,-1), (3,-5), (5,2) :-

$\rightarrow\frac{1}{2} \left|\begin{array}{ccc}\sf1&\sf - 5&\sf - 1\\\sf1&\sf 3&\sf - 5\\ \sf1 & \sf 5&\sf 2\end{array}\right| \\ \\ \sf now \: expanding \: by \: c_1 : \rightarrow \\ \\ \rightarrow \sf\frac{1}{2} \bigg((6 + 25) - ( - 10 + 5) + (25 + 3)\bigg)\\ \\ \rightarrow \sf\frac{1}{2} \bigg(31 + 5 + 28\bigg) \\ \\ \rightarrow \sf\frac{1}{2} \times 64 = 32 \: square \: units$

Area of triangle whose vertices are: (-5,-1), (3,-5), (5,2) = 32 square unit

Answer 2

Now , we have to find area of triangle whose vertices are: (-5,-1), (3,-5), (5,2) :-

$\begin{gathered}\rightarrow\frac{1}{2} \left|\begin{array}{ccc}\sf1&\sf - 5&\sf - 1\\\sf1&\sf 3&\sf - 5\\ \sf1 & \sf 5&\sf 2\end{array}\right| \\ \\ \sf now \: expanding \: by \: c_1 : \rightarrow \\ \\ \rightarrow \sf\frac{1}{2} \bigg((6 + 25) - ( - 10 + 5) + (25 + 3)\bigg)\\ \\ \rightarrow \sf\frac{1}{2} \bigg(31 + 5 + 28\bigg) \\ \\ \rightarrow \sf\frac{1}{2} \times 64 = 32 \: square \: units\end{gathered}$

Area of triangle whose vertices are: (-5,-1), (3,-5), (5,2) = 32 square unit