Question

The area of the triangle with vertices (-1, 2), (2, 4), (0, 0) is​

Answer 1

We have to find out the area of the triangle with vertices (-1, 2), (2, 4), (0, 0) .

Now , we know that :-

$\rightarrow \tt Area \: of \: triangle = \sf{\dfrac{1}{2} \times \left|\begin{array}{c c c} \tt x_{1} & \tt y_{1} & \tt 1 \\ \tt x_{2} & \tt y_{2} & \tt 1 \\ \tt x_{3} & \tt y_{3} & \tt 1\end{array}\right|}$

here value of :-

$\tt x_1 = - 1 \\ \\\tt x_2 = 2 \\ \\ \tt x_3 =0$

$\tt y_1 = 2 \\ \\\tt y_2 = 4 \\ \\ \tt y_3 =0$

Now put the above Values to find out the area of given triangle :-

$\rightarrow \tt Area \: of \: triangle = \sf{\dfrac{1}{2} \times \left|\begin{array}{c c c} \tt - {1} & \tt 2& \tt 1 \\ \tt{2} & \tt 4& \tt 1 \\ \tt 0 & \tt 0& \tt 1\end{array}\right|}$

Now , expanding determinant from Column 1:-

$\tt\rightarrow Area \: of \: triangle = \sf{\dfrac{1}{2}} \times \bigg|\bigg( - 1(4 - 0) - 2(2 - 0) + 0(2 - 4)\bigg)\bigg|$

[We have to put mod because area of triangle cannot be zero.]

$\tt\rightarrow Area \: of \: triangle = \sf{\dfrac{1}{2}} \times \bigg|\bigg( -4 - 4 + 0\bigg)\bigg|$

$\tt\rightarrow Area \: of \: triangle = \sf{\dfrac{1}{2}} \times 8 = 4 \: square \: unit$

$\tt\rightarrow Area \: of \: triangle = 4 \: square \: unit$

Answer :

area of the triangle with vertices (-1, 2), (2, 4), (0, 0) = 4 square unit

$\tt \large\underline{ \red{Additional-Information}}$

If three collinear points are given then :-

$\longrightarrow \sf{\dfrac{1}{2} \times \left|\begin{array}{c c c} \tt x_{1} & \tt y_{1} &\tt 1 \\ \tt x_{2} & \tt y_{2} & \tt 1 \\ \tt x_{3} & \tt y_{3} & \tt 1\end{array}\right|}=0$