QUESTION

# Find the area of the triangle, whose vertices are (2, 0), (1, 2) and (-1, 6). What do you observe?

Hint: The formula that we can use to find the area of a triangle with vertices A $\left( {{x}_{1}},{{y}_{1}} \right)$ , B $\left( {{x}_{2}},{{y}_{2}} \right)$ and C $\left( {{x}_{3}},{{y}_{3}} \right)$ is as follows
$Area=~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]$
If area of triangle with vertices A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$) is zero, then$Area=~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]$ = 0 and the points A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$) are collinear.

Complete Step-by-Step solution:
As mentioned in the question, we have to find the area of the triangle whose coordinates are given in the question.
Now, we can find the area of this triangle by using the formula that is mentioned in the hint as follows
As the coordinates are A $\left( 2,0 \right)$ , B $\left( 1,2 \right)$ and C $\left( -1,6 \right)$ which is given in the question, we can get the area of the triangle as
\begin{align} & =~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right] \\ & =\dfrac{1}{2}\left[ 2(2-6)+1(6-0)+-1(0-2) \right] \\ & =\dfrac{1}{2}\left[ -8+6+2 \right] \\ & =0 \\ \end{align}
As we get the area of this triangle as zero, we can use the information that is given in the hint which is
if the area of the triangle with vertices A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$) is zero, then the points are collinear.
Hence, the points are collinear.

Note: The students can make an error if they don’t know about the formula for calculation of the area of a triangle whose vertices are given that is mentioned in the hint as follows
The formula that we can use to find the area of a triangle with vertices A $\left( {{x}_{1}},{{y}_{1}} \right)$ , B $\left( {{x}_{2}},{{y}_{2}} \right)$ and C $\left( {{x}_{3}},{{y}_{3}} \right)$ is as follows
$Area=~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]$