Question

The area of triangle with vertices $A(x_1,y_1),B(x_2,y_2)andC(x_3,y_3)$ is:

• A. $\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
• B. $\frac{1}{2}\left[x_1\right(y_2-y_3)+x_3(y_3-y_1)+x_1(y_1-y_2\left)\right]$
• C. $\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_1(y_1-y_2\left)\right]$
• D. $Allofthese$

Answer 1

The correct option is A

$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

Let ABC be any triangle whose vertices are $A(x_1,y_1),B(x_2,y_2)andC(x_3,y_3)$. Draw AP, BQ and CR perpendiculars from A, B and C, respectively, to the x-axis. Clearly ABQP, APRC and BQRC are all trapezia.
Now from figure, it is clear that
Area of △ABC = Area of trapezium ABQP + Area of trapezium APRC - Area of trapezium BQRC.
Area of a trapezium $=\frac{1}{2}$ (Sum of parallel sides)(Distance between the parallel sides)
Therefore, Area of $\Delta ABC=\frac{1}{2}(BQ+AP)QP+\frac{1}{2}(AP+CR)PR-\frac{1}{2}(BQ+CR)QR$

$=\frac{1}{2}(y_2+y_1)(x_1-x_2)+\frac{1}{2}(y_3+y_1)(x_3-x_1)-\frac{1}{2}(y_2+y_3)(x_3-x_2)$

$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
Thus, the area of $\Delta ABC$ is given by the expression $\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$