Question

Say true or false.

The area of the triangle formed by the points $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is the numerical value of the expression

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

• A. True
• B. False

Answer 1

The correct option is A True

Let $ABC$ be a triangle whose vertices are $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$.
Draw the lines $AD,BE,CF$ perpendicular to $x$-axis.
From the figure, $ED=$ $x_1-x_2$; $DF=$ $x_3-x_1$; $EF=$ $x_3-x_2$
Area of the triangle $ABC=$ Area of the trapezium $ABED+$ Area of the trapezium $ADFC-$ Area of the trapezium $BEFC$
$=$ $\frac{1}{2}\left[\right(BE+AD\left)ED\right]$ $+$ $\frac{1}{2}\left[\right(AD+CF\left)DF\right]$ $-$ $\frac{1}{2}\left[\right(BE+CF\left)EF\right]$
$=$ $\frac{1}{2}\left[\right(y_2+y_1\left)\right(x_1-x_2\left)\right]$ $+$ $\frac{1}{2}\left[\right(y_1+y_3\left)\right(x_3-x_1\left)\right]$ $-$ $\frac{1}{2}\left[\right(y_2+y_3\left)\right(x_3-x_2\left)\right]$
$=$ $\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$