## Area of Triangle Drawn Between Three Points

Given a triangle drawn between three points in the
$xy$
plane we can find the area of the triangle by
1) Finding the lengths of the sides using the distance formula
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
.
2) Finding an angle in the triangle using the sine rule.
3) Finding the area of the triangle using the formula
$Area= \frac{1}{2} ab sin C$

For the triangle below

The sides are
$PQ= \sqrt{(4-1)^2+(9-2)^2}=\sqrt{58}$

$PR= \sqrt{(5-1)^2+(4-2)^2}=\sqrt{20}$

$QR= \sqrt{(4-5)^2+)9-4)^2}=\sqrt{26}$

The angle
$P$
is the solution to
$QR^2=QP^2+PR^2-2 (QP)(PR)cos P \rightarrow cos P=\frac{QP^2+PR^2-QR^2}{2 (QP)(PR)}= \frac{58+20=26}{2 \times \sqrt{58} \times \sqrt{20}}=0.7634$

Then
$P=cos^{-1}(0.7634)=40.24^o$
.
Finally
$Area= \frac{1}{2} (PQ)(PR)sin P= \frac{1}{2} \sqrt{58} \sqrt{20} sin 40.24=11.00 \; units^2$
.