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

area of triangle 1

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  
\[QR^2=QP^2+PR^2-2 (QP)(PR)cos P \rightarrow cos P=\frac{QP^2+PR^2-QR^2}{2 (QP)(PR)}= \frac{58+20=26}{2 \times \sqrt{58} \times \sqrt{20}}=0.7634\]

area of triangle 2

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

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