Area of Triangle Drawn Between Three Points

Given a triangle drawn between three points in the  
  plane we can find the area of the triangle by
1) Finding the lengths of the sides using the distance formula  
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  
  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

\[Area= \frac{1}{2} (PQ)(PR)sin P= \frac{1}{2} \sqrt{58} \sqrt{20} sin 40.24=11.00 \; units^2\]

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