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Evaluate a Third-Order Determinant
-The Area of a Triangle with Vertices

question 52

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Evaluate a Third-Order Determinant
-The area of a triangle with vertices (x1,y1) ,(x2,y2) \left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right) , and (x3,y3) \left( x _ { 3 } , y _ { 3 } \right) is
 Area =±12x1y11x2y21x3y31\text { Area } = \pm \frac { 1 } { 2 } \left| \begin{array} { l l l } x _ { 1 } & y _ { 1 } & 1 \\x _ { 2 } & y _ { 2 } & 1 \\x _ { 3 } & y _ { 3 } & 1\end{array} \right| \text {, }
where the symbol ±\pm indicates that the appropriate sign should be chosen to yield a positive area. Use this formula to find the area of a triangle whose vertices are (6,10) ,(8,8) ( 6,10 ) , ( 8 , - 8 ) , and (9,4) ( - 9 , - 4 ) .


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