The area of a triangle with vertices $\left( x _ { 1 } , y _ { 1 } \right),$ $\left( x _ { 2 } , y _ { 2 } \right),$ and $\left( x _ { 3 } , y _ { 3 } \right)$ can be given as the absolute value of the determinant $\frac { 1 } { 2 } \left| \begin{array} { l l l }
x _ { 1 } & y _ { 1 } & 1 \
x _ { 2 } & y _ { 2 } & 1 \
x _ { 3 } & y _ { 3 } & 1
\end{array} \right|$ Use this formula to find the area of each triangle whose coordinates are $( - 6,2 ),$ $( 0,3 ),$ and $( - 5,6 ).$
A)11.0
B)1.0
C)15.5
D)11.5
E)6.0

Question

The area of a triangle with vertices $\left( x _ { 1 } , y _ { 1 } \right),$ $\left( x _ { 2 } , y _ { 2 } \right),$ and $\left( x _ { 3 } , y _ { 3 } \right)$ can be given as the absolute value of the determinant $\frac { 1 } { 2 } \left| \begin{array} { l l l } 
x _ { 1 } & y _ { 1 } & 1 \
x _ { 2 } & y _ { 2 } & 1 \
x _ { 3 } & y _ { 3 } & 1
\end{array} \right|$ Use this formula to find the area of each triangle whose coordinates are $( - 6,2 ),$ $( 0,3 ),$ and $( - 5,6 ).$
A)11.0
B)1.0
C)15.5
D)11.5
E)6.0

Quizplus · Accepted Answer

The area of the triangle is calculated using the determinant formula given. Substituting the coordinates into the determinant formula and calculating gives an area of 11.5.

The Area of a Triangle with Vertices (x1,y1),\left( x _ { 1 } , y _ { 1 } \right),(x1​,y1​),

The Area of a Triangle with Vertices $\left( x _ { 1 } , y _ { 1 } \right),$