Solve the problem.
-The area of a triangle with vertices $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)$, and $\left( x _ { 3 } , y _ { 3 } \right)$ is
$$\text { Area } = \pm \frac { 1 } { 2 } \left| \begin{array} { l l l }
x _ { 1 } & y _ { 1 } & 1 \
x _ { 2 } & y _ { 2 } & 1 \
x _ { 3 } & y _ { 3 } & 1
\end{array} \right|$$
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 $( 2,10 ) , ( 6 , - 4 )$, and $( - 3 , - 9 )$.
A) 73
B) 146
C) 1
D) 2

Question

Solve the problem.
-The area of a triangle with vertices $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)$, and $\left( x _ { 3 } , y _ { 3 } \right)$ is
$$\text { Area } = \pm \frac { 1 } { 2 } \left| \begin{array} { l l l } 
x _ { 1 } & y _ { 1 } & 1 \
x _ { 2 } & y _ { 2 } & 1 \
x _ { 3 } & y _ { 3 } & 1
\end{array} \right|$$
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 $( 2,10 ) , ( 6 , - 4 )$, and $( - 3 , - 9 )$.
A) 73
B) 146
C) 1
D) 2

Quizplus · Accepted Answer

The area of the triangle is calculated using the determinant formula:  Area = $ \frac{1}{2} \left| \begin{array}{ccc} 2 & 10 & 1 \ 6 & -4 & 1 \ -3 & -9 & 1 \end{array} \right| $.  Calculating the determinant gives:  = $ \frac{1}{2} \left( 2(-4 \cdot 1 - 1 \cdot -9) - 10(6 \cdot 1 - 1 \cdot -3) + 1(6 \cdot -9 - (-4) \cdot -3) \right) $  = $ \frac{1}{2} \left( 2(-4 + 9) - 10(6 + 3) + 1(-54 - 12) \right) $  = $ \frac{1}{2} \left( 2 \cdot 5 - 10 \cdot 9 - 66 \right) $  = $ \frac{1}{2} \left( 10 - 90 - 66 \right) $  = $ \frac{1}{2} \times 146 $  = 73.

Solve the Problem (x1,y1),(x2,y2)\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)(x1​,y1​),(x2​,y2​)

Solve the Problem $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)$