A triangle with vertices at (x1, y1), (x2, y2), and (x3, y3) has area equal to the absolute value of D, where $$D = \frac { 1 } { 2 } \left| \begin{array} { l l l }
x _ { 1 } & y _ { 1 } & 1 \
x _ { 2 } & y _ { 2 } & 1 \
x _ { 3 } & y _ { 3 } & 1
\end{array} \right|$$ Find the area of the triangle having vertices at P, Q, and R.
-$$\mathrm { P } ( 0 , - 4 ) , \mathrm { Q } ( 0,2 ) , \mathrm { R } ( 6 , - 2 )$$
A) 10
B) 0
C) 36
D) 18

Question

A triangle with vertices at (x1, y1), (x2, y2), and (x3, y3) has area equal to the absolute value of D, where $$D = \frac { 1 } { 2 } \left| \begin{array} { l l l } 
x _ { 1 } & y _ { 1 } & 1 \
x _ { 2 } & y _ { 2 } & 1 \
x _ { 3 } & y _ { 3 } & 1
\end{array} \right|$$ Find the area of the triangle having vertices at P, Q, and R.
-$$\mathrm { P } ( 0 , - 4 ) , \mathrm { Q } ( 0,2 ) , \mathrm { R } ( 6 , - 2 )$$
A) 10
B) 0
C) 36
D) 18

Quizplus · Accepted Answer

The area of the triangle is calculated using the determinant formula: $$D = \frac { 1 } { 2 } \left| \begin{array} { l l l } 0 & -4 & 1 \0 & 2 & 1 \6 & -2 & 1\end{array} \right|$$. Evaluating the determinant gives 36, and the area is $\frac{1}{2} \times 36 = 18$.

A Triangle with Vertices at (X1, Y1), (X2, Y2), and (X3