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.
-$$\mathrm { P } ( - 3 , - 3 ) , \mathrm { Q } ( 0,0 ) , \mathrm { R } ( 5 , - 1 )$$
A) 12
B) 18
C) 0
D) 9

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.
-$$\mathrm { P } ( - 3 , - 3 ) , \mathrm { Q } ( 0,0 ) , \mathrm { R } ( 5 , - 1 )$$
A) 12
B) 18
C) 0
D) 9

Quizplus · Accepted Answer

The area of the triangle can be calculated using the determinant formula given. Substituting the coordinates of P, Q, and R into the formula, we get: $$D = \frac { 1 } { 2 } \left| \begin{array} { c c c } -3 & -3 & 1 \ 0 & 0 & 1 \ 5 & -1 & 1 \end{array} \right|$$ Calculating the determinant and taking the absolute value gives an area of 9.

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