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.
-$\mathrm { P } ( - 3 , - 3 ) , \mathrm { Q } ( 0,5 ) , \mathrm { R } ( 4,2 )$
A) 38
B) 41
C) 0
D) $\frac { 41 } { 2 }$

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.
-$\mathrm { P } ( - 3 , - 3 ) , \mathrm { Q } ( 0,5 ) , \mathrm { R } ( 4,2 )$
A) 38
B) 41
C) 0
D) $\frac { 41 } { 2 }$

Quizplus · Accepted Answer

The area of the triangle is calculated using the determinant formula given. Substituting the coordinates of P, Q, and R into the formula, we get: $$D = \frac { 1 } { 2 } \left| \begin{array} { c c c } -3 & -3 & 1 \ 0 & 5 & 1 \ 4 & 2 & 1 \end{array} \right|$$ Calculating the determinant and taking its absolute value gives an area of $\frac { 41 } { 2 }$.

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