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.
-$P ( 0 , - 4 ) , Q ( 0,5 ) , R ( 5,1 )$
A) 3
B) 45
C) $\frac { 45 } { 2 }$
D) 0

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.
-$P ( 0 , - 4 ) , Q ( 0,5 ) , R ( 5,1 )$
A) 3
B) 45
C) $\frac { 45 } { 2 }$
D) 0

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 determinant formula, we get: $$D = \frac { 1 } { 2 } \left| \begin{array} { c c c } 0 & -4 & 1 \ 0 & 5 & 1 \ 5 & 1 & 1 \end{array} \right|$$ Calculating the determinant and taking the absolute value gives us $\frac { 45 } { 2 }$.

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