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.
-$$\mathrm { P } ( - 3,2 ) , \mathrm { Q } ( 1 , - 3 ) , \mathrm { R } ( 5,1 )$$
A) 32
B) 18
C) 36
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.
-$$\mathrm { P } ( - 3,2 ) , \mathrm { Q } ( 1 , - 3 ) , \mathrm { R } ( 5,1 )$$
A) 32
B) 18
C) 36
D) 0

Quizplus · Accepted Answer

The area of the triangle can be found by substituting the coordinates of the vertices into the determinant formula and taking the absolute value of half of the result. Substituting the given points into the determinant formula gives:$$D = \frac { 1 } { 2 } \left| \begin{array} { c c c } -3 & 2 & 1 \1 & -3 & 1 \5 & 1 & 1\end{array} \right|$$Calculating the determinant:$$D = \frac { 1 } { 2 } [(-3)(-3)(1) + 2(1)(5) + 1(1)(1) - 1(-3)(5) - (-3)(1)(1) - 2(1)(1)]$$$$D = \frac { 1 } { 2 } [9 + 10 + 1 + 15 + 3 - 2]$$$$D = \frac { 1 } { 2 } [36]$$$$D = 18$$Therefore, the area of the triangle is 18.

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