Evaluate a Third-Order Determinant
-Determinants are used to show that three points lie on the same line (are collinear). If
$$\left| \begin{array} { l l l }
x _ { 1 } & y _ { 1 } & 1 \
x _ { 2 } & y _ { 2 } & 1 \
x _ { 3 } & y _ { 3 } & 1
\end{array} \right| = 0 ,$$
then the points $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)$, and $\left( x _ { 3 } , y _ { 3 } \right)$ are collinear. If the determinant does not equal 0 , then the points are not collinear. Are the points $( - 10,5 ) , ( 0 , - 6 )$, and $( - 20,18 )$ collinear?
A) No
B) Yes

Question

Evaluate a Third-Order Determinant
-Determinants are used to show that three points lie on the same line (are collinear). If
$$\left| \begin{array} { l l l } 
x _ { 1 } & y _ { 1 } & 1 \
x _ { 2 } & y _ { 2 } & 1 \
x _ { 3 } & y _ { 3 } & 1
\end{array} \right| = 0 ,$$
then the points $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)$, and $\left( x _ { 3 } , y _ { 3 } \right)$ are collinear. If the determinant does not equal 0 , then the points are not collinear. Are the points $( - 10,5 ) , ( 0 , - 6 )$, and $( - 20,18 )$ collinear?
A) No
B) Yes

Quizplus · Accepted Answer

The determinant for the points $(-10, 5), (0, -6),$ and $(-20, 18)$ is calculated as follows:$$\left| \begin{array}{ccc}-10 & 5 & 1 \0 & -6 & 1 \-20 & 18 & 1\end{array} \right|= (-10)(-6)(1) + (5)(1)(-20) + (1)(0)(18) - (1)(-6)(-20) - (1)(5)(0) - (-10)(1)(18)= 60 - 100 + 0 + 120 - 0 + 180= 260.$$Since the determinant is not equal to 0, the points are not collinear.

Evaluate a Third-Order Determinant -Determinants Are Used to Show That ∣x1y11x2y21x3y31∣=0,\left| \begin{array} { l l l } x _ { 1 } & y _ { 1 } & 1 \\x _ { 2 } & y _ { 2 } & 1 \\x _ { 3 } & y _ { 3 } & 1\end{array} \right| = 0 ,​x1​x2​x3​​y1​y2​y3​​111​​=0,

Evaluate a Third-Order Determinant
-Determinants Are Used to Show That $\left| \begin{array} { l l l } x _ { 1 } & y _ { 1 } & 1 \\x _ { 2 } & y _ { 2 } & 1 \\x _ { 3 } & y _ { 3 } & 1\end{array} \right| = 0 ,$