Question Statement

By finding the area of the triangle, show that the points $A(6,0)$, $B(-3,6)$ and $C(3,2)$ are collinear.

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Background and Explanation

Three points are collinear if they lie on the same straight line. When three points form a triangle with zero area, they must be collinear — they cannot enclose any region.

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is given by the determinant formula:

$\text{Area}=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Collinearity condition: Three points are collinear if and only if this area equals zero.

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Solution

To prove that $A(6,0)$, $B(-3,6)$, and $C(3,2)$ are collinear, we calculate the area of $△ABC$.

Step 1: Set up the determinant

Substitute the coordinates into the area formula:

$\text{Area}=\frac{1}{2}\left|\begin{array}{rrr}6& 0& 1\\ -3& 6& 1\\ 3& 2& 1\end{array}\right|$

Step 2: Expand along the first row

$=\frac{1}{2}\left(6\left|\begin{array}{cc}6& 1\\ 2& 1\end{array}\right|-0\left|\begin{array}{rr}-3& 1\\ 3& 1\end{array}\right|+1\left|\begin{array}{rr}-3& 6\\ 3& 2\end{array}\right|\right)$

Step 3: Evaluate the $2\times 2$ determinants

$\left|\begin{array}{cc}6& 1\\ 2& 1\end{array}\right|=(6)(1)-(1)(2)=6-2=4$

$\left|\begin{array}{rr}-3& 6\\ 3& 2\end{array}\right|=(-3)(2)-(6)(3)=-6-18=-24$

Step 4: Substitute and simplify

$\text{Area}=\frac{1}{2}(6(4)-0+1(-24))$

$=\frac{1}{2}(24-24)$

$=\frac{1}{2}(0)=0\text{ sq. units}$

Conclusion: Since Area $=0$, the points $A$, $B$, and $C$ do not form a triangle. They lie on the same straight line and are therefore collinear. $■$

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Key Formulas Used

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Summary of Steps

1. Write the $3\times 3$ determinant with the three points' coordinates and a column of 1s
2. Expand along the first row using cofactor expansion
3. Evaluate each $2\times 2$ minor
4. Substitute: $\frac{1}{2}(24-24)=0$
5. Conclude collinearity since Area $=0$