Q.Show that the points $(a + 5,\, a - 4)$, $(a - 2,\, a + 3)$ and $(a,\, a)$ do not lie on a straight line for any value of $a$.

Concept — collinearity test. Three points are collinear exactly when the triangle they form has zero area. Compute that area; if it is a non-zero constant, the points can never be collinear.

Let $A(a+5,\,a-4)$, $B(a-2,\,a+3)$, $C(a,\,a)$.

The bracketed differences are $(a+3)-a=3$, $a-(a-4)=4$, $(a-4)-(a+3)=-7$, so …

The idea

Three points lie on a straight line precisely when the triangle they span is degenerate, i.e. has zero area. So instead of comparing slopes, compute the area and show it is a non-zero constant.

Set up

Let

The area of $\triangle ABC$ is

Work the steps

1. Compute the $y$-differences:

2. Substitute:

3. Expand inside the bars:

4. Hence …

Method: Area-of-Triangle Determinant Test for Collinearity

Use this method whenever you must decide whether three given points (possibly involving a variable like $a$) can ever lie on a single straight line.

Steps

Step 1: Set up the area determinant

For points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$, the area of the triangle they form is

which is (up to sign and the $\tfrac12$ factor) the expansion of $\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}$. Three points are collinear exactly when this area — equivalently, this determinant — equals $0$; a strictly positive area means the points genuinely form a triangle and can never be collinear.

Step 2: Substitute the given coordinates and simplify the $y$-differences first

Compute $y_2-y_3$, $y_3-y_1$, $y_1-y_2$ before substituting into the area formula — these differences are usually where variable terms cancel, so simplifying them first keeps the later algebra clean.

Step 3: Expand and collect terms in the variable …

Common Mistakes

Mistake 1: Sign/distribution errors when expanding the area expression

Why it's wrong: substituting $(a+5,a-4)$, $(a-2,a+3)$, $(a,a)$ into $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$ involves several negative signs; dropping or mis-signing one term (e.g. writing $a(a-4)$ instead of $a(-7)$) makes the $a$-terms fail to cancel, and a student ends up with an expression that looks like it does depend on $a$.

Mistake 2: Misreading what "the area is a nonzero constant" proves …