Q.Show that points A $(a, b+c)$, B $(b, c+a)$, C $(c, a+b)$ are collinear.

Concept: Collinearity Condition — three points are collinear if the area of the triangle formed by them is zero.

Step 1: Use the determinant formula for area:

Step 2: Substitute the given points:

Step 3: Simplify each term:

Step 4: Sum the terms:

Since the area is zero, the points are collinear.

The key idea here is the Collinearity Condition: three points are collinear if and only if the area of the triangle formed by them is zero. For points given as coordinates, the area can be computed using a determinant. If that determinant equals zero, the points lie on a single straight line.

Let's apply this to points A $(a, b+c)$, B $(b, c+a)$, C $(c, a+b)$.

1. Set up the area determinant The area of triangle ABC is given by:

For collinearity, we need this determinant to be zero. So we evaluate:

2. Simplify the determinant A common trick: subtract the first row from the second and third rows. This doesn't change the determinant's value but simplifies entries.

Simplify the second and third rows:

• Row 2: $b-a$, and $c+a-b-c = a-b$
• Row 3: $c-a$, and $a+b-b-c = a-c$ So:

3. Expand along the third column Since the third column has only one non-zero entry (the 1 in the first row), expansion is easy:

(The sign is positive because the cofactor for position (1,3) is $(-1)^{1+3}=1$.)

4. Evaluate the 2×2 determinant

Notice that $(a-b) = -(b-a)$ and $(c-a) = -(a-c)$. So:

5. Conclusion Since $D = 0$, the area of triangle ABC is zero. Therefore, points A, B, C are collinear.

Method: Proving Three Points Are Collinear (Algebraically, via Zero Area)

The general technique for a "show that these points are collinear" proof question — the 2D determinant form of the collinearity condition.

Steps

Step 1: Set up the area/collinearity determinant symbolically

Using the given (possibly symbolic) coordinates, write

(equivalently the $3\times3$ determinant with a column of $1$'s).

Step 2: Expand and simplify every term fully

Substitute the actual coordinate expressions and expand each product completely — don't stop at a partially simplified expression.

Step 3: Collect and cancel like terms

Group similar terms together; for a genuine collinearity identity, everything should cancel to leave exactly $0$.

Step 4: Conclude collinearity from $D=0$

Since the area is $\tfrac12|D|$, showing $D=0$ for all values of the parameters (not just one numeric check) proves the three points always lie on a straight line.

Common Mistakes

Mistake 1: Stopping at a simplified-but-nonzero-looking expression and wrongly concluding the points aren't collinear

Why it's wrong: an intermediate algebraic form (before full cancellation) can look nonzero even though it always reduces to $0$ — stopping early risks a false "not collinear" conclusion. Correct approach: keep expanding and collecting terms all the way through; a genuine collinearity identity must cancel completely to $0$.

Mistake 2: "Verifying" collinearity by substituting a few specific numeric values for the parameters instead of proving it algebraically

Why it's wrong: a numeric spot-check (e.g. trying $a=1,b=2,c=3$) only shows the claim holds for that one case — it is not a valid proof that the determinant is $0$ for every value of $a,b,c$. Correct approach: keep the coordinates symbolic throughout and show the determinant reduces identically to $0$ as an algebraic identity.