Q.If the points (𝒙𝟏, 𝒚𝟏), (𝒙𝟐, 𝒚𝟐) and (𝒙𝟏 + 𝒙𝟐, 𝒚𝟏 + 𝒚𝟐) are collinear, then 𝒙𝟏𝒚𝟐 is equal to
(A) 𝒙𝟐𝒚𝟏
(B) 𝒙𝟏𝒚𝟏
(C) 𝒙𝟐𝒚𝟐
(D) 𝒙𝟏𝒙𝟐

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

Step 1: For points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_1+x_2, y_1+y_2)$, set the determinant to zero:

Step 2: Expand the determinant:

Step 3: Simplify each term:

• First term: $x_1(-y_1) = -x_1y_1$
• Second term: $-y_1(-x_1) = +x_1y_1$ …

The problem gives three points: $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_1 + x_2, y_1 + y_2)$. They are collinear — meaning they lie on a single straight line. The most direct way to handle this is through the area condition: three points are collinear if and only if the area of the triangle formed by them is zero.

Why does this work? Because if points are on the same line, you cannot form a triangle with non-zero area — the "triangle" collapses into a line segment. The area formula for a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is:

Setting this to zero (ignoring the absolute value and the factor $\frac12$) gives the collinearity condition:

This is the standard determinant form. Let's apply it step by step.

1. Label the points

   Let $(x_1, y_1) = (x_1, y_1)$, $(x_2, y_2) = (x_2, y_2)$, and $(x_3, y_3) = (x_1 + x_2, y_1 + y_2)$.

2. Write the collinearity condition

3. Substitute the coordinates

   • $y_3 = y_1 + y_2$
   • $x_3 = x_1 + x_2$

   So:

4. Simplify each term

   • First term: $x_1(y_2 - y_1 - y_2) = x_1(-y_1) = -x_1 y_1$
   • Second term: $x_2(y_1 + y_2 - y_1) = x_2(y_2) = x_2 y_2$
   • Third term: $(x_1 + x_2)(y_1 - y_2) = x_1 y_1 - x_1 y_2 + x_2 y_1 - x_2 y_2$

   Adding them:

5. Cancel like terms $-x_1 y_1$ and $+x_1 y_1$ cancel. $x_2 y_2$ and $-x_2 y_2$ cancel. We are left with: …

Method: Area-Determinant Test for Collinear Points

This method proves or exploits collinearity of three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ by using the determinant form of the triangle-area formula — the standard CBSE route to any "are these points collinear" or "find the relation forced by collinearity" question.

Steps

Step 1: Write the collinearity condition as a determinant equal to zero

Three points are collinear exactly when the triangle they form has zero area, which is written as

Step 2: Substitute the given coordinates, including any expressed as sums/combinations

Plug in the actual coordinates for each point. If a point is given in terms of the others (e.g. $(x_1+x_2,\,y_1+y_2)$), substitute those expressions directly into the third row rather than treating them as new independent variables.

Step 3: Use a row operation to expose the hidden relation …

Common Mistakes

Mistake 1: Not substituting the sum-coordinates before expanding

Why it's wrong: leaving $x_1+x_2$ and $y_1+y_2$ unexpanded, as if they were independent third-point coordinates, makes the determinant algebra much harder to simplify and often causes terms that should cancel to be missed. Correct approach: substitute $x_3=x_1+x_2$ and $y_3=y_1+y_2$ directly into the $3\times3$ determinant before expanding, so the like terms cancel cleanly.

Mistake 2: Losing track of the cofactor sign pattern while expanding

Why it's wrong: expanding a $3\times3$ determinant along a row uses alternating $+,-,+$ signs on the cofactors — dropping or flipping one sign turns the correct relation $x_1y_2=x_2y_1$ into its negative, matching the wrong option. Correct approach: expand term by term keeping the $+,-,+$ pattern explicit, and sanity-check with simple numbers (e.g. $x_1=1,y_1=1,x_2=2,y_2=2$). …