Sample 2024-25 Class 12 Maths Q6 Solution: x₁y₂ when (x₁, y₁), (x₂, y₂), (x₁ + x₂, y₁ + y₂) are collinear

The question

If the points $(x_1, y_1)$ ⁠, $(x_2, y_2)$ and $(x_1 + x_2, y_1 + y_2)$ are collinear, then $x_1y_2$ is equal to

(A) $x_2y_1$

(B) $x_1y_1$

(C) $x_2y_2$

(D) $x_1x_2$

Exam answer

For collinear points the area of the triangle is 0:

\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_1 + x_2 & y_1 + y_2 & 1 \end{vmatrix} = 0

Why?

The area of the triangle with these three vertices is $\frac12|\Delta|$ ⁠, where $\Delta$ is the determinant above (each row is $x$ ⁠, $y$ ⁠, 1 for one point).

Three collinear points make no triangle, so the area is 0, and then $\Delta = 0$ ⁠.

Applying $R_3 \to R_3 - R_1 - R_2$ ⁠:

\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ 0 & 0 & -1 \end{vmatrix} = 0

Why?

Subtracting multiples of other rows from a row does not change a determinant. The third row is the sum of the first two in its first two places, so they become 0. The last place becomes

1 - 1 - 1 = -1

Place by place, new $R_3$ = old $R_3 - R_1 - R_2$ ⁠:

\begin{aligned} &(x_1 + x_2) - x_1 - x_2 = 0 \\ &(y_1 + y_2) - y_1 - y_2 = 0 \\ &1 - 1 - 1 = -1 \end{aligned}

Expanding along $R_3$ ⁠:

\Rightarrow -(x_1y_2 - x_2y_1) = 0

Why?

Only the last entry of $R_3$ is not 0. Its sign is $(-1)^{3 + 3} = +1$ ⁠, so the determinant is $-1$ times the minor

\begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} = x_1y_2 - x_2y_1

Written out along $R_3$ ⁠, with signs $+$ ⁠, $-$ ⁠, $+$ ⁠:

\begin{aligned} &0\begin{vmatrix} y_1 & 1 \\ y_2 & 1 \end{vmatrix} - 0\begin{vmatrix} x_1 & 1 \\ x_2 & 1 \end{vmatrix} \\ &\quad + (-1)\begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} \\ &= 0 - 0 + (-1)(x_1y_2 - x_2y_1) \\ &= -(x_1y_2 - x_2y_1) \end{aligned}

$\therefore\ x_1y_2 = x_2y_1$

Why?

\begin{aligned} -(x_1y_2 - x_2y_1) &= 0 \\ x_1y_2 - x_2y_1 &= 0 \quad (\times\,(-1)) \\ x_1y_2 &= x_2y_1 \quad (+\,x_2y_1) \end{aligned}

Answer: (A) $x_2y_1$

Where this comes from

Chapter 4, Determinants: Collinear points: area of a triangle is zero.

Question type: Area given, or points on one line

How to spot it: click to see

Type 4. Area given, or points on one line: use $\pm$ or 0

Spot it: the area is given and a vertex has a $k$ in it; or "collinear"; or "find the equation of the line using determinants".

Here: The three points are collinear, so their area determinant is 0. $R_3 \to R_3 - R_1 - R_2$ leaves $0, 0, -1$ in $R_3$ ⁠, and expanding gives $x_1y_2 = x_2y_1$ ⁠.

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Q6: x1y2x_1y_2x1​y2​ when (x1,y1)(x_1, y_1)(x1​,y1​)⁠, (x2,y2)(x_2, y_2)(x2​,y2​)⁠, (x1+x2,y1+y2)(x_1 + x_2, y_1 + y_2)(x1​+x2​,y1​+y2​) are collinear

Q6: $x_1y_2$ when $(x_1, y_1)$ ⁠, $(x_2, y_2)$ ⁠, $(x_1 + x_2, y_1 + y_2)$ are collinear