Q.(i) Find equation of line joining $(1, 2)$ and $(3, 6)$ using determinants.

Concept: Area of a triangle formed by three points is zero when the points are collinear — this gives the equation of a line through two points.

(i) For points $(1,2)$, $(3,6)$, and a general point $(x,y)$, the area condition is:

Expanding the determinant:

$x(2-6) - y(1-3) + 1(6-6) = -4x + 2y = 0$ …

Why determinants give the equation of a line

The key idea is geometric. Three points are collinear (lie on the same straight line) if and only if the area of the triangle they form is zero.

If we have two fixed points $A(x_1, y_1)$ and $B(x_2, y_2)$, and a variable point $P(x, y)$, then $P$ lies on the line through $A$ and $B$ exactly when the area of $\triangle PAB$ is zero.

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

Setting this area to zero (and ignoring the absolute value, since we only care about the condition) gives the equation of the line.

(i) Line joining $(1, 2)$ and $(3, 6)$

1. Set up the collinearity condition.

Let $P(x, y)$ be any point on the line. For $P$, $(1, 2)$, and $(3, 6)$ to be collinear, the determinant of their coordinates must be zero:

2. Expand the determinant.

Using expansion along the first row (or any row/column):

3. Compute each $2 \times 2$ determinant.

• First minor: $(2)(1) - (1)(6) = 2 - 6 = -4$
• Second minor: $(1)(1) - (1)(3) = 1 - 3 = -2$
• Third minor: $(1)(6) - (2)(3) = 6 - 6 = 0$

Substituting:

4. Simplify.

Divide through by $-2$:

Method: Finding the Equation of a Line Through Two Points Using a Determinant

Use this whenever asked to find a line's equation "using determinants."

Steps

Step 1: Introduce a general point $(x,y)$ as the third vertex

For $P(x,y)$ to lie on the line through the two given points $A$ and $B$, the triangle $PAB$ must have zero area — that IS the collinearity condition.

Step 2: Write the $3\times3$ determinant and set it to $0$

Step 3: Expand along the first row, tracking the sign pattern

$x\begin{vmatrix} y_1 & 1 \\ y_2 & 1 \end{vmatrix} - y\begin{vmatrix} x_1 & 1 \\ x_2 & 1 \end{vmatrix} + 1\begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} = 0.$ …

Common Mistakes

Mistake 1: Forgetting the alternating $+,-,+$ sign pattern when expanding along the first row

Why it's wrong: dropping the minus sign on the middle ($y$) term flips its coefficient, turning a correct equation like $-4x+2y=0$ into the wrong $-4x-2y=0$. Correct approach: always write out the checkerboard sign explicitly for each term of a first-row expansion before substituting the minors.

Mistake 2: Leaving the equation unsimplified or failing to sanity-check it against the two given points …