Q.Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants and find $k$ if $D(k, 0)$ is a point such that area of triangle $ABD$ is $3$ sq units.

Concept: Vector Equation of Line (via determinant form for collinearity).

Step 1 – Equation of line AB

For any point $P(x, y)$ on line $AB$, the area of triangle $ABP$ is zero. Using the determinant formula:

Expanding along the second row:

So the line is $y = 3x$.

Step 2 – Area condition for triangle ABD

Area of $\triangle ABD$ with $A(1,3)$, $B(0,0)$, $D(k,0)$ is given as 3 sq units. …

Why use determinants for a line?

The idea is beautiful: three points are collinear if the area of the triangle they form is zero. So if we want the equation of the line through two fixed points $A$ and $B$, we take a general point $P(x, y)$ on that line and demand that the area of triangle $ABP$ be zero. That gives us the line's equation — no slope formula needed, no memorised forms. Determinants make this automatic.

For area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, the formula is:

We'll use this twice: once to find the line, once to find $k$.

1. Equation of line $AB$

Let $P(x, y)$ be any point on line $AB$. For $A(1,3)$, $B(0,0)$, and $P(x,y)$ to be collinear, the area of triangle $ABP$ must be zero:

The absolute value and the factor $\frac{1}{2}$ don't matter for zero, so we just set the determinant to zero:

Expand along the second row (which has two zeros — clever choice):

So:

That's $1 \cdot y - 3 \cdot x = 0$, or:

That's the line through $A$ and $B$. Notice: we never computed a slope — the determinant did it for us.

2. Finding $k$ such that area of $\triangle ABD$ is $3$ sq units

Now $D(k, 0)$ is a point on the $x$-axis. We want the area of triangle $ABD$ to be exactly $3$ (square units). Using the same determinant formula:

Multiply both sides by $2$:

Again expand along the second row (two zeros): …

Method: Using the Collinearity Determinant to Find a Line's Equation, Then Solve an Area Condition

This method finds the equation of a line through two known points by demanding that a general point on the line make the triangle-area determinant vanish, then reuses the same determinant (now nonzero) to solve for an unknown coordinate that gives a target area.

Steps

Step 1: Set up the general collinearity condition

For three points to be collinear, the "triangle" they form must have zero area, i.e. the determinant

Step 2: Substitute the two known points and a general point $(x,y)$

Put the two given points in rows 1 and 2, and a variable point $(x,y)$ in row 3 (or wherever is most convenient).

Step 3: Expand along the row/column with the most zeros

If one of the known points is the origin (or has a zero coordinate), expanding along that row collapses most terms immediately.

Step 4: Set the expansion to zero and simplify

This produces a linear equation in $x$ and $y$ — the equation of the line.

Step 5: Reuse the same determinant setup for the area condition …

Common Mistakes

Mistake 1: Reporting only $k=-2$ and missing $k=2$

Why it's wrong: solving $|-3k|=6$ by writing $-3k=6$ alone (without also considering $-3k=-6$) finds only one root; the absolute value equation genuinely has two solutions. Correct approach: whenever an absolute value equals a positive number, split into the two cases (or apply $\pm$) and check both.

Mistake 2: Confusing the "line through A and B" condition (area $=0$) with the "area $=3$" condition, and setting the wrong equation to zero …