Family of lines passing through intersection of two given lines

Question

Find the equation of the line passing through the intersection of $2x + 3y - 5 = 0$ and $3x - 2y + 1 = 0$ , and also passing through the point $(1, 1)$ .

(JEE Main 2022, similar pattern)

Solution — Step by Step

Any line passing through the intersection of $L_1: 2x + 3y - 5 = 0$ and $L_2: 3x - 2y + 1 = 0$ can be written as:

L_1 + \lambda L_2 = 0

(2x + 3y - 5) + \lambda(3x - 2y + 1) = 0

This is the family of lines through their intersection. Different values of $\lambda$ give different lines in this family.

The line passes through $(1, 1)$ . Substitute $x = 1, y = 1$ :

(2 + 3 - 5) + \lambda(3 - 2 + 1) = 0

0 + \lambda(2) = 0

\lambda = 0

Substituting $\lambda = 0$ :

\mathbf{2x + 3y - 5 = 0}

Interesting — the required line is $L_1$ itself. This means the point $(1, 1)$ actually lies on $L_1$ . Quick check: $2(1) + 3(1) - 5 = 0$ . Confirmed.

Why This Works

The equation $L_1 + \lambda L_2 = 0$ represents every line through the intersection of $L_1$ and $L_2$ (except $L_2$ itself, which corresponds to $\lambda \to \infty$ ). Why? Because any point satisfying both $L_1 = 0$ and $L_2 = 0$ also satisfies $L_1 + \lambda L_2 = 0$ for every $\lambda$ .

This “family” approach is powerful because it avoids finding the intersection point explicitly. You directly get the required line by applying one additional condition (passing through a point, or having a given slope, or being parallel to an axis).

Alternative Method

Find the intersection point by solving the two equations simultaneously:

From $2x + 3y = 5$ and $3x - 2y = -1$ : multiply first by 2 and second by 3, then add: $4x + 9x + 6y - 6y = 10 - 3$ , so $13x = 7$ , $x = 7/13$ , $y = (5 - 14/13)/3 = 51/(13 \times 3) = 17/13$ .

Intersection: $(7/13, 17/13)$ . Then find the line through $(7/13, 17/13)$ and $(1, 1)$ . This works but involves uglier fractions.

The family-of-lines method is always faster when the additional condition is simple (a point, a slope, or parallel/perpendicular to an axis). It skips the messy intersection calculation entirely. Use it as your default approach for JEE.

Common Mistake

The family $L_1 + \lambda L_2 = 0$ does NOT include $L_2$ itself (that would need $\lambda = \infty$ ). If the problem’s answer turns out to be $L_2$ , write it separately. In practice, when $\lambda = 0$ gives the answer (as here), it means the answer is $L_1$ , and when $\lambda$ is finite and non-zero, it is a new line. But if you suspect the answer might be $L_2$ , check it directly.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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