JEE Corner — Straight Lines for JEE Main and Advanced

Concurrency and the Family of Lines

The chapter so far gives you the machinery; JEE tests four standard extensions of it. This section builds each one from what you already know.

1. Concurrency of three lines. Three lines $a_ix + b_iy + c_i = 0$ ($i = 1, 2, 3$) are concurrent when they pass through one common point. Two equivalent tests:

Key Point (Concurrency): Either solve two of the lines and substitute the point into the third, or use the determinant test:

$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$$

The determinant condition is necessary; in exam practice (non-degenerate, non-parallel lines) it settles the question in one computation.

2. Family of lines through an intersection. For lines $L_1 = 0$ and $L_2 = 0$, every equation

$$L_1 + \lambda L_2 = 0$$

is a line through their intersection point $P$ (substituting $P$ makes both $L_1$ and $L_2$ vanish, hence the whole expression). Varying $\lambda$ sweeps out all lines through $P$ except $L_2$ itself.

Why it saves time: you impose the extra condition (passes through a given point, parallel or perpendicular to a given line, given distance from a point) directly on $L_1 + \lambda L_2 = 0$ and solve for $\lambda$ — no need to compute the intersection point first.

Worked micro-example: the line through the intersection of $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$, perpendicular to $6x - 7y + 8 = 0$: write $(2 + 3\lambda)x + (4\lambda - 3)y + (4 - 5\lambda) = 0$ and demand slope $-\frac{7}{6}$. That gives $6(2 + 3\lambda) = 7(4\lambda - 3)$, so $\lambda = \frac{33}{10}$, and the line is $119x + 102y = 125$ — the same answer the junction-point route reaches with more work.

A useful extreme case: among all lines of the family through $P$, the one farthest from a given point $Q$ is the line through $P$ perpendicular to $PQ$ (any other member passes closer). The one closest would pass through $Q$ itself if allowed.

Foot, Image, and Distance Along a Direction

3. Foot of the perpendicular and image in a line — the JEE formulas. For the line $Ax + By + C = 0$ and point $P(x_1, y_1)$, set

$$t = \frac{Ax_1 + By_1 + C}{A^2 + B^2}$$

Key Point (Foot and image): The foot of the perpendicular $M$ and the image $P'$ of $P$ are

$$M = (x_1 - At,\; y_1 - Bt), \qquad P' = (x_1 - 2At,\; y_1 - 2Bt)$$

The direction $(A, B)$ is perpendicular to the line, and $t$ measures how far (in that normalized direction) $P$ sits from the line; going $-t$ lands on the line, going $-2t$ lands the mirror image. This one formula replaces the two-equation setup (perpendicular slope + midpoint) — both give the same answer, but the formula is a 30-second computation.

Check it: for $(-1, 3)$ and $3x - 4y - 16 = 0$: $t = \frac{-3 - 12 - 16}{25} = -\frac{31}{25}$, so the foot is $\left(-1 + \frac{93}{25},\; 3 - \frac{124}{25}\right) = \left(\frac{68}{25}, -\frac{49}{25}\right)$ — exactly the answer the two-equation route produced earlier.

4. Distance measured along a direction (the parametric idea). A point at distance $r$ from $(x_1, y_1)$ along the direction making angle $\theta$ with the positive $x$-axis is

$$(x_1 + r\cos\theta,\; y_1 + r\sin\theta)$$

To find how far you travel from $P$ along direction $\theta$ before hitting a line $L$, substitute this point into $L = 0$ and solve for $r$. This is the clean way to handle every distance-along-a-direction problem, and it is the seed of the parametric form of a line used heavily in JEE conics.

5. Two more exam staples.

The area of the triangle formed by three lines: find the three vertices by solving the lines in pairs, then use $\text{Area} = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Special case worth memorising: $y = m_1x + c_1$, $y = m_2x + c_2$, $x = 0$ give area $\frac{(c_1 - c_2)^2}{2|m_1 - m_2|}$.

Reflection problems: a ray from $P$ reflecting off a mirror line to reach $Q$ travels as if it went straight from the image $P'$ to $Q$ — reflection questions are image questions in disguise.

[JEE Tip] In the foot/image formula, the most common exam error is dropping the $A^2 + B^2$ denominator in $t$ or forgetting the factor 2 for the image. Anchor it with the origin check: the image of the origin in $x + y - 2 = 0$ has $t = \frac{-2}{2} = -1$, so $P' = (2, 2)$ — symmetric, as it must be.

JEE-Style Solved Examples

Example 1: Concurrency check

Show that the lines $3x - 4y + 5 = 0$, $7x - 8y + 5 = 0$ and $4x + 5y - 45 = 0$ are concurrent, and find the common point.

Solution:

Step 1 — Intersect the first two. Double the first: $6x - 8y + 10 = 0$; subtract from the second: $x - 5 = 0$, so $x = 5$ and then $y = 5$.

Step 2 — Test the third. $4(5) + 5(5) - 45 = 20 + 25 - 45 = 0$ ✓.

Step 3 — Conclude. All three pass through $(5, 5)$ — concurrent. ∎

Takeaway: Line up the coefficients (double, subtract) — elimination beats formula-hunting for 2×2 systems.

Example 2: Family of lines through a given point

Find the line through the intersection of $x + 2y - 5 = 0$ and $3x - y - 1 = 0$ that passes through the origin.

Solution:

Step 1 — Write the family. $(x + 2y - 5) + \lambda(3x - y - 1) = 0$.

Step 2 — Impose the condition. At $(0, 0)$: $-5 - \lambda = 0$, so $\lambda = -5$.

Step 3 — Substitute back. $(x + 2y - 5) - 5(3x - y - 1) = 0$ gives $-14x + 7y = 0$, i.e. $y = 2x$.

Takeaway: The family method never computes the intersection point — the condition finds $\lambda$ directly.

Example 3: Farthest member of a family

Among all lines through the intersection $P$ of $x + y - 2 = 0$ and $3x - y - 2 = 0$, find the one at maximum distance from $Q(2, 3)$.

Solution:

Step 1 — Find $P$. Adding the equations: $4x - 4 = 0$, so $P = (1, 1)$.

Step 2 — Use the extreme-case fact. The farthest line through $P$ from $Q$ is perpendicular to $PQ$ — any other member cuts closer to $Q$.

Step 3 — Build it. $PQ$ slope $= 2$ → line slope $-\frac{1}{2}$: $y - 1 = -\frac{1}{2}(x - 1)$, i.e. $x + 2y - 3 = 0$; its distance from $Q$ is $PQ = \sqrt{5}$.

Takeaway: Farthest = perpendicular to $PQ$ at $P$; the maximum distance is $|PQ|$ itself.

Example 4: Foot of the perpendicular by formula

Find the foot of the perpendicular from $(2, 3)$ to the line $x + y - 11 = 0$.

Solution:

Step 1 — Compute $t$. $t = \frac{2 + 3 - 11}{1^2 + 1^2} = \frac{-6}{2} = -3$.

Step 2 — Apply the foot formula. $M = (x_1 - At, y_1 - Bt) = (2 - 1(-3), 3 - 1(-3)) = (5, 6)$.

Step 3 — Check. $5 + 6 - 11 = 0$ ✓ — the foot lies on the line.

Takeaway: One $t$, two coordinates — and the on-the-line check is instant.

Example 5: Image by formula

Find the image of $(4, -13)$ in the line $5x + y + 6 = 0$.

Solution:

Step 1 — Compute $t$. $t = \frac{5(4) + (-13) + 6}{25 + 1} = \frac{13}{26} = \frac{1}{2}$.

Step 2 — Apply the image formula (factor 2). $P' = (4 - 2 \cdot 5 \cdot \frac{1}{2},\ -13 - 2 \cdot 1 \cdot \frac{1}{2}) = (-1, -14)$.

Step 3 — Check the midpoint. $\left(\frac{4-1}{2}, \frac{-13-14}{2}\right) = \left(\frac{3}{2}, -\frac{27}{2}\right)$: $\frac{15}{2} - \frac{27}{2} + 6 = 0$ ✓.

Takeaway: Foot uses $t$; image uses $2t$ — the factor 2 is the exam's favourite thing to steal.

Example 6: Distance along a direction

A line is drawn through $A(2, 3)$ making $45°$ with the positive $x$-axis. How far from $A$ does it meet $x + y - 10 = 0$?

Solution:

Step 1 — Parametrise. A point at distance $r$ along the $45°$ direction: $\left(2 + \frac{r}{\sqrt{2}},\ 3 + \frac{r}{\sqrt{2}}\right)$.

Step 2 — Substitute into the line. $2 + \frac{r}{\sqrt{2}} + 3 + \frac{r}{\sqrt{2}} - 10 = 0$ gives $r\sqrt{2} = 5$.

Step 3 — Solve. $r = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$.

Takeaway: The parametric point $(x_1 + r\cos\theta, y_1 + r\sin\theta)$ turns "how far along" into a linear equation in $r$.

Example 7: Area of a triangle formed by three lines

Find the area of the triangle formed by $y = x$, $y = 2x$ and $y = 3x + 4$.

Solution:

Step 1 — Vertices from pairwise intersections. $y = x$ meets $y = 2x$ at $(0, 0)$; $y = x$ meets $y = 3x + 4$ at $(-2, -2)$; $y = 2x$ meets $y = 3x + 4$ at $(-4, -8)$.

Step 2 — Shoelace formula. $\text{Area} = \frac{1}{2}|0(-2 + 8) + (-2)(-8 - 0) + (-4)(0 + 2)|$.

Step 3 — Evaluate. $\frac{1}{2}|0 + 16 - 8| = 4$.

Takeaway: Three lines → three 2×2 solves → one shoelace. Keep the vertices exact; the moduli handle orientation.

Example 8: A family member parallel to an axis

For what $\lambda$ is $(2x - 3y + 4) + \lambda(3x + 4y - 5) = 0$ parallel to the $x$-axis, and what is the line?

Solution:

Step 1 — Collect coefficients. $(2 + 3\lambda)x + (4\lambda - 3)y + (4 - 5\lambda) = 0$.

Step 2 — Kill the $x$-term. $2 + 3\lambda = 0$: $\lambda = -\frac{2}{3}$.

Step 3 — Substitute back. $-\frac{17}{3}y + \frac{22}{3} = 0$, i.e. $y = \frac{22}{17}$.

Takeaway: Structural conditions (parallel to an axis, given slope) act on the COEFFICIENTS of the family — no geometry needed.

Example 9: Reflection as an image problem

A ray of light leaves $P(1, 2)$, strikes the mirror line $y = 0$ and passes through $Q(5, 3)$. Verify the reflection point using the image principle, and state the principle.

Solution:

Step 1 — State the principle. The reflected path $P \to A \to Q$ straightens into the segment $P'Q$, where $P'$ is the image of $P$ in the mirror — so $A = P'Q \cap \text{mirror}$.

Step 2 — Compute. $P' = (1, -2)$; line $P'Q$ has slope $\frac{3 + 2}{5 - 1} = \frac{5}{4}$.

Step 3 — Intersect with $y = 0$. $0 + 2 = \frac{5}{4}(x - 1)$ gives $x = \frac{13}{5}$: $A = \left(\frac{13}{5}, 0\right)$ — matching the direct computation.

Takeaway: Equal angles at the mirror = straight line through the image; every reflection question is an image question.

Example 10: Perpendicular distance as a minimum

Of all points on the line $3x + 4y - 25 = 0$, find the one closest to the origin, and the minimum distance.

Solution:

Step 1 — The closest point is the foot. $t = \frac{3(0) + 4(0) - 25}{9 + 16} = -1$.

Step 2 — Foot formula. $(0 - 3(-1),\ 0 - 4(-1)) = (3, 4)$.

Step 3 — Minimum distance. $\frac{|-25|}{5} = 5$ — and $\sqrt{3^2 + 4^2} = 5$ confirms it.

Takeaway: "Closest point on a line" = foot of the perpendicular; the distance formula gives the value without the point.