Line Through P(1, 2) Cutting Minimum Area Triangle – Straight Lines JEE Main 2026 PYQ

Nishant Kumar Gupta

Former Faculty · Allen · Aakash · Narayana

Maths mentor behind PadhoLikhoJEE and Quantum Chemistry Classes (Arrah, Bihar). Teaches the “why” behind every JEE trap — not just the trick.

Quick Summary

Question Type:

Coordinate geometry (optimisation with intercepts)

Chapter:

Coordinate Geometry — Straight Lines

Difficulty:

⭐⭐⭐ Medium

Time to Solve:

90 seconds

Key Concept:

For intercepts a, b with 1/a + 2/b = 1, AM–GM gives ab ≥ 8 with equality at a = 2, b = 4 — the minimum-area line is x/2 + y/4 = 1, i.e. 2x + y − 4 = 0

Correct Answer:

Option (2) – 2x + y − 4 = 0

Why:

At the minimum the fixed point P(1, 2) bisects the intercept segment AB: a = 2·1 and b = 2·2

The Question

JEE Main 2026 (8 April – Evening Shift) – Coordinate Geometry — Straight Lines

If a straight line drawn through the point P(1, 2) intersects the coordinate
axes at A and B such that the origin O, A, and
B form a triangle of minimum area, the equation of the line is:

1. (1) x + y − 7 = 0
2. (2) 2x + y − 4 = 0 ✓
3. (3) 2x + y + 14 = 0
4. (4) x + 2y − 5 = 0

Quick Answer

Correct Option: (2) – 2x + y − 4 = 0

Reasoning: the elegant one-line fact: at minimum area, the fixed point is the midpoint of the intercept segment:

• Midpoint shortcut: minimum area ⟺ P bisects AB → intercepts a = 2·x_P = 2 and b = 2·y_P = 4
• Intercept form: x/2 + y/4 = 1 → multiply by 4: 2x + y − 4 = 0
• AM–GM proof of the shortcut: 1 = 1/a + 2/b ≥ 2√(2/ab) ⟹ ab ≥ 8, equality when 1/a = 2/b ⟺ b = 2a ⟺ P is the midpoint
• Minimum area = ab/2 = 4 ✓ — and 2(1) + 2 − 4 = 0 confirms the line passes through P

Understanding the Concept

Setting Up the Intercept Frame

A line cutting the positive x-axis at A(a, 0) and the positive y-axis at B(0, b) has intercept
form x/a + y/b = 1, and the triangle it forms with the origin has area ab/2. Requiring the
line to pass through P(1, 2) imposes the constraint 1/a + 2/b = 1. The optimisation problem is
now: minimise ab subject to a weighted-reciprocal constraint — a shape that appears in dozens
of JEE questions, from “minimum length of intercept segment” to “minimum perimeter” variants.

AM–GM Does the Calculus For You

Apply AM–GM to the two constraint terms: 1 = 1/a + 2/b ≥ 2√((1/a)(2/b)) = 2√(2/ab). Squaring
and rearranging: ab ≥ 8, with equality exactly when 1/a = 2/b, i.e. b = 2a. Feeding b = 2a
back into the constraint: 1/a + 2/(2a) = 2/a = 1, so a = 2 and b = 4. The minimum area is
8/2 = 4 square units, and the line is x/2 + y/4 = 1 — multiplying by 4 gives 2x + y − 4 = 0.
No derivatives, no second-derivative test: the inequality carries the optimisation and its
equality condition delivers the minimiser in the same breath.

The Midpoint Gem

Look at the minimiser geometrically: a = 2 = 2·1 and b = 4 = 2·2, so the intercept segment
runs from (2, 0) to (0, 4) and its midpoint is (1, 2) — precisely P. This is a theorem, not a
coincidence: for the minimum-area (and minimum-length) intercept line through a first-quadrant
point, the point bisects the intercepted segment. The intuition: the family of lines through P
sweeps the intercept triangle from degenerate-flat to huge, and the balanced configuration —
where P sits exactly halfway — is where the product ab bottoms out. Recognising this converts
every such question into a 10-second answer: double the coordinates of the fixed point to get
the intercepts.

Option Forensics and Sanity Checks

Before committing, run the two 5-second checks. First, does the option’s line even pass through
P? Option (1) gives 1 + 2 − 7 ≠ 0 and option (3) gives 2 + 2 + 14 ≠ 0 — both die instantly,
leaving (2) and (4) to the optimisation. Second, does the surviving line have its intercepts
at (2, 0) and (0, 4)? Setting y = 0 in 2x + y − 4 = 0 gives x = 2; setting x = 0 gives y = 4.
Both checks align with the AM–GM minimiser. This pass-through filter plus the midpoint gem
makes the question effectively self-verifying — which is exactly why JEE recycles it with new
points every few years.

Detailed Step-by-Step Solution

Final Answer

Essential Facts for This Topic

Minimum-Area Intercept Toolkit

1. Intercept form: x/a + y/b = 1; triangle area with axes = |ab|/2
2. The midpoint gem: min-area (and min-length) intercept line through P(h, k) has intercepts a = 2h, b = 2k — line: x/h + y/k = 2
3. AM–GM engine: for h/a + k/b = 1: 1 ≥ 2√(hk/ab) ⟹ ab ≥ 4hk, equality iff h/a = k/b ⟺ P bisects AB
4. Calculus fallback: b = kh/(a − h) makes area = hk·a/(2(a − h)); d/da = 0 at a = 2h
5. Instant filters: candidate lines must pass through P (substitute!) and have positive intercepts for a first-quadrant triangle
6. General form for P(h, k): minimum-area line is x/h + y/k = 2, i.e. kx + hy = 2hk

Common Mistakes to Avoid

Key Concept Summary

What You Must Remember

1. Constraint from P: 1/a + 2/b = 1; objective: minimise ab/2
2. AM–GM bound: ab ≥ 8, equality iff 1/a = 2/b (i.e. b = 2a)
3. Minimiser: a = 2, b = 4 — exactly twice the coordinates of P(1, 2): the midpoint gem
4. Line: x/2 + y/4 = 1 → 2x + y − 4 = 0; minimum area = 4
5. Instant filters: options (1) and (3) fail the pass-through test at P — kill them on sight
6. General template: through P(h, k): x/h + y/k = 2, intercepts (2h, 0), (0, 2k)

The Golden Rule for Intercept Optimisation

“For the minimum-area axes-triangle cut by a line through P(h, k), P bisects the intercept — write x/h + y/k = 2 and you are done before AM–GM even warms up.”

Frequently Asked Questions

Q1: Why does the minimum occur when P is the midpoint of AB?

A: AM–GM equality requires the two constraint terms to be equal: 1/a = 2/b. With P = (1, 2) on the line, the section formula then shows P divides AB in the ratio equal parts — geometrically, P is the midpoint of the segment from (a, 0) to (0, b). Equivalently, a = 2h and b = 2k for P(h, k). The balanced configuration is where the product of intercepts bottoms out.

Q2: Can this be solved with calculus instead of AM–GM?

A: Yes: eliminate b using the constraint, b = 2a/(a − 1), so the area is f(a) = a²/(a − 1). Differentiating, f′(a) = a(a − 2)/(a − 1)², which vanishes at a = 2 (and the sign change confirms a minimum). Then b = 4, agreeing with the AM–GM route. The inequality method is preferred in the exam because the equality condition hands over the minimiser simultaneously.

Q3: What is the minimum area of triangle OAB?

A: The intercepts at the optimum are a = 2 and b = 4, so the area is ab/2 = 8/2 = 4 square units. The line x/2 + y/4 = 1 (i.e. 2x + y − 4 = 0) cuts exactly this minimal triangle from the axes.

Q4: How do I adapt the midpoint shortcut to any point P(h, k)?

A: The minimum-area line through P(h, k) in the first quadrant is x/h + y/k = 2, with intercepts (2h, 0) and (0, 2k) — double each coordinate of P. For P(1, 2) this instantly gives x/1 + y/2 = 2, i.e. 2x + y = 4, matching the full derivation.

Q5: Why do options like x + y − 7 = 0 appear at all?

A: They are pass-through decoys: plausible-looking lines that do not contain P(1, 2) (1 + 2 − 7 ≠ 0). Substituting the given point into every candidate is a five-second filter that eliminates such options without any optimisation — always run it first.

Prerequisites to Solve This Question

Before attempting this problem, you should be comfortable with:

1. Intercept form of a line: x/a + y/b = 1 and reading intercepts from any equation
2. Area of an axes triangle: Δ = |ab|/2 for intercepts a, b
3. AM–GM inequality: x + y ≥ 2√(xy) with the equality condition x = y
4. Constraint handling: converting ‘line passes through P’ into an equation in a, b
5. Section formula: midpoint of (a, 0) and (0, b) as (a/2, b/2) — the gem’s geometric core

After Solving This, You Can:

• Answer any ‘minimum-area intercept line through P(h, k)’ question in 15 seconds via x/h + y/k = 2
• Handle the minimum-length-of-AB variant — same midpoint gem, different objective
• Solve negative-quadrant versions (P in the second quadrant) by tracking signs of intercepts
• Cross-check any candidate line with the two instant filters (pass-through, intercept signs)
• Convert between intercept, slope, and general forms fluently under time pressure

Study Tips for This Topic

For JEE Main:

1. Memorise the gem, not just the method: x/h + y/k = 2 through P(h, k) answers the whole question family — derivation is your backup, not your plan
2. Filter options first: substitute P into all four options; half of them usually die before the optimisation starts
3. Budget 90 seconds: filter (15s) + constraint and AM–GM (40s) + line equation and verification (35s)

Common JEE Variants:

• “Minimum length of the intercept segment AB through P(h, k)” — same equality condition, objective a² + b²
• P in the second quadrant with axes-triangle in the second quadrant — intercepts carry signs; area = |ab|/2
• “Line through P makes minimum-area triangle with the axes in the first quadrant” for P(h, k), h, k > 0 — the general template
• Fixed-slope family instead of fixed-point family — optimise along parallel lines
• Tangent-line versions: line through P tangent to a curve, minimising the axes triangle — the gem generalises

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium (the midpoint gem makes it 15 seconds; without it, AM–GM or calculus is a solid 2-minute route)

JEE Main Frequency: High — minimum-area/minimum-length intercept questions cycle through JEE Main papers every few sessions

JEE Advanced Frequency: Medium — Advanced embeds the same optimisation inside tangent and normal problems

Topic Importance: High — a flagship example of inequality-driven optimisation in coordinate geometry

Written by Nishant Kumar Gupta
Former Faculty · Allen · Aakash · Narayana — Quantum Chemistry Classes, Arrah (Bihar)
Last Updated: 2 Sep 2026 ·
Question Source: JEE Main 2026 (8 April – Evening Shift) Previous Year Question ·
Topic: Straight Lines – Minimum-Area Intercept Triangle and AM–GM