Straight Lines: JEE Main PYQ, 2020–2025

Answer: (A)

The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂-x₁)² + (y₂-y₁)²]. Here x₂-x₁ = 2-(-2) = 4 and y₂-y₁ = 0-3 = -3. So AB = √(16+9) = √25 = 5 units.

Answer: (A)

The slope of AB is (6 − 2)/(4 − 1) = 4/3. The line through A is y − 2 = (4/3)(x − 1), which simplifies to 4x − 3y + 2 = 0. The altitude from C is the perpendicular distance from C(5, 1) to this line, which is |4(5) − 3(1) + 2|/√(16 + 9) = 19/5.

Answer: (A)

For two lines with slopes m₁ and m₂, tan θ = |(m₁ − m₂)/(1 + m₁m₂)|. The given line has slope 3, so |(m − 3)/(1 + 3m)| = 1. Case 1: m − 3 = 1 + 3m gives m = −2. Case 2: m − 3 = −1 − 3m gives m = 1/2. Through (2, −1), slope −2 gives y + 1 = −2(x − 2), i.e. 2x + y = 3, with x-intercept 3/2. Slope 1/2 gives x − 2y = 4, with x-intercept 4. The product is (3/2)(4) = 6.

Answer: (A)

A point on the x-axis is P(x, 0). Equating PA² = PB²: (x−1)² + (0−2)² = (x−3)² + (0−4)². Expanding, x² − 2x + 5 = x² − 6x + 25, so 4x = 20 and x = 5. The point is (5, 0).

Answer: (A)

Reflecting in y = x swaps the coordinates, so (3, −4) becomes (−4, 3). Reflecting in the x-axis changes the sign of y, so (3, −4) becomes (3, 4). The distance between (−4, 3) and (3, 4) is √((3+4)² + (4−3)²) = √(49 + 1) = √50 = 5√2.

Answer: (A)

Parallel lines have equal slopes. The slopes are −k/4 and −1/k, so k² = 4 and k = ±2. Parallel and distinct requires that the constants do not follow the same ratio as the coefficients. For k = 2 the first line is 2x + 4y = 4, which is x + 2y = 2, identical to the second line, so they coincide. For k = −2 the first line is −2x + 4y = 0, which is x − 2y = 0, and the second is x − 2y = 2. These are parallel but different. The only valid value is k = −2.

Answer: (D)

CA = CB = √45, so the triangle is isosceles and the line x = 0 (the perpendicular bisector of AB and the median and altitude from C) is its axis of symmetry. The centroid (0, 2), the circumcentre and the orthocentre all lie on this axis, since each is fixed by symmetry and so cannot leave the axis. The line is x = 0.

Answer: (A)

Apply the distance formula: PQ² = a²[(cos α − cos β)² + (sin α − sin β)²]. Expanding and using cos²θ + sin²θ = 1 gives a²[2 − 2(cos α cos β + sin α sin β)] = a²[2 − 2cos(α − β)]. Since 1 − cos θ = 2sin²(θ/2), PQ² = 4a² sin²((α − β)/2). Hence PQ = 2a|sin((α − β)/2)|. The cosine option comes from a sign slip in the identity, the sin(α − β) option ignores the half-angle, and the last option drops the factor 2.

Answer: (A)

For points on the same side of a line, PA + PB is smallest when P lies on the straight segment from A to the mirror image of B. The image of B(7, 5) in the x-axis is B′(7, −5). Since PB = PB′ for every P on the x-axis, PA + PB = PA + PB′ ≥ AB′ = √((7 − 1)² + (−5 − 3)²) = √100 = 10. Equality holds when P is the point where AB′ meets the x-axis.

Answer: (A)

Any point on the y-axis has x = 0. If it divides AB in the ratio m : n, its x-coordinate is (m·4 + n·(−2))/(m + n). Setting this to 0 gives 4m = 2n, so m : n = 1 : 2.