Straight Lines: JEE Advanced PYQ, 2007–2026

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)

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: (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: (C)

The circumcentre lies on the perpendicular bisector of AB, which is x = 3. On the line y = x − 2 it is therefore (3, 1). Equating the distances to A and C gives 3² + 1² = (3 − 2)² + (1 − k)². So 10 = 1 + (1 − k)², which gives (k − 1)² = 9 and k = 4 or −2. Since k > 0, k = 4. Check: the distance from (3, 1) to C(2, 4) is √10, which equals the distance to A and to B.

Answer: (A)

Writing the line as y − 4 = m(x − 2) misses the vertical line. The vertical line x = 2 is at distance 2 from the origin, so it works. For the others, mx − y + 4 − 2m = 0 gives |4 − 2m|/√(m²+1) = 2. Squaring: 16 − 16m + 4m² = 4m² + 4, so m = 3/4, and the line is 3x − 4y + 10 = 0. The second line makes angle θ with the x-axis where tan θ = 3/4, so it makes angle 90° − θ with the vertical line. Hence the required tangent is cot θ = 4/3.

Answer: (A)

The centroid is ((0+6+3)/3, (0+0+h)/3) = (3, h/3). By symmetry the orthocentre lies on x = 3. The slope of BC is h/(3−6) = −h/3, so the altitude from A(0, 0) has slope 3/h and equation y = 3x/h. At x = 3 this gives y = 9/h, so H = (3, 9/h). Setting 9/h = h/3 gives h² = 27, so h = 3√3. Then each side has length 6, which is consistent with an equilateral triangle, the case in which all the centres coincide.

Answer: (D)

The centroid G is the intersection of the two medians: solving y = x and x + 2y = 6 gives G = (2, 2). G divides the median AD in the ratio 2 : 1, so D = A + (3/2)(G − A) = (2, 5) + (3/2)(0, −3) = (2, 1/2). Then B + C = 2D = (4, 1). Let B = (b, b) on y = x and C = (6 − 2c, c) on x + 2y = 6. Then b + 6 − 2c = 4 and b + c = 1, which gives c = 1 and b = 0. So B = (0, 0) and C = (4, 1). The area is ½|2(0 − 1) + 0(1 − 5) + 4(5 − 0)| = ½·18 = 9. As a check, the median from B goes through the midpoint (3, 3) of AC, and the median from C goes through the midpoint (1, 5/2) of AB.