Previous year questions: Straight Lines

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Six items are real JEE Main 2026 questions from the official National Testing Agency papers and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the source and the marks. Stems were checked against the official PDF pages, and every numerical answer was re-derived and matched to the official key. Start each question by writing the line or the points in a standard form, then use the condition that the question supplies.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

The mid-points of the sides of a triangle ABC are (5 by 2, 7), (5 by 2, 3) and (4, 5). If its incentre is (h, k), find 3h plus k.

Answer: Let the mid-points be M1, M2, M3. The vertices are A equal to M1 plus M3 minus M2, which is (4, 9); B equal to M1 plus M2 minus M3, which is (1, 5); and C equal to M2 plus M3 minus M1, which is (4, 1). The side lengths are a equal to BC equal to 5, b equal to CA equal to 8 and c equal to AB equal to 5. The incentre is (aA plus bB plus cC) divided by (a plus b plus c), which is (48 by 18, 90 by 18) equal to (8 by 3, 5). So 3h plus k is 8 plus 5 equal to 13. The official key is 13.

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Let the line L1: x plus 3 equal to 0 meet the lines L2: x minus y equal to 0 and L3: 3x plus y equal to 0 at A and B respectively. Let the bisector of the obtuse angle between L2 and L3 meet L1 at C. Find BC squared to AC squared.

Answer: The intersection points are A equal to (minus 3, minus 3) and B equal to (minus 3, 9). The angle bisectors of L2 and L3 satisfy the equality of distances to the two lines: the absolute value of x minus y over root 2 equals the absolute value of 3x plus y over root 10. Squaring and simplifying gives x squared plus 4xy minus y squared equal to 0, whose solutions are the two bisectors. The obtuse bisector has slope 2 plus root 5, so its intersection with x equal to minus 3 is C equal to (minus 3, minus 6 minus 3 root 5). Then AC squared is (3 plus 3 root 5) squared and BC squared is (15 plus 3 root 5) squared. Their ratio simplifies to 5, so BC squared to AC squared is 5 is to 1. The official key is 5 is to 1.

Q3 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Let the vertex A of triangle ABC be (1, 2) and the mid-point of side AB be (5, minus 1). If the centroid is (3, 4) and the circumcentre is (alpha, beta), find 21 times (alpha plus beta).

Answer: From the mid-point, B is (9, minus 4). The centroid condition gives C: A plus B plus C equals three times the centroid, so C equals (9, 12) minus (10, minus 2), which is (minus 1, 14). The perpendicular bisector of AB has equation 4x minus 3y equal to 23, and the perpendicular bisector of AC has equation minus x plus 6y equal to 48. Solving gives y equal to 215 by 21 and x equal to 94 by 7. Then alpha plus beta is 497 by 21 and 21 times that is 497. The official key is 497.

Q4 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

In an equilateral triangle PQR, the vertex P is at (3, 5) and the side QR lies along the line x plus y equal to 4. If the orthocentre is (alpha, beta), find 9 times (alpha plus beta).

Answer: In an equilateral triangle the orthocentre, centroid and incentre coincide, so the orthocentre lies on the altitude from P. The distance of P from the line QR is the height h, which is the absolute value of 3 plus 5 minus 4 over root 2, that is 2 root 2. The orthocentre is at two-thirds of the height from P along the perpendicular towards QR: the unit direction from P to the line is (minus 1, minus 1) by root 2, so the orthocentre is (3, 5) plus (4 by 3) times (minus 1, minus 1), which is (5 by 3, 11 by 3). Then alpha plus beta is 16 by 3 and 9 times that is 48. The official key is 48.

Q5 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

The line x minus y equal to 4 meets the circle (x minus 4) squared plus (y plus 3) squared equal to 9 at Q and R. If P(alpha, beta) is a point on the circle with PQ equal to PR, find (6 alpha plus 8 beta) squared.

Answer: The centre of the circle is (4, minus 3) and its radius is 3. The points equidistant from Q and R lie on the perpendicular through the centre, which is the line through the centre perpendicular to x minus y equal to 4, that is the line with direction (1, minus 1). The perpendicular foot of the centre on the given chord line is the mid-point of QR, and P is on the circle along the same line: P equals (4, minus 3) plus or minus 3 times (1, minus 1) by root 2. Then 6 alpha plus 8 beta equals 6 times (4 plus or minus 3 by root 2) plus 8 times (minus 3 minus or minus 3 by root 2), and the roots terms cancel to give 6 by root 2, whose square is 18. The official key is 18.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

The lines 4x plus 3y minus 1 equal to 0 and 3x plus 4y minus 1 equal to 0 meet at a point. A variable line through that point cuts the coordinate axes at P and Q. Obtain the locus of the mid-point of PQ.

Answer: The two given lines meet at (1 by 7, 1 by 7). Let the variable line meet the axes at (a, 0) and (0, b), so its equation is x by a plus y by b equal to 1. Because it passes through (1 by 7, 1 by 7), we get 1 by a plus 1 by b equal to 7. If the mid-point of PQ is (h, k), then a is 2h and b is 2k, so 1 by 2h plus 1 by 2k equals 7, which gives h plus k equal to 14hk. Replacing h and k by x and y gives the locus x plus y minus 14xy equal to 0. The official key is x plus y minus 14xy equal to 0.

Q7 (NCERT Exemplar, 3 marks)

A line cuts off an intercept minus 3 from the y-axis and has tangent of inclination 3 by 5. Find its equation.

Answer: The slope is 3 by 5 and the y-intercept is minus 3, so the slope-intercept form gives y equal to 3x by 5 minus 3. Multiplying by 5 gives 5y equal to 3x minus 15, that is 5y minus 3x plus 15 equal to 0. Source: NCERT Exemplar Class 11 Mathematics, Chapter 10, Exercise 10.3, Objective Type, Q22.

Q8 (NCERT Exemplar, 3 marks)

Find the equation of the line through (3, 2) perpendicular to y equal to x.

Answer: The given line has slope 1, so the perpendicular has slope minus 1. Through (3, 2) the point-slope form gives y minus 2 equal to minus (x minus 3), that is x plus y equal to 5. Source: NCERT Exemplar Class 11 Mathematics, Chapter 10, Exercise 10.3, Objective Type, Q24.

Q9 (NCERT Exemplar, 3 marks)

If the line x by a plus y by b equal to 1 passes through (2, minus 3) and (4, minus 5), find the pair (a, b).

Answer: Substituting the two points gives 2 by a minus 3 by b equal to 1 and 4 by a minus 5 by b equal to 1. Subtracting twice the first from the second gives 1 by b equal to minus 1, so b is minus 1, and then 2 by a plus 3 equals 1, so a is minus 1. The pair is (minus 1, minus 1). Source: NCERT Exemplar Class 11 Mathematics, Chapter 10, Exercise 10.3, Objective Type, Q27.

Q10 (NCERT Exemplar, 3 marks)

Find the distance of the point of intersection of 2x minus 3y plus 5 equal to 0 and 3x plus 4y equal to 0 from the line 5x minus 2y equal to 0.

Answer: Solving the two lines gives y equal to 15 by 17 and x equal to minus 20 by 17. Substituting in the distance formula with the line 5x minus 2y equal to 0 gives the numerator as 100 plus 30 over 17, that is 130 by 17, and the denominator as the square root of 29. So the distance is 130 divided by 17 root 29. Source: NCERT Exemplar Class 11 Mathematics, Chapter 10, Exercise 10.3, Objective Type, Q28.