CBSE Class 11 Mathematics Straight Lines JEE Previous Year Questions Set 01

Answer: (a) isosceles and right angled

Explanation:
1. Let A(4, 0), B(−1, −1), C(3, 5). AB² = 25 + 1 = 26, AC² = 1 + 25 = 26, BC² = 16 + 36 = 52.
2. AB = AC, so the triangle is isosceles.
3. AB² + AC² = 52 = BC², so it is right angled at A.

Teacher's Note:
1. Work with squares of sides to check both isosceles and Pythagoras conditions together.
2. Slopes give a quick check: m(AB) = 1/5 and m(AC) = −5, product −1.

Answer: (b) y − 2 = 0

Explanation:
1. √3x − y = 0 has inclination 60° and √3x + y = 0 has inclination 120°; they make 60° at O and are symmetric about the y-axis.
2. For ΔOAB to be equilateral, AB must make 60° with both lines, so AB is horizontal.
3. Horizontal line through (2, 2): y − 2 = 0.

Teacher's Note:
1. The two given lines are mirror images in the y-axis, so the third side of the equilateral triangle must be perpendicular to the y-axis.
2. A horizontal line makes 60° with lines inclined at 60° and 120°.

Answer: (d) y(cos α + sin α) + x(cos α − sin α) = a

Explanation:
1. The vertices adjacent to O are A(a cos α, a sin α) and C(−a sin α, a cos α); the required diagonal is AC.
2. Test (d) at A: a sin α(cos α + sin α) + a cos α(cos α − sin α) = a(sin²α + cos²α) = a. ✓
3. At C: a cos α(cos α + sin α) − a sin α(cos α − sin α) = a(cos²α + sin²α) = a. ✓ So AC is (d).

Teacher's Note:
1. The second side through O is at angle α + 90°, giving the vertex (−a sin α, a cos α).
2. Checking both end points of the diagonal in an option is a quick way to identify the right equation.

Answer: (a) 2x + 3y = 9

Explanation:
1. Let C(h, k). Centroid = \( \left( \frac{2 - 2 + h}{3}, \frac{-3 + 1 + k}{3} \right) = \left( \frac{h}{3}, \frac{k - 2}{3} \right) \).
2. It lies on 2x + 3y = 1: \( \frac{2h}{3} + (k - 2) = 1 \), i.e. 2h + 3k = 9.
3. Locus of C: 2x + 3y = 9.

Teacher's Note:
1. Express the centroid in terms of the moving vertex and substitute in the given line.
2. Replace (h, k) by (x, y) only at the end.

Answer: (d) \( \frac{x}{2} - \frac{y}{3} = 1 \) and \( \frac{x}{-2} + \frac{y}{1} = 1 \)

Explanation:
1. Let intercepts be a and b = −1 − a; through (4, 3): \( \frac{4}{a} + \frac{3}{-1 - a} = 1 \).
2. 4(−1 − a) + 3a = a(−1 − a) gives a² = 4, so a = 2, b = −3 or a = −2, b = 1.
3. Lines: \( \frac{x}{2} - \frac{y}{3} = 1 \) and \( \frac{x}{-2} + \frac{y}{1} = 1 \).

Teacher's Note:
1. The intercept form always has 1 on the right side, which rules out (a) and (b).
2. Check with the point: 4/2 − 3/3 = 1 and 4/(−2) + 3 = 1.

Answer: (d) below the x-axis at a distance of (3/2) from it

Explanation:
1. Multiply the first equation by b and the second by a and subtract: 2b²y + 3b² + 2a²y + 3a² = 0.
2. (a² + b²)(2y + 3) = 0, and a² + b² ≠ 0, so y = −3/2.
3. The required line is y = −3/2, below the x-axis at a distance 3/2.

Teacher's Note:
1. For a line parallel to the x-axis, only the y-coordinate of the intersection point is needed.
2. Eliminate x to find y directly.

Answer: (b) \( \left( 1, \frac{7}{3} \right) \)

Explanation:
1. Other vertices: B = 2(−1, 2) − (1, 1) = (−3, 3) and C = 2(3, 2) − (1, 1) = (5, 3).
2. Centroid = \( \left( \frac{1 - 3 + 5}{3}, \frac{1 + 3 + 3}{3} \right) \).
3. = (1, 7/3).

Teacher's Note:
1. If M is the mid-point of AB, then B = 2M − A.
2. Then the centroid is the average of the three vertices.

Answer: (b) 4x + 3y = 24

Explanation:
1. If A bisects the intercept, the intercepts are 2 × 3 = 6 and 2 × 4 = 8.
2. Line: \( \frac{x}{6} + \frac{y}{8} = 1 \).
3. 4x + 3y = 24.

Teacher's Note:
1. When (x₁, y₁) bisects the intercept, the line is x/x₁ + y/y₁ = 2.
2. Check: 4(3) + 3(4) = 24.

Answer: (b) \( \left( \frac{1}{2}, 3 \right) \)

Explanation:
1. Both rays are in the first quadrant, so a > 0, and the point must lie above y = x/2 and below y = 3x.
2. \( \frac{a}{2} < a^2 < 3a \); dividing by a > 0 gives \( \frac{1}{2} < a < 3 \).
3. So a ∈ (1/2, 3).

Teacher's Note:
1. Inside the angle between two rays from the origin means between the two lines for the same x.
2. Dividing by a is allowed only because a is positive here.

Answer: (c) {−1, 3}

Explanation:
1. The right angle is at B. BC is horizontal (y = 1), so AB is vertical: h = 1.
2. Area = \( \frac{1}{2} \times BC \times AB = \frac{1}{2} \times 1 \times |k - 1| = 1 \).
3. |k − 1| = 2, so k = 3 or k = −1.

Teacher's Note:
1. The hypotenuse is opposite the right angle, so the right angle is at B.
2. A side perpendicular to a horizontal side is vertical.

Answer: (a) −4

Explanation:
1. Mid-point \( \left( \frac{1 + k}{2}, \frac{7}{2} \right) \); slope of PQ = \( \frac{-1}{k - 1} \), so the bisector has slope k − 1.
2. Bisector: \( y - \frac{7}{2} = (k - 1)\left( x - \frac{1 + k}{2} \right) \); at x = 0, \( y = \frac{7}{2} - \frac{k^2 - 1}{2} = \frac{8 - k^2}{2} \).
3. \( \frac{8 - k^2}{2} = -4 \) gives k² = 16, k = ±4; from the options k = −4.

Teacher's Note:
1. The perpendicular bisector passes through the mid-point and has slope −1/(slope of PQ).
2. Set x = 0 to read off the y-intercept.

Answer: (a) exactly one value of p

Explanation:
1. Lines perpendicular to a common line are parallel, so their slopes are equal.
2. Slopes: p(p² + 1) and \( -\frac{(p^2 + 1)^2}{p^2 + 1} = -(p^2 + 1) \).
3. p(p² + 1) = −(p² + 1) gives p = −1 (as p² + 1 ≠ 0): exactly one value.

Teacher's Note:
1. Two lines perpendicular to the same line are parallel to each other.
2. Cancel p² + 1 safely because it is never zero.

Answer: (d) \( \frac{23}{\sqrt{17}} \)

Explanation:
1. \( \frac{13}{5} + \frac{32}{b} = 1 \) gives b = −20, so L: 4x − y = 20 (slope 4).
2. K parallel: \( -\frac{3}{c} = 4 \), c = −3/4, so K: 4x − y + 3 = 0.
3. Distance = \( \frac{|-20 - 3|}{\sqrt{17}} = \frac{23}{\sqrt{17}} \).

Teacher's Note:
1. Write both lines with the same x and y coefficients before using the distance formula.
2. Slope of x/a + y/b = 1 is −b/a.

Answer: (c) \( (1, \infty) \)

Explanation:
1. Adding: (1 + a)x = 1 + |a|, so \( x = \frac{1 + |a|}{1 + a} \) and y = |a| − x.
2. For a ≥ 0: x = 1 and y = a − 1, which is positive only for a > 1.
3. For −1 < a < 0: \( y = -\frac{a^2 + 1}{1 + a} < 0 \), not allowed. Hence a ∈ (1, ∞).

Teacher's Note:
1. Treat a ≥ 0 and a < 0 separately to remove the modulus.
2. First quadrant needs both x > 0 and y > 0.

Answer: (c) 2x + 3y = 1

Explanation:
1. Let the centroid be (h, k); then the third vertex is C(3h, 3k + 2).
2. C lies on 2x + 3y = 9: 6h + 9k + 6 = 9, i.e. 2h + 3k = 1.
3. Locus: 2x + 3y = 1.

Teacher's Note:
1. Write the moving vertex in terms of the centroid: C = 3G − A − B.
2. Then substitute into the given line.

Answer: (c) 6

Explanation:
1. Point = \( \left( \frac{3 \times 2 + 2 \times 1}{5}, \frac{3 \times 4 + 2 \times 1}{5} \right) = \left( \frac{8}{5}, \frac{14}{5} \right) \).
2. k = 2 × 8/5 + 14/5 = 30/5.
3. k = 6.

Teacher's Note:
1. Apply the section formula with m : n = 3 : 2 measured from (1, 1).
2. Substitute the point into the line to get k.

Answer: (c) 2x + 9y + 7 = 0

Explanation:
1. S = mid-point of QR = (13/2, 1).
2. Slope of PS = \( \frac{1 - 2}{13/2 - 2} = -\frac{2}{9} \).
3. Line through (1, −1): \( y + 1 = -\frac{2}{9}(x - 1) \), i.e. 2x + 9y + 7 = 0.

Teacher's Note:
1. The median from P goes to the mid-point of the opposite side QR.
2. A parallel line only needs the slope of PS and the given point.

Answer: (d) 3bc − 2ad = 0

Explanation:
1. In the fourth quadrant and equidistant from the axes, the point is (t, −t).
2. First line: 4at − 2at + c = 0 gives \( t = -\frac{c}{2a} \); second: 5bt − 2bt + d = 0 gives \( t = -\frac{d}{3b} \).
3. \( \frac{c}{2a} = \frac{d}{3b} \), i.e. 3bc − 2ad = 0.

Teacher's Note:
1. Equidistant from both axes in the fourth quadrant means y = −x.
2. Equate the two expressions for t.

Answer: (d) 2x − 5y = 0

Explanation:
1. R is on the x-axis and on x − 2y = 2, so R = (2, 0).
2. PQ is horizontal through P, so Q is on y = 3 and on x − 2y = 2: Q = (8, 3).
3. Centroid = \( \left( \frac{5 + 8 + 2}{3}, \frac{3 + 3 + 0}{3} \right) = (5, 2) \), which satisfies 2x − 5y = 0.

Teacher's Note:
1. Use each given condition to fix one vertex at a time.
2. Then test the centroid in each option.

Answer: (b) \( \frac{4\sqrt{3}}{15} \)

Explanation:
1. Altitude = distance of (1, 2) from 3x + 4y − 9 = 0 = \( \frac{|3 + 8 - 9|}{5} = \frac{2}{5} \).
2. Side = \( \frac{2}{\sqrt{3}} \times \text{altitude} = \frac{4}{5\sqrt{3}} \).
3. = \( \frac{4\sqrt{3}}{15} \).

Teacher's Note:
1. The vertex opposite the base is at a perpendicular distance equal to the altitude.
2. Altitude = (√3/2) × side for an equilateral triangle.

Answer: (b) 3x + 2y = 18

Explanation:
1. Let the line meet the axes at P(a, 0) and Q(0, b). A is nearer to the x-axis, so PA : AQ = 1 : 2.
2. A = \( \left( \frac{2a}{3}, \frac{b}{3} \right) = (4, 3) \), so a = 6, b = 9.
3. Line: \( \frac{x}{6} + \frac{y}{9} = 1 \), i.e. 3x + 2y = 18.

Teacher's Note:
1. The point nearer to the x-axis is closer to the x-intercept end P.
2. Use the section formula with ratio 1 : 2 from P to Q.

Answer: (b) \( \frac{5}{\sqrt{13}} \)

Explanation:
1. L ⊥ 5x − y = 1: x + 5y = c, with intercepts c and c/5; area \( \frac{c^2}{10} = 5 \) gives c = ±5√2.
2. L is parallel to x + 5y = 0, so the distance is \( \frac{|c|}{\sqrt{1 + 25}} = \frac{5\sqrt{2}}{\sqrt{26}} \).
3. = \( \frac{5}{\sqrt{13}} \).

Teacher's Note:
1. L and x + 5y = 0 have the same coefficients, so use the parallel-lines distance formula.
2. Simplify √2/√26 = 1/√13.

Answer: (b) 780

Explanation:
1. Interior points satisfy x ≥ 1, y ≥ 1 and x + y < 41, i.e. x + y ≤ 40.
2. For x = 1, 2, …, 39 the number of values of y is 40 − x.
3. Total = 39 + 38 + … + 1 = \( \frac{39 \times 40}{2} = 780 \).

Teacher's Note:
1. Interior excludes points on the sides, so use strict inequalities x > 0, y > 0, x + y < 41.
2. Count column by column for each integer x.

Answer: (b) lie on a straight line

Explanation:
1. Slope between (0, 8/3) and (1, 3) = \( \frac{3 - 8/3}{1} = \frac{1}{3} \).
2. Slope between (1, 3) and (82, 30) = \( \frac{27}{81} = \frac{1}{3} \).
3. Equal slopes through the common point (1, 3), so the points are collinear.

Teacher's Note:
1. Check collinearity before classifying a triangle.
2. The line is y = x/3 + 8/3; both other points satisfy it.

Answer: (b) \( \left( \frac{1}{3}, -\frac{8}{3} \right) \)

Explanation:
1. The given sides meet at A(1, 2); the opposite vertex is C = 2(−1, −2) − (1, 2) = (−3, −6).
2. Let B(b, b + 1) on x − y + 1 = 0; then D = 2(−1, −2) − B = (−2 − b, −5 − b) must lie on 7x − y − 5 = 0.
3. −14 − 7b + 5 + b − 5 = 0 gives b = −7/3, so B(−7/3, −4/3) and D(1/3, −8/3).

Teacher's Note:
1. The diagonals of a rhombus bisect each other, so each vertex is reflected through the centre to get the opposite one.
2. The two given sides meet at a vertex of the rhombus.

Answer: (c) 4 : 1

Explanation:
1. The lines are 4x + 3y − 10 = 0 and 4x + 3y + 5/2 = 0, which are parallel and on opposite sides of O.
2. For any line through O, OA : OB equals the ratio of the perpendicular distances of O from them: 10 : 5/2.
3. OA : OB = 4 : 1.

Teacher's Note:
1. Make the coefficients of x and y the same in both equations first.
2. By similar triangles, distances along any line through O are in the ratio of perpendicular distances.

Answer: (b) \( 2\sqrt{3} - 2 \)

Explanation:
1. Vertices adjacent to O: A(2 cos 30°, 2 sin 30°) = (√3, 1) and C(−2 sin 30°, 2 cos 30°) = (−1, √3).
2. Fourth vertex B = A + C = (√3 − 1, 1 + √3).
3. Sum of x-coordinates = 0 + √3 − 1 + (√3 − 1) = 2√3 − 2.

Teacher's Note:
1. The second side through O is at 30° + 90° = 120°.
2. In a square OABC the fourth vertex is the sum of the two adjacent vertices (as position vectors).

Answer: (d) 9

Explanation:
1. B on x + y = 5: B(b, 5 − b); mid-point of AB on x = 4 gives b = 7, so B(7, −2).
2. C on x = 4: C(4, c); mid-point of AC, (5/2, (2 + c)/2), on x + y = 5 gives c = 3, so C(4, 3).
3. Area = \( \frac{1}{2} |1(-2 - 3) + 7(3 - 2) + 4(2 + 2)| = \frac{1}{2} \times 18 = 9 \).

Teacher's Note:
1. Each vertex lies on its own median, and the median from one vertex passes through the mid-point of the opposite side.
2. Use the area formula once all three vertices are known.

Answer: (d) 9 : 1

Explanation:
1. A(λ/3, 0), B(0, λ). OP is perpendicular to AB, so P = \( \left( \frac{3\lambda}{10}, \frac{\lambda}{10} \right) \) (on the line and on y = x/3).
2. P lies on AB between B (x = 0) and A (x = λ/3), so BP : PA equals the ratio of the changes in x: \( \frac{3\lambda}{10} : \left( \frac{\lambda}{3} - \frac{3\lambda}{10} \right) \).
3. = \( \frac{3\lambda}{10} : \frac{\lambda}{30} = 9 : 1 \).

Teacher's Note:
1. Points on a line divide it in the same ratio as their x-coordinates (or y-coordinates) do.
2. In a right triangle, the foot of the altitude divides the hypotenuse in the ratio of the squares of the adjacent legs: OB² : OA² = 9 : 1.

Answer: (a) collinear

Explanation:
1. (−a, −b) = −1·(a, b) and (a², ab) = a·(a, b), so each point is a multiple of (a, b).
2. All of them lie on the line through the origin with direction (a, b), i.e. bx − ay = 0.
3. Hence the four points are collinear.

Teacher's Note:
1. Points of the form t(a, b) all lie on the line through the origin and (a, b).
2. Check with the equation bx − ay = 0: for (a², ab) it gives a²b − a²b = 0.

Answer: (a) isosceles

Explanation:
1. Vertices: x + y = 0 with 3x + y = 4 gives (2, −2); x + y = 0 with x + 3y = 4 gives (−2, 2); 3x + y = 4 with x + 3y = 4 gives (1, 1).
2. Distances from (1, 1): to (2, −2) is √10, to (−2, 2) is √10; the third side is √32.
3. Two equal sides, and 10 + 10 ≠ 32, so the triangle is isosceles (not right angled).

Teacher's Note:
1. Find all three vertices, then compare the side lengths.
2. The figure is symmetric about y = x, which already suggests an isosceles triangle.

Answer: (d) a straight line passing through the point \( \left( \frac{\pi}{2}, -\sin^2 1 \right) \) and parallel to the x-axis.

Explanation:
1. \( \cos x \cos(x + 2) = \frac{1}{2}[\cos(2x + 2) + \cos 2] \) and \( \cos^2(x + 1) = \frac{1}{2}[1 + \cos(2x + 2)] \).
2. Difference = \( \frac{1}{2}(\cos 2 - 1) = -\sin^2 1 \), a constant.
3. So the graph is the horizontal line y = −sin²1, which passes through (π/2, −sin²1).

Teacher's Note:
1. Use 2 cos A cos B = cos(A + B) + cos(A − B) and cos²θ = (1 + cos 2θ)/2.
2. A constant function has a horizontal straight-line graph (slope 0).

Answer: (d) rhombus

Explanation:
1. Slopes of the diagonals: −1/3 and 3.
2. Product = −1, so the diagonals are perpendicular.
3. A parallelogram with perpendicular diagonals is a rhombus.

Teacher's Note:
1. Perpendicular diagonals alone do not make a square; equal diagonals are also needed.
2. Nothing is given about the diagonal lengths, so the safe answer is rhombus.

Answer: (c) a = 2, b = 3

Explanation:
1. Diagonals PR and QS bisect each other.
2. Mid-point of PR = (3, 9/2) = mid-point of QS = \( \left( \frac{4 + a}{2}, \frac{6 + b}{2} \right) \).
3. a = 2, b = 3.

Teacher's Note:
1. In parallelogram PQRS the diagonals are PR and QS.
2. Equivalently S = P + R − Q.

Answer: (c) 3

Explanation:
1. Take A₀(1, 0), A₁(1/2, √3/2), A₂(−1/2, √3/2), A₄(−1/2, −√3/2).
2. A₀A₁ = \( \sqrt{\frac{1}{4} + \frac{3}{4}} = 1 \); A₀A₂ = \( \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} \); A₀A₄ = √3.
3. Product = 1 × √3 × √3 = 3.

Teacher's Note:
1. Vertices of a regular hexagon in a unit circle are (cos 60°k, sin 60°k).
2. A side of a regular hexagon equals the radius of its circumcircle.

Answer: (a) 2

Explanation:
1. 3x + 4(mx + 1) = 9 gives \( x = \frac{5}{3 + 4m} \).
2. x is an integer when 3 + 4m ∈ {±1, ±5}, i.e. m = −1/2, −1, 1/2, −2.
3. Integer values: m = −1 and m = −2, so 2 values.

Teacher's Note:
1. For 5/(3 + 4m) to be an integer, 3 + 4m must be a divisor of 5.
2. Then keep only those m that are themselves integers.

Answer: (b) 3 : 4

Explanation:
1. Write the lines as 4x + 2y − 9 = 0 and 4x + 2y + 12 = 0; they lie on opposite sides of O.
2. For any line through O, OP : OQ = ratio of perpendicular distances of O = 9 : 12.
3. OP : OQ = 3 : 4.

Teacher's Note:
1. Use the same x, y coefficients for both lines before comparing constants.
2. Similar triangles make the ratio independent of the direction of the line through O.

Answer: (c) \( \left( 3, \frac{4}{3} \right) \)

Explanation:
1. The point inside a triangle that divides it into three triangles of equal area is the centroid.
2. Centroid = \( \left( \frac{0 + 3 + 6}{3}, \frac{0 + 4 + 0}{3} \right) \).
3. R = (3, 4/3).

Teacher's Note:
1. Each median splits a triangle into two equal areas, which is why the centroid gives three equal parts.
2. Check: area OQR = ½ × 6 × 4/3 = 4, one third of area OPQ = 12.

Answer: (b) \( y - \sqrt{3}x + 2 + 3\sqrt{3} = 0 \)

Explanation:
1. √3x + y = 1 has slope −√3, i.e. inclination 120°.
2. L has inclination 120° ± 60° = 180° or 60°; 180° gives a horizontal line, which does not meet the x-axis, so the slope is √3.
3. L: y + 2 = √3(x − 3), i.e. \( y - \sqrt{3}x + 2 + 3\sqrt{3} = 0 \).

Teacher's Note:
1. Use inclinations: the required line is at θ ± 60° where θ is the given line's inclination.
2. The condition 'intersects the x-axis' removes the horizontal line y = −2.

Answer: (a) a + b − c > 0; (c) a − b + c > 0

Explanation:
1. By symmetry the lines meet on y = x: \( x = y = -\frac{c}{a + b} \).
2. Distance from (1, 1) = \( \sqrt{2}\left( 1 + \frac{c}{a + b} \right) = \sqrt{2} \cdot \frac{a + b + c}{a + b} < 2\sqrt{2} \), so a + b + c < 2(a + b), i.e. a + b − c > 0.
3. Also a > b gives a − b + c > 0. So (a) and (c) are true.

Teacher's Note:
1. Interchanging x and y swaps the two lines, so their intersection lies on y = x.
2. Option (c) follows directly from the given order a > b, c > 0.

Answer: (a) 3x + 2y ≥ 0; (c) 2x − 3y − 12 ≤ 0

Explanation:
1. A linear expression takes its extreme values over a triangle at the vertices, so test the three vertices.
2. 3x + 2y: 9, 15, 1 (all ≥ 0) ✓; 2x − 3y − 12: −19, −2, −20 (all ≤ 0) ✓.
3. 2x + y − 13 is negative at (1, 3) and −2x + y is negative at (5, 0), so (b) and (d) fail.

Teacher's Note:
1. If all three vertices satisfy a linear inequality, every point of the triangle satisfies it.
2. One failing vertex is enough to reject an option.

Answer: 6

Explanation:
1. In the first quadrant, \( d_1 = \frac{|x - y|}{\sqrt{2}} \), \( d_2 = \frac{x + y}{\sqrt{2}} \). For x ≥ y, d₁ + d₂ = √2x; for y ≥ x, d₁ + d₂ = √2y.
2. So the region is where 2 ≤ √2 max(x, y) ≤ 4, i.e. √2 ≤ max(x, y) ≤ 2√2.
3. It is the square [0, 2√2]² minus the square [0, √2)²: area = 8 − 2 = 6.

Teacher's Note:
1. Remove the modulus by splitting the quadrant along y = x.
2. The condition max(x, y) ≤ k describes a square with corner at the origin.

Answer: 18

Explanation:
1. Let L: y − 2 = m(x − 8) with m < 0. Then OP = 8 − 2/m and OQ = 2 − 8m.
2. OP + OQ = 10 + (−8m) + \( \left( -\frac{2}{m} \right) \) ≥ 10 + 2\( \sqrt{(-8m)\left( -\frac{2}{m} \right)} \) = 10 + 8 = 18 (AM ≥ GM).
3. Equality when −8m = −2/m, i.e. m = −1/2; minimum OP + OQ = 18.

Teacher's Note:
1. With m negative, −8m and −2/m are both positive, so AM ≥ GM applies.
2. At m = −1/2 the intercepts are 12 and 6, and 12 + 6 = 18.