JEE Mathematics Pair of Straight Lines MCQs Set 05

Answer: (c) 3x² − 3y² + 8xy − 20x − 10y + 25 = 0

Explanation:
1. PQ and PR make 45° with QR, whose slope is −2. For a line of slope m through P: |(m + 2)/(1 − 2m)| = 1.
2. (m + 2) = (1 − 2m) gives m = −1/3, and (m + 2) = −(1 − 2m) gives m = 3.
3. The lines are y − 1 = 3(x − 2), i.e. 3x − y − 5 = 0, and y − 1 = −(x − 2)/3, i.e. x + 3y − 5 = 0. Product: (3x − y − 5)(x + 3y − 5) = 3x² + 8xy − 3y² − 20x − 10y + 25 = 0.

Teacher's Note:
1. The legs of a right isosceles triangle make 45° with the hypotenuse.
2. Multiply the two linear factors to get the pair.

Answer: (c) \( \left(\frac65,-\frac{12}5\right) \)

Explanation:
1. The expression factors as −(x − 2y − 6)(x + 3y + 6) = 0, so the lines are x − 2y − 6 = 0 and x + 3y + 6 = 0.
2. Subtracting: 5y + 12 = 0, so y = −12/5, and x = 2y + 6 = 6/5.

Teacher's Note:
1. Factorise 6y² − xy − x² as (2y − x)(3y + x) and fit the linear and constant terms.
2. Solve the two linear equations.

Answer: (d) \( \tan^{-1}\frac{83}{19} \)

Explanation:
1. Here a = 33, 2h = −71, b = −14, so h² − ab = 5041/4 + 462 = 6889/4 and √(h² − ab) = 83/2.
2. tanθ = 2√(h² − ab)/|a + b| = 83/19.

Teacher's Note:
1. Use tanθ = 2√(h² − ab)/|a + b| for the acute angle.
2. Check: the factors are (3x − 7y)(11x + 2y), with slopes 3/7 and −11/2.

Answer: (d) \( \tan^{-1}\frac1{13} \)

Explanation:
1. x² − 7xy + 12y² = (x − 3y)(x − 4y), so the lines are x = 3y and x = 4y with slopes 1/3 and 1/4.
2. tanθ = |(1/3 − 1/4)/(1 + 1/12)| = (1/12)/(13/12) = 1/13.

Teacher's Note:
1. Factorise the quadratic form into two lines first.
2. tanθ = |(m1 − m2)/(1 + m1m2)|.

Answer: (d) 2

Explanation:
1. Here a = 12, 2h = -10, b = 2, 2g = 11, 2f = -5, c = k.
2. The condition abc + 2fgh − af² − bg² − ch² = 0 reduces to 2 − k = 0.
3. So k = 2.
4. Check: with k = 2 the expression factorises as (3x − y + 2)(4x − 2y + 1).

Teacher's Note:
1. Use the determinant condition for a general second degree equation to represent a pair of lines.
2. The condition is linear or quadratic in k; check each root.

Answer: (a) −3

Explanation:
1. Here a = 2, 2h = 1, b = -1, 2g = k, 2f = 6, c = -9.
2. The condition abc + 2fgh − af² − bg² − ch² = 0 reduces to k²/4 + 3k/2 + 9/4 = 0.
3. So k = -3.
4. Check: with k = −3 the expression factorises as (x + y − 3)(2x − y + 3).

Teacher's Note:
1. Use the determinant condition for a general second degree equation to represent a pair of lines.
2. Factorise the quadratic part as (x + y)(2x − y) first.

Answer: (c) 6x − 5y + 14 = 0 and 6x + 5y − 56 = 0

Explanation:
1. x² − 7x + 6 = (x − 1)(x − 6), so the sides are x = 1 and x = 6. y² − 14y + 40 = (y − 4)(y − 10), so the sides are y = 4 and y = 10.
2. The vertices are (1, 4), (6, 4), (6, 10) and (1, 10).
3. Diagonal through (1, 4) and (6, 10): slope 6/5, so 6x − 5y + 14 = 0. Diagonal through (1, 10) and (6, 4): slope −6/5, so 6x + 5y − 56 = 0.

Teacher's Note:
1. Factorise each quadratic to find the four lines.
2. Join opposite vertices to get the diagonals.

Answer: (b) 8h² = 9ab

Explanation:
1. Let A = (t, 0). The line x = t meets the pair where by² + 2hty + at² = 0, giving the y-coordinates y1 and y2 of B and C with y1 + y2 = −2ht/b and y1y2 = at²/b.
2. AB = BC with A at y = 0 means y2 = 2y1. Then 3y1 = −2ht/b and 2y1² = at²/b.
3. y1 = −2ht/(3b), so 2(4h²t²)/(9b²) = at²/b, giving 8h² = 9ab.

Teacher's Note:
1. B is the midpoint of A and C, so the ordinate of C is twice that of B.
2. Use the sum and product of the roots of the quadratic in y.

Answer: (d) 41x − 38y = 0

Explanation:
1. The bisector 3x − 4y = 0 (slope 3/4) and 4x + 3y = 0 come from 12x² − 7xy − 12y² = (3x − 4y)(4x + 3y).
2. The given line x = 2y has slope 1/2. The other line is its reflection in a bisector, so with the bisector of slope 3/4, the reflected slope m satisfies (3/4 − 1/2)/(1 + 3/8) = (m − 3/4)/(1 + 3m/4).
3. The left side is (1/4)/(11/8) = 2/11. Solving gives m = 41/38, so the other line is 41x − 38y = 0.

Teacher's Note:
1. The second line is the mirror image of the first in a bisector, so it makes the same angle with the bisector.
2. Check: 41x − 38y = 0 makes equal angles with the bisector 3x − 4y = 0 as x = 2y does.

Answer: (a) −12

Explanation:
1. Here a = 6, 2h = 1, b = k, 2g = -11, 2f = 43, c = -35.
2. The condition abc + 2fgh − af² − bg² − ch² = 0 reduces to −961k/4 − 2883 = 0.
3. So k = -12.
4. Check: with k = −12 the expression factorises as (2x + 3y − 7)(3x − 4y + 5).

Teacher's Note:
1. Use the determinant condition for a general second degree equation to represent a pair of lines.
2. Here k is the coefficient of y², so it appears in b.

Answer: (d) \( \tan^{-1}\frac35 \)

Explanation:
1. The equation factors as (x − 4y + 2)(x − y − 1) = 0.
2. Slopes: 1/4 and 1. tanθ = |(1 − 1/4)/(1 + 1/4)| = (3/4)/(5/4) = 3/5.

Teacher's Note:
1. The leading terms x² − 5xy + 4y² = (x − 4y)(x − y); add constants whose product is −2 and which give the linear terms x + 2y.
2. Use tanθ = |(m1 − m2)/(1 + m1m2)|.

Answer: (c) \( \frac{\pi}{2} \)

Explanation:
1. From 3x − y = 2, (3x − y)/2 = 1. Homogenise: 7x² + 8y² − 4xy + (2x − 4y)(3x − y)/2 − 8((3x − y)/2)² = 0.
2. (2x − 4y)(3x − y)/2 = 3x² − 7xy + 2y² and 8((3x − y)/2)² = 18x² − 12xy + 2y².
3. Sum: (7 + 3 − 18)x² + (−4 − 7 + 12)xy + (8 + 2 − 2)y² = −8x² + xy + 8y² = 0. The coefficients of x² and y² add to zero, so the lines are perpendicular: angle π/2.

Teacher's Note:
1. Homogenise the curve using the line in the form (3x − y)/2 = 1.
2. a + b = 0 means the pair of lines is perpendicular.

Answer: (b) \( \frac{3\sqrt{10}}{5} \)

Explanation:
1. x² + 6xy + 9y² = (x + 3y)², so the equation is (x + 3y)² + 4(x + 3y) − 5 = 0, i.e. (x + 3y + 5)(x + 3y − 1) = 0.
2. The lines x + 3y + 5 = 0 and x + 3y − 1 = 0 are parallel.
3. Distance = |5 − (−1)|/√(1 + 9) = 6/√10 = 3√10/5.

Teacher's Note:
1. Treat x + 3y as a single variable and factorise the quadratic.
2. Distance between parallel lines = |c1 − c2|/√(a² + b²).

Answer: (d) 12x² − 7xy − 12y² = 0

Explanation:
1. Here a = 4, 2h = −24 so h = −12, and b = 11.
2. Bisectors: (x² − y²)/(a − b) = xy/h, that is (x² − y²)/(−7) = xy/(−12).
3. 12(x² − y²) = 7xy, so 12x² − 7xy − 12y² = 0.

Teacher's Note:
1. The pair of bisectors of ax² + 2hxy + by² = 0 is (x² − y²)/(a − b) = xy/h.
2. Cross-multiply and simplify.

Answer: (b) −15

Explanation:
1. Here a = 6, 2h = 11, b = -10, 2g = 1, 2f = 31, c = k.
2. The condition abc + 2fgh − af² − bg² − ch² = 0 reduces to −361k/4 − 5415/4 = 0.
3. So k = -15.
4. Check: with k = −15 the expression factorises as (2x + 5y − 3)(3x − 2y + 5).

Teacher's Note:
1. Use the determinant condition for a general second degree equation to represent a pair of lines.
2. Substitute the value back and factorise to confirm.

Answer: (a) m1m2 = 3/8, θ = sin⁻¹(2/(5√5)) and the sum of the coordinates of P is −1

Explanation:
1. The expression factors as (3x + 4y + 5)(x + 2y + 3) = 0 (check: 3x² + 10xy + 8y² + 14x + 22y + 15).
2. Slopes: m1 = −3/4 and m2 = −1/2, so m1 + m2 = −5/4 and m1m2 = 3/8. tanθ = |(−3/4 + 1/2)/(1 + 3/8)| = 2/11, so sinθ = 2/√125 = 2/(5√5).
3. Intersection: 3x + 4y = −5 and x + 2y = −3 give P = (1, −2), with coordinate sum −1. So m1m2 = 3/8, θ = sin⁻¹(2/(5√5)) and the sum is −1 all hold.

Teacher's Note:
1. Factorise the second-degree expression into two linear factors.
2. tanθ = |(m1 − m2)/(1 + m1m2)|.

Answer: (b) 14

Explanation:
1. Here a = 12, 2h = 1, b = -6, 2g = -29, 2f = 8, c = k.
2. The condition abc + 2fgh − af² − bg² − ch² = 0 reduces to 2023/2 − 289k/4 = 0.
3. So k = 14.
4. Check: with k = 14 the expression factorises as (3x − 2y − 2)(4x + 3y − 7).

Teacher's Note:
1. Use the determinant condition for a general second degree equation to represent a pair of lines.
2. Compare the product of the constants of the two factors with k.

Answer: (a) \( 90^\circ \)

Explanation:
1. The expression factors as −(x + 3y + 9)(3x − y + 2) = 0.
2. The slopes are −1/3 and 3, whose product is −1, so the lines are perpendicular.
3. Alternatively a + b = −3 + 3 = 0 gives a right angle.

Teacher's Note:
1. For ax² + 2hxy + by² the lines are perpendicular when a + b = 0.
2. The quadratic part alone decides the angle.

Answer: (d) \( (1,-2)\ \text{and}\ \frac15 \)

Explanation:
1. The equation factors as (x + y + 1)(2x + 3y + 4) = 0 (expand to check).
2. Intersection: x + y = −1 and 2x + 3y = −4 give y = −2, x = 1: the point (1, −2).
3. Slopes −1 and −2/3: tanθ = |(−1 + 2/3)/(1 + 2/3)| = (1/3)/(5/3) = 1/5.

Teacher's Note:
1. Treat the equation as a quadratic in x, or guess the factorisation from the leading terms 2x² + 5xy + 3y² = (x + y)(2x + 3y).
2. The constants must multiply to 4 and the linear terms must give 6x + 7y.