JEE Mathematics Pair of Straight Lines MCQs Set 04

Answer: (b) 60°

Explanation:
1. The lines are y = ±√3 x, with inclinations 60° and 120°.
2. The angle between them is 60°.

Teacher's Note:
1. Factorise as (y - √3 x)(y + √3 x).
2. The acute angle between lines at 60° and 120° is 60°.

Answer: (c) 90°

Explanation:
1. xy = 0 represents the two coordinate axes, which are perpendicular.

Teacher's Note:
1. A homogeneous equation with no x² or y² term gives the coordinate axes.
2. a + b = 0 confirms perpendicularity.

Answer: (d) 0° (the lines coincide)

Explanation:
1. x² + 4xy + 4y² = (x + 2y)², so the two lines coincide.
2. The angle is 0° (h² = ab: 4 = 4).

Teacher's Note:
1. A perfect-square form means coincident lines.
2. Check h² = ab.

Answer: (a) (x - y + 1)(2x + y - 3) = 0

Explanation:
1. Check abc + 2fgh - af² - bg² - ch² = 0 holds, so it factorises.
2. Expanding (x - y + 1)(2x + y - 3) = 2x² - xy - y² - x + 4y - 3, which matches.

Teacher's Note:
1. Factorise the quadratic part first: 2x² - xy - y² = (2x + y)(x - y).
2. Then fix the constants by matching the linear terms.

Answer: (d) a² + ac + bd + d² = 0

Explanation:
1. Slopes m satisfy d m³ + c m² + b m + a = 0 with roots m1, m2, m3 and m1 m2 = -1.
2. Then m3 = -a/(d m1 m2) = a/d. Sum of roots: m1 + m2 + a/d = -c/d and pair sum: -1 + (a/d)(m1 + m2) = b/d.
3. Eliminating m1 + m2 = -(c + a)/d gives -1 - a(a + c)/d² = b/d, i.e. a² + ac + bd + d² = 0.

Teacher's Note:
1. Use Vieta's relations for the cubic in the slope.
2. Substitute m1 m2 = -1 to eliminate the roots.

Answer: (a) 9/5 sq. units

Explanation:
1. 3x² - 8xy - 3y² = (3x + y)(x - 3y), lines y = -3x and y = x/3, which are perpendicular.
2. They meet x + 2y = 3 at (-3/5, 9/5) and (9/5, 3/5); both are at distance sqrt(90)/5 from the origin.
3. Area = ½ × (90/25) = 9/5.

Teacher's Note:
1. Factorise the pair of lines, then find where each meets the third line.
2. Equal legs confirm the triangle is isosceles.

Answer: (c) \( 2\sqrt2 \)

Explanation:
1. x² + 2xy - 3y² = (x - y)(x + 3y). The point (2, 2) lies on x - y = 0.
2. The normal there is perpendicular to x - y = 0 through (2, 2): x + y = 4.
3. Distance from the origin = 4/√2 = 2√2.

Teacher's Note:
1. Factorise the curve into lines and see which line contains the point.
2. The normal to a line is the perpendicular line through the point.

Answer: (d) \( \left(\tfrac83,\tfrac83\right) \)

Explanation:
1. 12x² - 20xy + 7y² = (2x - y)(6x - 7y): lines y = 2x and y = 6x/7.
2. These meet 2x - 3y + 4 = 0 at (1, 2) and (7, 6); the third vertex is the origin.
3. Centroid = ((0 + 1 + 7)/3, (0 + 2 + 6)/3) = (8/3, 8/3).

Teacher's Note:
1. Factorise the pair of lines through the origin first.
2. The origin is the third vertex.

Answer: (a) \( \sqrt5 \)

Explanation:
1. (2x + y)² - 3(2x + y) - 4 = (2x + y - 4)(2x + y + 1).
2. Distance between 2x + y - 4 = 0 and 2x + y + 1 = 0 is 5/√5 = √5.

Teacher's Note:
1. Treat 2x + y as one variable and factorise the quadratic.
2. Then use the distance between parallel lines.

Answer: (c) \( a=5,\ c=-\tfrac{61}{34} \)

Explanation:
1. Perpendicular lines: a + b = 0, so a = 5 (since b = -5).
2. The pair-of-lines condition abc + 2fgh - af² - bg² - ch² = 0 with h = 3, g = -2, f = 3 gives 5(-5)c + 2(3)(-2)(3) - 5(9) + 5(4) - 9c = 0.
3. -34c - 61 = 0, so c = -61/34.

Teacher's Note:
1. Use a + b = 0 for perpendicular lines and the determinant condition for a pair of lines.
2. Take care with the signs of g and f.

Answer: (d) x² - y² = 0

Explanation:
1. Here a = b = 0 and h = 1/2, so (x² - y²)/(a - b) = xy/h becomes (x² - y²)/0 = 2xy, which gives x² - y² = 0.
2. The bisectors of the coordinate axes are y = x and y = -x.

Teacher's Note:
1. The bisectors of the coordinate axes are y = ±x.
2. Their combined equation is x² - y² = 0.

Answer: (a) xy = 0

Explanation:
1. The lines y = x and y = -x have the coordinate axes as bisectors.
2. Using (x² - y²)/(a - b) = xy/h with h = 0 gives xy = 0.

Teacher's Note:
1. The bisectors of y = ±x are the axes.
2. Use the general bisector formula with h = 0.

Answer: (b) 11x² - 14xy + 3y² = 0

Explanation:
1. Homogenise using (2x - y)/3 = 1: x² - y² - xy + (3x - 6y)(2x - y)/3 + 18((2x - y)/3)² = 0.
2. This gives x² - y² - xy + (2x² - 5xy + 2y²) + (8x² - 8xy + 2y²) = 0.
3. So 11x² - 14xy + 3y² = 0.

Teacher's Note:
1. To homogenise, make every term degree 2 using the line equation written as 1.
2. Multiply the linear terms by (line/constant) and the constant by (line/constant)².

Answer: (d) \( \tfrac12 \)

Explanation:
1. Homogenise with x + y = 1: x² + y² + (x - 2y)(x + y) - m(x + y)² = 0.
2. Coefficient of x²: 1 + 1 - m = 2 - m. Coefficient of y²: 1 - 2 - m = -1 - m.
3. Right angle: sum of these is zero, so 1 - 2m = 0 and m = 1/2.

Teacher's Note:
1. Use a + b = 0 on the homogenised equation.
2. Expand carefully when homogenising.

Answer: (d) \( \tfrac{11}{3} \)

Explanation:
1. For ax² + 2hxy + by² = 0, m1 m2 = a/b.
2. Here a = 11 and b = 3, so m1 m2 = 11/3.

Teacher's Note:
1. Slopes m satisfy b m² + 2h m + a = 0.
2. Product of roots = a/b.