JEE Mathematics Pair of Straight Lines MCQs Set 03

Answer: (a) m \in R

Explanation:
1. For \( ax^2+2hxy+by^2=0 \), the lines are real and distinct when \( h^2-ab\gt0 \).
2. Here \( \tfrac{m^2}{4}+12\gt0 \) for every real m.

Teacher's Note:
1. Check the sign of \( h^2-ab \).
2. Since a and b have opposite signs, ab is negative.

Answer: (c) -2

Explanation:
1. Perpendicular lines need a + b = 0.
2. So 2 + m = 0, m = -2; the lines are then real.

Teacher's Note:
1. Use the coefficient condition a + b = 0.
2. Verify with the discriminant.

Answer: (d) none of these

Explanation:
1. \( x^3+y^3=(x+y)(x^2-xy+y^2) \).
2. The second factor is zero only at the origin, so the equation is a line together with a single point.

Teacher's Note:
1. Factorise the sum of cubes.
2. Check whether the quadratic factor has real lines.

Answer: (c) \( \pi/2-\theta \)

Explanation:
1. a = b = 1 and \( h=-\csc\theta \).
2. \( \tan\phi=\dfrac{2\sqrt{h^2-ab}}{a+b}=\sqrt{\csc^2\theta-1}=\cot\theta \), so \( \phi=\pi/2-\theta \).

Teacher's Note:
1. Use the angle formula for a pair of lines.
2. \( \csc^2\theta-1=\cot^2\theta \).

Answer: (a) 9 : 8

Explanation:
1. Coefficients are 1/a, 1/h, 1/b. Slopes m and 2m satisfy \( 3m=-2H/B \) and \( 2m^2=A/B \).
2. Eliminating m gives \( 8H^2=9AB \), so \( \dfrac8{h^2}=\dfrac9{ab} \) and \( ab:h^2=9:8 \).

Teacher's Note:
1. Use sum and product of slopes.
2. Substitute the reciprocal coefficients.

Answer: (b) right-angled

Explanation:
1. For \( y^2-4xy-x^2 \), a + b = -1 + 1 = 0, so its two lines are perpendicular.
2. The third line x + y = 1 is not parallel to a bisector of that pair, so the triangle is right-angled but not isosceles.

Teacher's Note:
1. Test a + b = 0 for the pair.
2. Compare the third line with the bisectors \( y=\pm x \)-type lines.

Answer: (d) none of these

Explanation:
1. \( y(y-2x)(y+2x)=0 \): the three lines y = 0, y = 2x and y = -2x all pass through the origin.
2. Concurrent lines do not enclose a triangle.

Teacher's Note:
1. Factorise completely.
2. Look for a common point.

Answer: (b) \( \tfrac43 \) unit\(^2\)

Explanation:
1. y = x meets the line at (2/3, 2/3); y = -x meets it at (-2, 2).
2. Area = \( \tfrac12\left|\tfrac23\cdot2-\tfrac23\cdot(-2)\right|=\tfrac12\cdot\tfrac83=\tfrac43 \).

Teacher's Note:
1. Find the two vertices besides the origin.
2. Use the determinant area formula.

Answer: (c) \( (-\tfrac73,\tfrac23) \)

Explanation:
1. The lines \( y=m_1x \), \( y=m_2x \) with \( 2m^2+7m+1=0 \) meet y = 1 at \( (1/m_1,1) \) and \( (1/m_2,1) \); the third vertex is the origin.
2. \( \tfrac1{m_1}+\tfrac1{m_2}=\dfrac{m_1+m_2}{m_1m_2}=\dfrac{-7/2}{1/2}=-7 \).
3. Centroid = \( (-7/3,\,2/3) \).

Teacher's Note:
1. Find the three vertices.
2. Use Vieta's relations for the slopes.

Answer: (a) (-1, 0)

Explanation:
1. The pair factorises as \( (2x+y-1)(x-y+1)=0 \).
2. Vertices: (0, 1), (2, -3) and (-1, 0).
3. The altitude from (0, 1) is perpendicular to x + y + 1 = 0; the altitude from (-1, 0) is perpendicular to 2x + y - 1 = 0. They meet at (-1, 0), so the triangle has a right angle there.

Teacher's Note:
1. Factorise the pair first.
2. The right-angle vertex is the orthocentre.

Answer: (c) \( 2x^2-xy-y^2-4x+y+2=0 \)

Explanation:
1. Shift the origin to (1, 0): replace x by x - 1.
2. \( 2(x-1)^2-(x-1)y-y^2=2x^2-4x+2-xy+y-y^2 \).

Teacher's Note:
1. Parallel lines have the same homogeneous part.
2. Translate the equation.

Answer: (c) \( \lambda=\mu \)

Explanation:
1. Parallel lines need \( h^2=ab \): \( \tfrac{(\lambda+\mu)^2}4=\lambda\mu \).
2. So \( (\lambda-\mu)^2=0 \) and \( \lambda=\mu \).

Teacher's Note:
1. Use the parallel-lines condition.
2. Expand and factorise.

Answer: (b) 3a + 2h = 0

Explanation:
1. Sum of slopes = \( -2h/b \); product = \( a/b \).
2. \( -2h/b=3a/b \) gives \( 3a+2h=0 \).

Teacher's Note:
1. Use the quadratic in m.
2. Cancel b.

Answer: (a) \( \sqrt3x^2-xy=0 \)

Explanation:
1. Slopes satisfy \( \sqrt3m^2-4m+\sqrt3=0 \): \( m=\sqrt3,\ 1/\sqrt3 \), i.e. angles 60\(^\circ\) and 30\(^\circ\).
2. After rotating by 30\(^\circ\) the angles are 90\(^\circ\) and 60\(^\circ\): lines x = 0 and \( y=\sqrt3x \).
3. Combined: \( x(\sqrt3x-y)=0 \), i.e. \( \sqrt3x^2-xy=0 \).

Teacher's Note:
1. Find the inclinations of the lines.
2. Add 30\(^\circ\) to each.

Answer: (c) y = |x - 2|

Explanation:
1. Reflection in x = 1 replaces x by 2 - x.
2. So \( y=|2-x|=|x-2| \).

Teacher's Note:
1. Reflect the point (x, y) to (2 - x, y).
2. Substitute in the equation.

Answer: (b) \( \sqrt3x^2-xy-2\sqrt3x+y+\sqrt3=0 \)

Explanation:
1. The equation is \( (x-1)^2-y^2=0 \): lines y = x - 1 (45\(^\circ\)) and y = -(x - 1) (135\(^\circ\)).
2. The 135\(^\circ\) line turned 45\(^\circ\) clockwise becomes 90\(^\circ\): x = 1. The 45\(^\circ\) line turned 15\(^\circ\) anticlockwise becomes 60\(^\circ\): \( y=\sqrt3(x-1) \).
3. Product: \( (x-1)(\sqrt3x-\sqrt3-y)=0 \), giving \( \sqrt3x^2-xy-2\sqrt3x+y+\sqrt3=0 \).

Teacher's Note:
1. Find the original inclinations.
2. Rotate each line about (1, 0).

Answer: (b) three lines, two of which coincide and are perpendicular to the third

Explanation:
1. \( x^3+x^2y-xy^2-y^3=(x+y)(x^2-y^2)=(x-y)(x+y)^2 \).
2. So the lines are x - y = 0 and the repeated line x + y = 0, which are perpendicular.

Teacher's Note:
1. Group the terms in pairs.
2. Factorise completely.

Answer: (a) (1, -2)

Explanation:
1. \( 2x^2-3xy-2y^2=(2x+y)(x-2y) \); the lines are perpendicular and (2, 1) lies on x - 2y = 0.
2. The other diagonal is 2x + y = 0; the centre is the origin, so the adjacent vertices are at distance \( \sqrt5 \) on it: (1, -2) and (-1, 2).
3. Of the options, (1, -2) qualifies.

Teacher's Note:
1. Factorise the pair of diagonals.
2. Adjacent vertices lie on the other diagonal.

Answer: (a) one pair bisects the angles between the other pair

Explanation:
1. The bisectors of \( ax^2+2hxy+by^2=0 \) are given by \( \dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h} \).
2. With b = -a: \( \dfrac{x^2-y^2}{2a}=\dfrac{xy}h \), i.e. \( hx^2-2axy-hy^2=0 \), the second pair.

Teacher's Note:
1. Write the bisector equation of the first pair.
2. Compare with the second equation.

Answer: (c) \( \pm2 \)

Explanation:
1. Equally inclined pairs have the same bisectors, so \( \dfrac{a-b}{h} \) is equal for both.
2. \( \dfrac{1-2}{\lambda}=\dfrac{\lambda}{-4} \) gives \( \lambda^2=4 \).

Teacher's Note:
1. Equate \( (a-b)/h \).
2. Solve for \( \lambda \).