JEE Mathematics Straight Lines MCQs Set 09

Answer: (c) 1

Explanation:
1. Concurrency requires \( \begin{vmatrix}1&a&a\\b&1&b\\c&c&1\end{vmatrix}=0 \), i.e. \( 1-ab-bc-ca+2abc=0 \).
2. The sum equals \( \frac{3abc-2(ab+bc+ca)+(a+b+c)}{(a-1)(b-1)(c-1)} \), and the denominator expands to \( abc-(ab+bc+ca)+(a+b+c)-1 \).
3. Numerator − denominator \( =2abc-(ab+bc+ca)+1=0 \), so the sum is 1.

Teacher's Note:
1. Write the concurrency determinant.
2. Compare the numerator and denominator of the combined fraction.

Answer: (b) y = 2x

Explanation:
1. The distance from (3, 1) to a line through (1, 2) is greatest when the line is perpendicular to the segment joining (3, 1) and (1, 2).
2. That segment has slope \( \frac{2-1}{1-3}=-\frac12 \), so the line has slope 2.
3. Through (1, 2): \( y-2=2(x-1) \), i.e. y = 2x.

Teacher's Note:
1. The distance cannot exceed the distance between the two points.
2. Equality holds when the line is perpendicular to the joining segment.

Answer: (a) (0, −1)

Explanation:
1. Taking each point in turn as the opposite vertex, the fourth vertex is \( P+Q-R \).
2. The possibilities are \( (-6+3+2,\,-4+5-1)=(-1,0) \), \( (-6-2-3,\,-4+1-5)=(-11,-8) \) and \( (3-2+6,\,5+1+4)=(7,10) \).
3. (0, −1) is not among them.

Teacher's Note:
1. Diagonals of a parallelogram bisect each other.
2. Try each vertex as the one opposite the missing vertex.

Answer: (b) \( \sqrt{10} \)

Explanation:
1. The mid-point of AC is \( \left(\frac{-1+5}2,\frac{3+1}2\right)=(2,2) \).
2. The median is the distance from B(1, −1) to (2, 2): \( \sqrt{1^2+3^2}=\sqrt{10} \).

Teacher's Note:
1. Find the mid-point of AC.
2. Use the distance formula.

Answer: (b) (1, 1)

Explanation:
1. The triangle has vertices (0, 0), (4, 0), (0, 3): a right triangle with sides 3, 4, 5.
2. Its inradius is \( r=\frac{3+4-5}2=1 \).
3. The incentre is at distance 1 from both axes: (1, 1).

Teacher's Note:
1. Inradius of a right triangle = (sum of legs − hypotenuse)/2.
2. The incentre is r from each side.

Answer: (a) \( \left(\frac12,\frac13\right) \)

Explanation:
1. Divide the condition by 6: \( \frac a2+\frac b3+c=0 \).
2. Compare with \( ax+by+c=0 \): it is satisfied by \( x=\frac12,\ y=\frac13 \) for all such a, b, c.
3. The fixed point is \( \left(\frac12,\frac13\right) \).

Teacher's Note:
1. Write the condition in the form a·x₀ + b·y₀ + c = 0.
2. Divide by the coefficient of c.

Answer: (b) x + y − 3 = 0

Explanation:
1. Adding the equations: 4x − 4 = 0, so x = 1 and y = 2: the point is (1, 2).
2. A line perpendicular to \( x-y+1=0 \) has the form \( x+y+k=0 \); through (1, 2): k = −3, giving \( x+y-3=0 \).
3. The perpendicular to the other line, x − 3y + 5 = 0, is not among the options.

Teacher's Note:
1. Find the point of intersection first.
2. Perpendicular to x − y + 1 = 0 means x + y + k = 0.

Answer: (c) p = 2

Explanation:
1. Equal slopes: \( \frac{3-1}{(2p+1)-(p+1)}=\frac{2p-3}{(2p+2)-(2p+1)} \).
2. So \( \frac2p=2p-3 \), i.e. \( 2p^2-3p-2=0 \) and \( (p-2)(2p+1)=0 \).
3. The roots are p = 2 and p = −1/2; of the options, p = 2 works.

Teacher's Note:
1. Collinear points have equal slopes between consecutive pairs.
2. Solve the resulting quadratic and compare with the options.

Answer: (c) 5522

Explanation:
1. The triangle with vertices O, \( \left(\frac c{m_1},c\right) \) and \( \left(\frac c{m_2},c\right) \) has area \( \frac{c^2|m_1-m_2|}{2|m_1m_2|} \).
2. Here \( (m_1-m_2)^2=(\sqrt3+2)^2-4(\sqrt3-1)=11 \) and \( m_1m_2=\sqrt3-1 \). So the area \( =\frac{\sqrt{11}}{2(\sqrt3-1)}c^2=\frac{\sqrt{11}(\sqrt3+1)}{4}c^2=\left(\frac{\sqrt{33}}4+\frac{\sqrt{11}}4\right)c^2 \).
3. Thus \( a^2+b^2=\frac{33}{16}+\frac{11}{16}=\frac{11}4 \) and \( 2008\cdot\frac{11}4=5522 \).

Teacher's Note:
1. Area of the triangle from y = c and two lines through the origin.
2. Use (m₁ − m₂)² = (m₁ + m₂)² − 4m₁m₂.

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

Explanation:
1. The lines y = −2 and y = mx meet at \( x=-\frac2m \).
2. This point lies on \( x=a+m \): \( -\frac2m=a+m \), so \( a=-\left(m+\frac2m\right) \).
3. For real m (m ≠ 0), \( \left|m+\frac2m\right|\ge2\sqrt2 \) by AM-GM, so the least |a| is \( 2\sqrt2 \).

Teacher's Note:
1. Find the intersection of the last two lines.
2. |m + 2/m| ≥ 2√2.

Answer: (c) \( \sqrt3x+y=0 \)

Explanation:
1. QP points along the negative x-axis (angle 180°) and QR makes the angle \( \tan^{-1}\sqrt3=60^\circ \) with the x-axis, so \( \angle PQR=120^\circ \).
2. The bisector makes the angle \( 60^\circ+60^\circ=120^\circ \) with the positive x-axis, i.e. slope \( \tan120^\circ=-\sqrt3 \).
3. Through the origin: \( y=-\sqrt3x \), i.e. \( \sqrt3x+y=0 \).

Teacher's Note:
1. Find the directions of QP and QR.
2. The bisector direction is halfway between them.

Answer: (a) an equilateral triangle

Explanation:
1. For vertices with integer coordinates, the area \( \frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| \) is rational.
2. An equilateral triangle of side s has area \( \frac{\sqrt3}4s^2 \), where \( s^2 \) is an integer (distance formula), so the area would be irrational.
3. Hence no equilateral triangle can have all vertices with integral coordinates. Right-angled and isosceles ones can (e.g. (0, 0), (1, 0), (0, 1)).

Teacher's Note:
1. Compare the rationality of the area computed in two ways.
2. Give examples for the other two types.

Answer: (a) \( a_1a_2+b_1b_2 \)

Explanation:
1. By the cosine rule in triangle OPQ: \( PQ^2=OP^2+OQ^2-2\,OP\cdot OQ\cos\angle POQ \).
2. So \( OP\cdot OQ\cos\angle POQ=\frac{(a_1^2+b_1^2)+(a_2^2+b_2^2)-[(a_1-a_2)^2+(b_1-b_2)^2]}{2}=a_1a_2+b_1b_2 \).

Teacher's Note:
1. Use the cosine rule.
2. Expand PQ² with the distance formula.

Answer: (b) A, B and C are collinear points

Explanation:
1. With common ratio r: \( A=(x_1,y_1),\ B=(rx_1,ry_1),\ C=(r^2x_1,r^2y_1) \).
2. Each point is a multiple of A: they lie on the line through the origin and A (the determinant of the three points is zero), so they are collinear.

Teacher's Note:
1. Write the points in terms of (x₁, y₁) and r.
2. Check the area of the triangle.

Answer: (a) x + 2y − 3 = 0

Explanation:
1. A parallel line is \( x+2y+k=0 \), and distances are proportional to the differences in the constants (−5 and 9).
2. The line nearer to L₁, dividing in the ratio 1 : 6: \( k=-5+\frac17(9-(-5))=-5+2=-3 \).
3. So the line is x + 2y − 3 = 0.

Teacher's Note:
1. Distances between parallel lines are proportional to the difference of constants.
2. Dividing 1 : 6 internally from L₁ means one-seventh of the way.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The first two lines meet at (b, b) (subtracting them gives x = y). They are mirror images of each other in y = x, so y = x is the bisector of the angle between them at this vertex.
2. The line y = x is perpendicular to the third side x + y = 0, so by R the triangle is isosceles (R is true: the bisector then also bisects the base, giving two congruent triangles).
3. Thus A is true and R explains it.

Teacher's Note:
1. Find the bisector of the angle between the first two lines.
2. Compare it with the third side.