JEE Mathematics Straight Lines MCQs Set 27

Answer: (a) \( \frac65,\ \frac12,\ -3 \)

Explanation:
1. No triangle if the lines are concurrent or two of them are parallel.
2. Concurrent: \( \begin{vmatrix}1&2&3\\3&-1&1\\\lambda&1&2\end{vmatrix}=0\Rightarrow5\lambda-6=0\Rightarrow\lambda=\frac65 \).
3. Parallel to \( x+2y=3 \) (slope \( -\frac12 \)): \( -\lambda=-\frac12\Rightarrow\lambda=\frac12 \). Parallel to \( 3x-y=1 \) (slope 3): \( -\lambda=3\Rightarrow\lambda=-3 \).

Teacher's Note:
1. Check concurrency and parallelism separately.
2. The first two lines are not parallel to each other.

Answer: (d) \( 8x+5y+60=0 \)

Explanation:
1. Let the intercept points be \( X(a,0) \) and \( Y(0,b) \). \( A \) divides \( XY \) in \( 1:2 \): \( \left(\frac{2a}3,\frac b3\right)=(-5,-4) \), so \( a=-\frac{15}2,\ b=-12 \).
2. Line: \( \frac{x}{-15/2}+\frac y{-12}=1\Rightarrow 8x+5y+60=0 \).

Teacher's Note:
1. Use the section formula for the points on the axes.
2. Convert intercept form to general form.

Answer: (b) \( (8,-8) \)

Explanation:
1. If \( R \) divides \( PQ \) in the ratio \( \lambda:1 \), then \( 5=\frac{6\lambda+2}{\lambda+1}\Rightarrow\lambda=3 \).
2. The harmonic conjugate divides \( PQ \) externally in the ratio \( 3:1 \): \( \left(\frac{18-2}{3-1},\frac{-6-10}{3-1}\right)=(8,-8) \).

Teacher's Note:
1. Find the ratio in which R divides PQ.
2. The harmonic conjugate divides externally in the same ratio.

Answer: (a) 0

Explanation:
1. Concurrency: \( \begin{vmatrix}\lambda^2+\lambda-1&-1&3\\ \lambda&-1&2\\1&-1&1\end{vmatrix}=0 \) (lines written as \( mx-y+c=0 \)), which simplifies to \( \lambda^2-\lambda=0 \), so \( \lambda=0 \) or 1.
2. For \( \lambda=1 \) all three slopes equal 1 and the lines are distinct parallel lines, not concurrent. Only \( \lambda=0 \) works (the point \( (1,2) \)).

Teacher's Note:
1. Reject the value that makes the lines parallel.
2. Verify by finding the common point.

Answer: (b) \( x^2+y^2=a^2+b^2 \)

Explanation:
1. Square both equations and add: \( x^2(\cos^2\alpha+\sin^2\alpha)+y^2(\sin^2\alpha+\cos^2\alpha)=a^2+b^2 \) (the cross terms cancel).

Teacher's Note:
1. Square and add to remove alpha.

Answer: (a) \( (3,4) \) and \( (6,8) \)

Explanation:
1. The points divide the segment in the ratios \( 1:2 \) and \( 2:1 \).
2. \( P=\left(\frac{9}{3},\frac{12}{3}\right)=(3,4) \) and \( Q=\left(\frac{18}{3},\frac{24}{3}\right)=(6,8) \).

Teacher's Note:
1. Trisection points are one third and two thirds of the way along.
2. Use the section formula or simply add one third of the displacement.

Answer: (d) \( \left(\frac13,-\frac23\right) \)

Explanation:
1. With the chord \( hx+ky=1 \): \( 3x^2+3y^2-2x(hx+ky)+4y(hx+ky)=0 \).
2. Sum of the coefficients of \( x^2 \) and \( y^2 \): \( (3-2h)+(3+4k)=0\Rightarrow h-2k=3 \), i.e. \( \frac h3-\frac{2k}3=1 \).
3. Hence the chords pass through \( \left(\frac13,-\frac23\right) \).

Teacher's Note:
1. Use the same method as for the previous curve.
2. Write the condition in the form h x0 + k y0 = 1.

Answer: (c) \( 4x^3-4xy^2-3ax^2+ay^2=0 \)

Explanation:
1. Let \( P=(h,k) \), \( \angle PAB=\theta \), so \( \tan\theta=\frac kh \). The exterior angle at \( B \) is \( \pi-3\theta \): \( \tan(\pi-3\theta)=\frac k{h-a} \), i.e. \( \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=-\frac k{h-a} \).
2. Substituting \( \tan\theta=\frac kh \): \( \frac{3h^2-k^2}{h(h^2-3k^2)}=-\frac1{h-a} \).
3. Cross-multiplying: \( (3h^2-k^2)(h-a)=-h(h^2-3k^2) \Rightarrow4h^3-4hk^2-3ah^2+ak^2=0 \).

Teacher's Note:
1. Use tan 3 theta in terms of tan theta.
2. The exterior angle at B gives the slope of PB.

Answer: (a) \( (1+3\sqrt3,\ 3-\sqrt3) \)

Explanation:
1. \( X=x\cos\theta+y\sin\theta=2\cdot\frac12+6\cdot\frac{\sqrt3}2=1+3\sqrt3 \).
2. \( Y=-x\sin\theta+y\cos\theta=-2\cdot\frac{\sqrt3}2+6\cdot\frac12=3-\sqrt3 \).

Teacher's Note:
1. Use X = x cos t + y sin t and Y = -x sin t + y cos t.

Answer: (d) 1

Explanation:
1. With \( a=h,\ b=4h,\ c=-2,\ 2h'=-5,\ 2g=1,\ 2f=2 \), the condition \( abc+2fgh'-af^2-bg^2-ch'^2=0 \) becomes \( -8h^2-2h+10=0 \), i.e. \( 4h^2+h-5=0 \).
2. \( (h-1)(4h+5)=0\Rightarrow h=1 \) or \( -\frac54 \). For both values the lines are real (since \( h'^2 \gt ab \)). Only \( h=1 \) is an integer.

Teacher's Note:
1. Apply the determinant condition.
2. Count only integer roots.

Answer: (c) 3

Explanation:
1. For \( \frac xa+\frac yb=1 \) through \( (2,4) \): \( 2b+4a=ab \), and area \( \frac12|ab|=16\Rightarrow ab=\pm32 \).
2. \( ab=32 \): \( a^2-8a+16=0 \), one solution \( a=4 \).
3. \( ab=-32 \): \( a^2+8a-16=0 \), two real solutions.
4. Total 3 lines.

Teacher's Note:
1. Use intercept form and the two sign cases of ab.
2. Count the real roots in each case.

Answer: (a) \( (3x-1)^2+9y^2=a^2+b^2 \)

Explanation:
1. \( 3x-1=a\cos t+b\sin t \) and \( 3y=a\sin t-b\cos t \).
2. Squaring and adding: \( (3x-1)^2+9y^2=a^2+b^2 \).

Teacher's Note:
1. Eliminate t by squaring and adding.

Answer: (b) 14

Explanation:
1. Points on the line: \( \left(-3+\frac r{\sqrt2},\frac r{\sqrt2}\right) \).
2. Substituting: \( \frac r{\sqrt2}=\left(-3+\frac r{\sqrt2}\right)^2+\left(-3+\frac r{\sqrt2}\right)+1\Rightarrow\frac{r^2}2-\frac{6r}{\sqrt2}+7=0 \).
3. Product of the roots \( r_1r_2=\frac{7}{1/2}=14 \).

Teacher's Note:
1. Use the parametric form of a line with distance r from A.
2. Product of roots = c/a.

Answer: (a) \( x^2+y^2-6x-6y=0 \)

Explanation:
1. For the vertex \( (h,k) \): \( \frac{k}{h-6}\cdot\frac{k-6}{h}=-1 \Rightarrow k^2-6k=-h^2+6h \).
2. Hence \( x^2+y^2-6x-6y=0 \) (the circle on the hypotenuse as diameter, excluding its end points).

Teacher's Note:
1. The legs are perpendicular: product of slopes is -1.
2. It is the circle on the hypotenuse as diameter.

Answer: (b) \( 43x-29y=71 \)

Explanation:
1. Family: \( (2x+3y-1)+\lambda(3x-4y-6)=0 \). Through \( (3,2) \): \( 11+\lambda(-5)=0\Rightarrow\lambda=\frac{11}5 \).
2. \( 5(2x+3y-1)+11(3x-4y-6)=0\Rightarrow43x-29y-71=0 \).

Teacher's Note:
1. Use the family of lines through the intersection.
2. Substitute the point to find lambda.

Answer: (b) \( \frac{256}{75} \)

Explanation:
1. The area of the triangle is 12 (sides 6, 5, 5 with height 4 on the side 6). Splitting at \( P \): \( 3p_1+\frac52p_2+\frac52p_3=12 \).
2. AM-GM: \( 4=\frac{3p_1+\frac52p_2+\frac52p_3}3\ge\left(\frac{75}4p_1p_2p_3\right)^{1/3} \), so \( p_1p_2p_3\le\frac{64\cdot4}{75}=\frac{256}{75} \).
3. Equality holds at an interior point, so this is the maximum.

Teacher's Note:
1. Use the area as a linear relation among the distances.
2. Apply AM-GM to the three terms.

Answer: (b) \( 3x-4y=0 \)

Explanation:
1. \( c^2=a^2+b^2 \), so \( \angle C=90^\circ \) and \( O,A,C,B \) are concyclic (circle with diameter \( AB \)).
2. Angles on the same arc are equal: \( \angle COB=\angle CAB \). In the triangle, \( \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{100+36-64}{120}=\frac35 \), so \( \cos\angle COB=\frac35 \) and \( \angle COA=90^\circ-\angle COB \) has \( \tan\angle COA=\frac34 \).
3. The angle that \( OC \) makes with the x-axis is fixed with tangent \( \frac34 \), so \( C \) lies on \( y=\frac34x \), i.e. \( 3x-4y=0 \).

Teacher's Note:
1. A right angle at C makes O, A, C, B concyclic.
2. Angles in the same segment are equal.

Answer: (a) \( x+3y-11=0 \)

Explanation:
1. The given line has slope 3, so the perpendicular has slope \( -\frac13 \).
2. \( y-3=-\frac13(x-2)\Rightarrow x+3y-11=0 \).

Teacher's Note:
1. Slopes of perpendicular lines multiply to -1.

Answer: (b) 2

Explanation:
1. The mid-point \( \left(\frac{\lambda^2+\lambda+1}2,\frac{2\lambda-1}2\right) \) lies on the mirror: \( 3\lambda^2-7\lambda+2=0\Rightarrow\lambda=\frac13,2 \).
2. The join of the point and its image must be perpendicular to the mirror: \( \lambda^2-\lambda-2=0\Rightarrow\lambda=-1,2 \).
3. Only \( \lambda=2 \) satisfies both conditions.

Teacher's Note:
1. Use both the mid-point condition and the perpendicularity condition.
2. Take the common root.

Answer: (a) \( \left(\frac{21}2,\frac83\right) \)

Explanation:
1. Equidistant from \( (5,12) \) and \( (16,12) \): \( x=\frac{21}2 \).
2. Equidistant from \( (0,0) \) and \( (5,12) \): \( 10x+24y=169 \), so \( 105+24y=169 \) and \( y=\frac83 \).

Teacher's Note:
1. The perpendicular bisector of the horizontal side is a vertical line.

Answer: (c) \( (1,-2) \)

Explanation:
1. Let the chord be \( hx+ky=1 \) and homogenise the curve: \( 3x^2-y^2-2x(hx+ky)+4y(hx+ky)=0 \).
2. Right angle at the origin: sum of the coefficients of \( x^2 \) and \( y^2 \) is zero: \( (3-2h)+(-1+4k)=0\Rightarrow h-2k=1 \).
3. So \( h\cdot1+k\cdot(-2)=1 \): every chord passes through \( (1,-2) \).

Teacher's Note:
1. Homogenise using the chord equation.
2. The condition h - 2k = 1 says the chord passes through (1, -2).

Answer: (d) \( 1:3 \) internally

Explanation:
1. If the ratio is \( m:n \), the ordinate is \( \frac{9m+5n}{m+n}=6 \).
2. So \( 3m=n \), i.e. \( m:n=1:3 \), and both parts are positive, so the division is internal.

Teacher's Note:
1. Apply the section formula to the y-coordinate only.
2. A positive ratio means internal division.

Answer: (a) \( p=2,\ \alpha=240^\circ \)

Explanation:
1. Write \( x+\sqrt3y=-4 \) and divide by \( \sqrt{1+3}=2 \) with the sign changed to make the right side positive: \( -\frac12x-\frac{\sqrt3}2y=2 \).
2. So \( \cos\alpha=-\frac12,\ \sin\alpha=-\frac{\sqrt3}2 \), giving \( \alpha=240^\circ \) and \( p=2 \).

Teacher's Note:
1. p must be positive.
2. Both cosine and sine are negative, so the angle is in the third quadrant.

Answer: (a) \( x+y+\frac1{\sqrt2}=0 \)

Explanation:
1. The slope of \( BC \) is \( \frac{-3+1}{-1+3}=-1 \), so the line is \( x+y+c=0 \) with \( \frac{|c|}{\sqrt2}=\frac12\Rightarrow c=\pm\frac1{\sqrt2} \).
2. The triangle lies in the region \( x+y \lt 0 \), so a line cutting \( OB \) and \( OC \) has \( x+y=-\frac1{\sqrt2} \), i.e. \( c=+\frac1{\sqrt2} \).

Teacher's Note:
1. The line is x + y + c = 0.
2. Use the position of the triangle to fix the sign of c.

Answer: (b) \( 5x+5y-3=0 \)

Explanation:
1. The other line is the reflection of \( x-7y+5=0 \) in the bisector. The lines meet at \( \left(-\frac1{10},\frac7{10}\right) \).
2. Reflect the point \( (-5,0) \) of the given line in \( x+3y-2=0 \): the expression equals \( -7 \), so the image is \( (-5,0)+\frac{14}{10}(1,3)=\left(-\frac{18}5,\frac{21}5\right) \).
3. The other line passes through \( \left(-\frac1{10},\frac7{10}\right) \) and \( \left(-\frac{18}5,\frac{21}5\right) \): slope \( -1 \), so \( 5x+5y=3 \).

Teacher's Note:
1. Reflect the given line in the bisector.
2. Both lines make equal angles with the bisector.

Answer: (b) \( \frac92 \) square units

Explanation:
1. The given line has slope 1, so the required lines have slopes 0 and infinity: \( y=4 \) and \( x=3 \).
2. They meet \( x-y=2 \) at \( (6,4) \) and \( (3,1) \). The triangle with vertices \( (3,4),(6,4),(3,1) \) is right-angled with legs 3 and 3.
3. Area \( =\frac12\cdot3\cdot3=\frac92 \).

Teacher's Note:
1. Lines at 45 degrees to a line of slope 1 are horizontal and vertical.
2. The triangle is right-angled at (3, 4).

Answer: (d) \( a\lambda(x-\alpha)+b\lambda(y-\beta)+a\alpha+b\beta+c=0 \)

Explanation:
1. Along the line, \( P=(\alpha+r\cos\theta,\ \beta+r\sin\theta) \) and \( Q=(\alpha+r_1\cos\theta,\ \beta+r_1\sin\theta)=(h,k) \) with \( r=\lambda r_1 \).
2. So \( P=(\alpha+\lambda(h-\alpha),\ \beta+\lambda(k-\beta)) \). It lies on the fixed line: \( a(\alpha+\lambda(h-\alpha))+b(\beta+\lambda(k-\beta))+c=0 \).
3. Replacing (h, k) by (x, y): a straight line.

Teacher's Note:
1. Express P in terms of Q using the constant ratio.
2. Substitute P in the equation of the fixed line.

Answer: (c) \( \dfrac{C^2}{\sqrt3\,(A^2+B^2)} \)

Explanation:
1. A line \( y=mx \) through the origin making \( 60^\circ \) with \( Ax+By+C=0 \) satisfies \( 3(B-Am)^2=(mB+A)^2 \); with \( m=\frac yx \) this is exactly \( (A^2-3B^2)x^2+8ABxy+(B^2-3A^2)y^2=0 \). Hence the triangle is equilateral.
2. The altitude from the origin is \( p=\frac{|C|}{\sqrt{A^2+B^2}} \), and an equilateral triangle of altitude \( p \) has area \( \frac{p^2}{\sqrt3} \).
3. Area \( =\frac{C^2}{\sqrt3(A^2+B^2)} \).

Teacher's Note:
1. The pair of lines makes 60 degrees with the third line.
2. Area of an equilateral triangle with altitude p is p^2 / sqrt3.

Answer: (b) \( x+7y+12=0 \) and \( x+7y-8=0 \)

Explanation:
1. Let the line be \( x+7y+\lambda=0 \). Distance between parallel lines: \( \frac{|\lambda-2|}{\sqrt{50}}=\sqrt2\Rightarrow|\lambda-2|=10 \).
2. \( \lambda=12 \) or \( -8 \).

Teacher's Note:
1. Distance between parallel lines is |c1 - c2|/sqrt(a^2 + b^2).