JEE Mathematics Straight Lines MCQs Set 30

Answer: (c) \( [1,\infty) \)

Explanation:
1. Solving, \( x=\dfrac{1+|a|}{1+a} \) and \( y=\dfrac{a|a|-1}{1+a} \). Both must be non-negative, which forces \( a\gt-1 \) and \( a|a|\ge1 \).
2. For \( -1\lt a\lt0 \), \( a|a|=-a^2\gt-1 \) fails; for \( a\ge0 \) we need \( a^2\ge1 \), so \( a\ge1 \). Hence \( a\in[1,\infty) \). Answer: (c)

Teacher's Note:
1. Both intercept-type coordinates must be non-negative.

Answer: (d) \( \dfrac{23}{\sqrt{17}} \)

Explanation:
1. \( \dfrac{13}5+\dfrac{32}b=1 \) gives \( b=-20 \), so \( L \): \( 4x-y-20=0 \). \( K \) has the same slope \( 4 \): \( -\dfrac3c=4 \), \( c=-\dfrac34 \), so \( K \): \( 4x-y+3=0 \).
2. Distance \( =\dfrac{|3+20|}{\sqrt{17}}=\dfrac{23}{\sqrt{17}} \). Answer: (d)

Teacher's Note:
1. Parallel lines \( ax+by+c_1=0 \) and \( ax+by+c_2=0 \) are at distance \( |c_1-c_2|/\sqrt{a^2+b^2} \).

Answer: (a) \( \left(\dfrac54,0\right) \)

Explanation:
1. For a point \( (x,y) \): \( \dfrac{(x-1)^2+y^2}{(x+1)^2+y^2}=\dfrac19 \), which gives \( 8x^2+8y^2-20x+8=0 \), i.e. \( x^2+y^2-\dfrac52x+1=0 \).
2. All three points lie on this circle, so the circumcentre is its centre \( \left(\dfrac54,0\right) \). Answer: (a)

Teacher's Note:
1. The locus of a point with a fixed distance ratio from two points is a circle.

Answer: (a) -4

Explanation:
1. The mid-point is \( \left(\dfrac{k+1}2,\dfrac72\right) \) and the bisector has slope \( -\dfrac1{\text{slope}(PQ)}=k-1 \). Its equation is \( y-\dfrac72=(k-1)\left(x-\dfrac{k+1}2\right) \).
2. At \( x=0 \), \( y=-4 \): \( -\dfrac{15}2=-\dfrac{k^2-1}2 \), so \( k^2=16 \); among the options, \( k=-4 \). Answer: (a)

Teacher's Note:
1. Write the bisector through the mid-point with the perpendicular slope.

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

Explanation:
1. \( QR \) has slope \( \sqrt3 \), so it makes \( 60^\circ \) with the positive \( x \)-axis, while \( QP \) points along the negative \( x \)-axis. Hence \( \angle PQR=120^\circ \).
2. The bisector makes \( 120^\circ \) with the positive \( x \)-axis (slope \( -\sqrt3 \)): \( \sqrt3x+y=0 \). Answer: (a)

Teacher's Note:
1. Find the direction angles of \( QP \) and \( QR \).

Answer: (b) \( 4x+3y=24 \)

Explanation:
1. The mid-point of the intercepts \( (a,0) \) and \( (0,b) \) is \( \left(\dfrac a2,\dfrac b2\right)=(3,4) \), so \( a=6 \), \( b=8 \).
2. The line is \( \dfrac x6+\dfrac y8=1 \), i.e. \( 4x+3y=24 \). Answer: (b)

Teacher's Note:
1. The mid-point of the intercepts is \( (a/2,b/2) \).

Answer: (b) \( \left(\dfrac12,3\right) \)

Explanation:
1. The point lies between the lines when \( \dfrac x2\lt y\lt3x \) with \( x=a\gt0 \): \( a^2\gt\dfrac a2 \) and \( a^2\lt3a \).
2. These give \( a\gt\dfrac12 \) and \( a\lt3 \), so \( a\in\left(\dfrac12,3\right) \). Answer: (b)

Teacher's Note:
1. Both inequalities require \( a\gt0 \) because of \( x\gt0 \).

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

Explanation:
1. Since \( \dfrac1a,\dfrac1b,\dfrac1c \) are in AP, \( \dfrac2b=\dfrac1a+\dfrac1c \), i.e. \( \dfrac1a-\dfrac2b+\dfrac1c=0 \).
2. This says the line is satisfied by \( x=1 \), \( y=-2 \), so it always passes through \( (1,-2) \). Answer: (b)

Teacher's Note:
1. Rewrite the HP condition as an AP of reciprocals.

Answer: (a) \( 2x+3y=9 \)

Explanation:
1. Let \( C=(x,y) \); the centroid is \( \left(\dfrac x3,\dfrac{y-2}3\right) \).
2. It lies on \( 2x+3y=1 \): \( \dfrac{2x}3+(y-2)=1 \), so \( 2x+3y=9 \). Answer: (a)

Teacher's Note:
1. Express the centroid in terms of the moving vertex.

Answer: (d) \( \dfrac x2-\dfrac y3=1 \) and \( \dfrac x{-2}+\dfrac y1=1 \)

Explanation:
1. With intercepts \( a \) and \( b=-(a+1) \): \( \dfrac xa-\dfrac y{a+1}=1 \). Through \( (4,3) \): \( \dfrac4a-\dfrac3{a+1}=1 \), so \( a^2=4 \), \( a=\pm2 \).
2. \( a=2 \), \( b=-3 \) gives \( \dfrac x2-\dfrac y3=1 \); \( a=-2 \), \( b=1 \) gives \( \dfrac x{-2}+\dfrac y1=1 \). Answer: (d)

Teacher's Note:
1. Use the intercept form with \( a+b=-1 \).

Answer: (c) 2

Explanation:
1. Comparing with \( ax^2+2hxy+by^2=0 \): \( a=1 \), \( h=-c \), \( b=-7 \). Then \( m_1+m_2=-\dfrac{2h}b=-\dfrac{2c}7 \) and \( m_1m_2=\dfrac ab=-\dfrac17 \).
2. \( -\dfrac{2c}7=4\left(-\dfrac17\right) \) gives \( c=2 \). Answer: (c)

Teacher's Note:
1. Use the sum and product of slopes of a pair of lines.

Answer: (d) -3

Explanation:
1. Put \( y=-\dfrac{3x}4 \) into the equation: \( 6x^2+\dfrac{3x^2}4+4c\cdot\dfrac{9x^2}{16}=0 \).
2. This gives \( \dfrac{27}4+\dfrac{9c}4=0 \), so \( c=-3 \). Answer: (d)

Teacher's Note:
1. A line of the pair satisfies the equation of the pair.

Answer: (a) lie on a straight line

Explanation:
1. With common ratio \( r \), the points are \( (a,b) \), \( (ar,br) \), \( (ar^2,br^2) \).
2. The slope between the first two and the last two is \( \dfrac ba \) in both cases, so the points are collinear. Answer: (a)

Teacher's Note:
1. Compute two slopes from the same point.

Answer: (a) \( \dfrac12\left(a_2^2+b_2^2-a_1^2-b_1^2\right) \)

Explanation:
1. Equating the squared distances from \( (x,y) \): \( (x-a_1)^2+(y-b_1)^2=(x-a_2)^2+(y-b_2)^2 \), which gives \( 2(a_2-a_1)x+2(b_2-b_1)y+a_1^2+b_1^2-a_2^2-b_2^2=0 \).
2. Dividing by \( -2 \): \( (a_1-a_2)x+(b_1-b_2)y+\dfrac12\left(a_2^2+b_2^2-a_1^2-b_1^2\right)=0 \), so \( c=\dfrac12\left(a_2^2+b_2^2-a_1^2-b_1^2\right) \). Answer: (a)

Teacher's Note:
1. The locus is the perpendicular bisector of the segment.

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

Explanation:
1. The centroid is \( x=\dfrac{a\cos t+b\sin t+1}3 \), \( y=\dfrac{a\sin t-b\cos t}3 \), so \( 3x-1=a\cos t+b\sin t \) and \( 3y=a\sin t-b\cos t \).
2. Squaring and adding, \( (3x-1)^2+(3y)^2=a^2+b^2 \). Answer: (b)

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

Answer: (d) \( pq=-1 \)

Explanation:
1. The bisectors of \( x^2-2pxy-y^2=0 \) are given by \( \dfrac{x^2-y^2}{1-(-1)}=\dfrac{xy}{-p} \), i.e. \( x^2-y^2+\dfrac2pxy=0 \).
2. This must be the second pair \( x^2-2qxy-y^2=0 \), so \( \dfrac2p=-2q \), i.e. \( pq=-1 \). Answer: (d)

Teacher's Note:
1. Use the bisector equation \( \dfrac{x^2-y^2}{a-b}=\dfrac{xy}h \).

Answer: (d) \( y(\cos\alpha+\sin\alpha)+x(\cos\alpha-\sin\alpha)=a \)

Explanation:
1. The two vertices adjacent to the origin are \( A(a\cos\alpha,a\sin\alpha) \) and \( B(-a\sin\alpha,a\cos\alpha) \); the diagonal not through the origin is \( AB \).
2. The line \( y(\cos\alpha+\sin\alpha)+x(\cos\alpha-\sin\alpha)=a \) is satisfied by both \( A \) and \( B \): at \( A \), \( a\sin\alpha\cos\alpha+a\sin^2\alpha+a\cos^2\alpha-a\sin\alpha\cos\alpha=a \); at \( B \) the same check gives \( a \). Answer: (d)

Teacher's Note:
1. Find the coordinates of the two vertices adjacent to the origin.

Answer: (a) isosceles and right angled

Explanation:
1. With \( A(4,0) \), \( B(-1,-1) \), \( C(3,5) \): \( AB^2=26 \), \( AC^2=1+25=26 \), \( BC^2=16+36=52 \).
2. \( AB=AC \) and \( BC^2=AB^2+AC^2 \), so the triangle is isosceles and right angled at \( A \). Answer: (a)

Teacher's Note:
1. Compare the squares of the side lengths.

Answer: (d) \( \left(1,\dfrac1{\sqrt3}\right) \)

Explanation:
1. All three sides equal \( 2 \), so the triangle is equilateral and the incentre coincides with the centroid.
2. The centroid is \( \left(\dfrac{1+0+2}3,\dfrac{\sqrt3}3\right)=\left(1,\dfrac1{\sqrt3}\right) \). Answer: (d)

Teacher's Note:
1. In an equilateral triangle the centres coincide.

Answer: (c) are concurrent with one line bisecting the angle between the other two

Explanation:
1. The lines \( 2x+11y=5 \) and \( 4x-3y=2 \) meet at \( \left(\dfrac{37}{50},\dfrac8{25}\right) \), which satisfies \( 24x+7y=20 \), so the lines are concurrent.
2. The bisectors of \( 24x+7y-20=0 \) and \( 4x-3y-2=0 \) are given by \( \dfrac{24x+7y-20}{25}=\pm\dfrac{4x-3y-2}5 \); the positive sign gives \( 2x+11y-5=0 \), so that line bisects the angle between the other two. Answer: (c)

Teacher's Note:
1. Check concurrency, then compare with the angle bisector.

Answer: (b) \( y-2=0 \)

Explanation:
1. The lines make angles \( 120^\circ \) and \( 60^\circ \) with \( OX \), so the angle between them (containing the \( x \)-axis direction) is \( 60^\circ \) and \( AB \) must be perpendicular to the bisector (the \( y \)-axis direction), i.e. horizontal.
2. The line \( y=2 \) meets them at \( \left(\mp\dfrac2{\sqrt3},2\right) \), with \( OA=OB=AB=\dfrac4{\sqrt3} \), so \( \triangle OAB \) is equilateral. Answer: (b)

Teacher's Note:
1. For an equilateral triangle the angle at \( O \) must be \( 60^\circ \).