JEE Mathematics Straight Lines MCQs Set 29

Answer: (c) I and II quadrants

Explanation:
1. Equidistant from the axes means \( y=x \) or \( y=-x \).
2. With \( y=x \): \( \left(\frac{15}8,\frac{15}8\right) \) (I quadrant). With \( y=-x \): \( \left(-\frac{15}2,\frac{15}2\right) \) (II quadrant). Answer: (c)

Teacher's Note:
1. Check both lines \( y=\pm x \).

Answer: (b) \( \dfrac13 \)

Explanation:
1. \( L_1 \) has slope \( \dfrac{4-2}{-3-1}=-\dfrac12 \), so \( h+2k=5 \). \( L_2 \) has slope \( 2 \): \( 3-k=2(4-h) \), so \( 2h-k=5 \).
2. Solving, \( h=3 \), \( k=1 \), so \( \dfrac kh=\dfrac13 \). Answer: (b)

Teacher's Note:
1. Perpendicular lines have slopes with product \( -1 \).

Answer: (c) \( \dfrac{1-\sqrt7}{1+\sqrt7} \)

Explanation:
1. The distance of \( P \) from \( x+y=7 \) is \( \sqrt2 \), so the angle \( \theta \) between the line and \( x+y=7 \) satisfies \( \tan\theta=\dfrac{\sqrt2}{\sqrt{16-2}}=\dfrac1{\sqrt7} \).
2. \( \left|\dfrac{m+1}{1-m}\right|=\dfrac1{\sqrt7} \). Taking the positive sign gives \( m=\dfrac{1-\sqrt7}{1+\sqrt7} \) (the other root, \( -\frac{\sqrt7+1}{\sqrt7-1} \), is not among the options). Answer: (c)

Teacher's Note:
1. Use the angle between two lines.

Answer: (d) \( \sqrt{\dfrac25} \)

Explanation:
1. Perpendicular: \( \left(-\dfrac1{a-1}\right)\left(-\dfrac2{a^2}\right)=-1 \), so \( a^3-a^2+2=0 \), \( (a+1)(a^2-2a+2)=0 \) and \( a=-1 \).
2. The lines are \( x-2y=1 \) and \( 2x+y=1 \), meeting at \( \left(\dfrac35,-\dfrac15\right) \), at distance \( \sqrt{\dfrac9{25}+\dfrac1{25}}=\sqrt{\dfrac25} \). Answer: (d)

Teacher's Note:
1. Use the product of slopes.

Answer: (b) \( \left(\dfrac13,2\right) \)

Explanation:
1. The other vertices are \( B=(2(-1)-1,\,2(1)-2)=(-3,0) \) and \( C=(2(2)-1,\,2(3)-2)=(3,4) \).
2. The centroid is \( \left(\dfrac{1-3+3}3,\dfrac{2+0+4}3\right)=\left(\dfrac13,2\right) \). Answer: (b)

Teacher's Note:
1. A vertex is twice the mid-point minus the other end.

Answer: (c) \( \left(\dfrac52,-1\right) \)

Explanation:
1. The line \( y=x-3 \) meets \( y=x^2-4x+3 \) where \( x^2-5x+6=0 \), so \( x=2,3 \), giving \( (2,-1) \) and \( (3,0) \).
2. Tangent slopes are \( 2(x-2) \): \( 0 \) at \( (2,-1) \) (line \( y=-1 \)) and \( 2 \) at \( (3,0) \) (line \( y=2x-6 \)). They meet where \( -1=2x-6 \), i.e. at \( \left(\dfrac52,-1\right) \). Answer: (c)

Teacher's Note:
1. Find the points first, then the two tangents.

Answer: (b) The lines are concurrent at the point \( \left(\dfrac34,\dfrac12\right) \)

Explanation:
1. Dividing \( 3p+2q+4r=0 \) by \( 4 \): \( p\cdot\dfrac34+q\cdot\dfrac12+r=0 \).
2. So every line \( px+qy+r=0 \) passes through \( \left(\dfrac34,\dfrac12\right) \). Answer: (b)

Teacher's Note:
1. Compare the condition with the equation of the line at a fixed point.

Answer: (a) 36

Explanation:
1. Vertices \( O \), \( (a,0) \), \( (0,b) \) with \( \dfrac12|a||b|=50 \), so \( |a||b|=100=2^2\cdot5^2 \), which has \( 9 \) factor pairs \( (|a|,|b|) \).
2. Each pair allows \( 2 \) signs for \( a \) and \( 2 \) for \( b \): \( 9\times4=36 \). Answer: (a)

Teacher's Note:
1. Count positive divisors of 100, then the sign choices.

Answer: (d) \( (2,2) \)

Explanation:
1. \( A=(8,0) \), \( B=(0,6) \), and the sides are \( OA=8 \), \( OB=6 \), \( AB=10 \) (opposite \( B,A,O \) respectively).
2. Incentre \( =\left(\dfrac{6\cdot8+8\cdot0+10\cdot0}{24},\dfrac{6\cdot0+8\cdot6+10\cdot0}{24}\right)=(2,2) \). Answer: (d)

Teacher's Note:
1. For a right triangle the inradius is \( (OA+OB-AB)/2=2 \).

Answer: (d) \( \dfrac{\sqrt5}2 \)

Explanation:
1. For \( P(t^2,t) \) on the curve, \( PA^2=\left(t^2-\dfrac32\right)^2+t^2=t^4-2t^2+\dfrac94=(t^2-1)^2+\dfrac54 \).
2. The minimum is \( \dfrac54 \) at \( t^2=1 \), so the shortest distance is \( \dfrac{\sqrt5}2 \). Answer: (d)

Teacher's Note:
1. Write the squared distance in terms of a parameter and complete the square.

Answer: (a) \( (3,6) \)

Explanation:
1. The two lines are perpendicular and meet at \( A(0,3) \), so the parallelogram is a rectangle with centre \( (2,4) \) and opposite vertex \( C=(4,5) \).
2. The other vertices lie on the given lines: \( B=(1,2) \) (on \( x+y=3 \)) and \( D=(3,6) \) (on \( x-y+3=0 \)). Answer: (a)

Teacher's Note:
1. The opposite vertex is the reflection of \( A \) in the centre.

Answer: (c) \( 5x-3y+1=0 \)

Explanation:
1. The diagonals of \( ABDC \) are \( AD \) and \( BC \), which bisect each other, so the common mid-point is the mid-point of \( BC \): \( P=\left(\dfrac52,\dfrac92\right) \).
2. The line \( AP \) has slope \( \dfrac{9/2-2}{5/2-1}=\dfrac53 \): \( y-2=\dfrac53(x-1) \), i.e. \( 5x-3y+1=0 \). Answer: (c)

Teacher's Note:
1. The diagonals of a parallelogram bisect each other.

Answer: (d) \( 5 \)

Explanation:
1. The given line has slope \( \dfrac23 \); the other has slope \( \dfrac{\beta-17}8 \).
2. \( \dfrac23\cdot\dfrac{\beta-17}8=-1 \) gives \( \beta-17=-12 \), so \( \beta=5 \). Answer: (d)

Teacher's Note:
1. Perpendicular lines have slopes with product \( -1 \).

Answer: (c) \( 3x+2y=xy \)

Explanation:
1. Let \( P(\alpha,0) \), \( Q(0,\beta) \); the line \( \dfrac x\alpha+\dfrac y\beta=1 \) passes through \( (2,3) \): \( \dfrac2\alpha+\dfrac3\beta=1 \).
2. \( R=(\alpha,\beta) \), so the locus is \( \dfrac2x+\dfrac3y=1 \), i.e. \( 3x+2y=xy \). Answer: (c)

Teacher's Note:
1. The fourth vertex of the rectangle has coordinates \( (\alpha,\beta) \).

Answer: (d) 780

Explanation:
1. Interior points satisfy \( x\gt0 \), \( y\gt0 \), \( x+y\lt41 \), i.e. \( x+y\le40 \).
2. For \( x=1,2,\dots,39 \) the number of \( y \) values is \( 39,38,\dots,1 \), so the total is \( \dfrac{39\cdot40}2=780 \). Answer: (d)

Teacher's Note:
1. Count lattice points column by column.

Answer: (c) circle of radius \( \sqrt2 \)

Explanation:
1. The lines of the family all pass through the intersection of \( 2x-3y+4=0 \) and \( x-2y+3=0 \), which is \( A(1,2) \).
2. The image \( P' \) of \( (2,3) \) satisfies \( AP'=AP=\sqrt{1+1}=\sqrt2 \), so the locus is the circle \( (x-1)^2+(y-2)^2=2 \) of radius \( \sqrt2 \). Answer: (c)

Teacher's Note:
1. A mirror image is at the same distance from every point on the mirror.

Answer: (c) \( 2ad-3bc=0 \)

Explanation:
1. In the fourth quadrant and equidistant from the axes, the point is \( (\alpha,-\alpha) \). The first line gives \( 4a\alpha-2a\alpha+c=0 \), so \( \alpha=-\dfrac c{2a} \).
2. The second gives \( 5b\alpha-2b\alpha+d=0 \), so \( \alpha=-\dfrac d{3b} \). Equating, \( 3bc=2ad \), i.e. \( 2ad-3bc=0 \). Answer: (c)

Teacher's Note:
1. A point equidistant from the axes in the fourth quadrant has coordinates \( (\alpha,-\alpha) \).

Answer: (b) \( 2-\sqrt2 \)

Explanation:
1. The triangle with these mid-points has vertices \( (0,0) \), \( (2,0) \) and \( (0,2) \), a right triangle with legs \( 2 \) and hypotenuse \( 2\sqrt2 \).
2. The inradius is \( \dfrac{2+2-2\sqrt2}2=2-\sqrt2 \), and the incentre is \( (r,r) \), so its \( x \)-coordinate is \( 2-\sqrt2 \). Answer: (b)

Teacher's Note:
1. A triangle is twice as large as its mid-point triangle.

Answer: (c) 6

Explanation:
1. The dividing point is \( \left(\dfrac{3\cdot2+2\cdot1}5,\dfrac{3\cdot4+2\cdot1}5\right)=\left(\dfrac85,\dfrac{14}5\right) \).
2. Then \( k=2\cdot\dfrac85+\dfrac{14}5=6 \). Answer: (c)

Teacher's Note:
1. Use the internal section formula.

Answer: (c) A is true, but R is false.

Explanation:
1. \( P=(-2,-2) \), \( Q=(1,-2) \), so \( OP=2\sqrt2 \) and \( OQ=\sqrt5 \). The acute-angle bisector at \( O \) is the internal bisector of \( \angle POQ \), which divides \( PQ \) in the ratio \( OP:OQ \), so the Assertion is true.
2. An angle bisector does not in general produce similar triangles, so the Reason is false. Answer: (c)

Teacher's Note:
1. Use the angle bisector theorem.

Answer: (c) \( y=\dfrac6x \)

Explanation:
1. The tangent at \( (x,y) \) has intercepts \( \left(x-y\dfrac{dx}{dy},0\right) \) and \( \left(0,y-x\dfrac{dy}{dx}\right) \). Bisection gives \( 2x=x-y\dfrac{dx}{dy} \), so \( \dfrac{dx}x+\dfrac{dy}y=0 \).
2. Integrating, \( xy=c \); through \( (2,3) \), \( c=6 \), so \( y=\dfrac6x \). Answer: (c)

Teacher's Note:
1. Form a differential equation from the mid-point condition.