JEE Mathematics Straight Lines MCQs Set 26

Answer: (c) \( 2\sqrt{37} \)

Explanation:
1. The vertices are \( A=(3,7)+(5,3)-(2,1)=(6,9) \), \( B=(3,7)+(2,1)-(5,3)=(0,5) \), \( C=(5,3)+(2,1)-(3,7)=(4,-3) \).
2. Sides: \( AB=2\sqrt{13},\ BC=4\sqrt5,\ AC=2\sqrt{37} \). The largest is \( 2\sqrt{37} \).

Teacher's Note:
1. Each vertex is the sum of two mid-points minus the third.
2. Compare squares: 52, 80, 148.

Answer: (c) \( (-6,\ 9-2\sqrt3) \)

Explanation:
1. First point: \( (1+5\sqrt2\cos135^\circ,\ 4+5\sqrt2\sin135^\circ)=(-4,9) \).
2. Second leg: \( (-4+4\cos(-120^\circ),\ 9+4\sin(-120^\circ))=(-4-2,\ 9-2\sqrt3) \).

Teacher's Note:
1. Use x = x1 + r cos(theta), y = y1 + r sin(theta) twice.
2. cos(-120) = -1/2 and sin(-120) = -sqrt3/2.

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

Explanation:
1. It passes through \( (-1,0) \) with slope \( \tan45^\circ=1 \): \( y=x+1 \), i.e. \( x-y+1=0 \).

Teacher's Note:
1. Use the point-slope form with the point (-1, 0).

Answer: (b) \( Y^2=\sqrt2\,X \)

Explanation:
1. A clockwise rotation by \( 45^\circ \) gives \( X=\frac{x-y}{\sqrt2},\ Y=\frac{x+y}{\sqrt2} \). (The xy-term vanishes only for \( \tan^2\theta=1 \), and the acute angle is \( 45^\circ \).)
2. Now \( x^2+2xy+y^2=(x+y)^2=2Y^2 \) and \( -2x+2y=-2(x-y)=-2\sqrt2X \).
3. The equation is \( 2Y^2-2\sqrt2X=0 \), i.e. \( Y^2=\sqrt2X \).

Teacher's Note:
1. Write the new coordinates for a clockwise rotation.
2. x + y and x - y are the key combinations.

Answer: (b) \( y=2x+1 \) or \( y=-2x+1 \)

Explanation:
1. The vertices are \( (k,k),\ (1,1),\ (2-k,k) \). The base along \( y=k \) has length \( |2-2k| \) and the height is \( |k-1| \).
2. Area \( =\frac12|2-2k||k-1|=(k-1)^2=4h^2 \), so \( k-1=\pm2h \).

Teacher's Note:
1. Find the three intersection points.
2. Take square roots with both signs.

Answer: (a) \( (5,10) \)

Explanation:
1. The orthocentre of the triangle is the incentre of the pedal triangle with vertices \( D(20,25),E(8,16),F(8,9) \). Its sides are \( EF=7,\ FD=20,\ DE=15 \).
2. Incentre \( =\frac{7D+20E+15F}{42}=\left(\frac{140+160+120}{42},\frac{175+320+135}{42}\right)=(10,15) \).
3. A side through a foot is perpendicular to the line joining that foot to \( H(10,15) \). \( AC \) passes through \( E(8,16) \) and is perpendicular to \( HE \) (slope \( -\frac12 \)): \( y=2x \). \( AB \) passes through \( F(8,9) \) and is perpendicular to \( HF \) (slope 3): \( x+3y=35 \).
4. These meet at \( A(5,10) \); similarly the other vertices are \( (50,-5) \) and \( (15,30) \).

Teacher's Note:
1. The pedal triangle's incentre is the orthocentre.
2. A side through a foot is perpendicular to the line joining the foot to the orthocentre.

Answer: (c) 18

Explanation:
1. Taking the lines as axes: \( |x|+|y|=3 \), a square with vertices \( (\pm3,0),(0,\pm3) \).
2. Its diagonals are 6 each: area \( =\frac12\cdot6\cdot6=18 \).

Teacher's Note:
1. The locus is |x| + |y| = 3.
2. It is a square (rhombus with equal diagonals).

Answer: (c) \( \tan^{-1}\frac17 \)

Explanation:
1. The lines are \( 3x-y+2=0 \) and \( 4x-2y+1=0 \), with slopes 3 and 2.
2. \( \tan\theta=\left|\frac{3-2}{1+6}\right|=\frac17 \).

Teacher's Note:
1. Factorise into two lines first.

Answer: (a) \( (7,8) \)

Explanation:
1. \( \left(\frac{0+5+16}3,\frac{0+12+12}3\right)=(7,8) \).

Teacher's Note:
1. Average the coordinates.

Answer: (b) \( 7x-4y+25=0,\ 7x-4y-3=0,\ 4x+7y-11=0 \)

Explanation:
1. The vertex \( (-3,1) \) lies on \( 4x+7y+5=0 \) (check: \( -12+7+5=0 \)). Two sides are perpendicular to this line: \( 7x-4y+\lambda=0 \).
2. Through \( (-3,1) \): \( -21-4+\lambda=0\Rightarrow\lambda=25 \). Through \( (1,1) \): \( 7-4+\lambda=0\Rightarrow\lambda=-3 \).
3. The fourth side is parallel to the given line through \( (1,1) \): \( 4x+7y-11=0 \).

Teacher's Note:
1. Opposite vertices: each lies on two of the sides.
2. Perpendicular lines have the form 7x - 4y + constant = 0.

Answer: (d) \( 4al+n=0 \)

Explanation:
1. The combined equation is \( 4alx^2+4amxy+ny^2=0 \); perpendicular lines need sum of the x^2 and y^2 coefficients zero: \( 4al+n=0 \).

Teacher's Note:
1. Use the condition a + b = 0 on the homogeneous equation.

Answer: (b) \( x+y=5 \) or \( x-7y+3=0 \)

Explanation:
1. Check the first line: \( x+y=5 \) meets \( 3x=y \) at \( Q=\left(\frac54,\frac{15}4\right) \); \( PQ=\sqrt{\left(\frac{11}4\right)^2+\left(\frac{11}4\right)^2}=\frac{11}{2\sqrt2} \).
2. Second line: \( x-7y+3=0 \) meets \( 3x=y \) at \( Q=\left(\frac3{20},\frac9{20}\right) \); \( PQ=\sqrt{\left(\frac{77}{20}\right)^2+\left(\frac{11}{20}\right)^2}=\frac{11\sqrt{50}}{20}=\frac{11}{2\sqrt2} \).
3. Parametric method: with direction angle \( \theta \), \( 3\cos\theta-\sin\theta=-2\sqrt2 \) leads to \( 7\tan^2\theta+6\tan\theta-1=0 \), so \( \tan\theta=-1 \) or \( \frac17 \), giving exactly these two lines.

Teacher's Note:
1. Write Q in parametric form from P at distance PQ.
2. A quadratic in tan theta gives two lines.

Answer: (c) \( \frac27a^2 \)

Explanation:
1. The pairs of parallel sides are \( 4y-3x=a,\ 3a \) (separation \( \frac{2a}5 \)) and \( 3y-4x=-a,\ -2a \) (separation \( \frac a5 \)).
2. The angle \( \theta \) between the families has \( \tan\theta=\left|\frac{\frac43-\frac34}{1+1}\right|=\frac7{24} \), so \( \sin\theta=\frac7{25} \).
3. Area \( =\frac{\frac{2a}5\cdot\frac a5}{\sin\theta}=\frac{2a^2}{25}\cdot\frac{25}7=\frac27a^2 \).

Teacher's Note:
1. Area = product of the two separations divided by sin of the angle between the families.
2. The slopes are 3/4 and 4/3.

Answer: (d) \( \lambda=2,\ \left(-\frac32,-\frac52\right) \)

Explanation:
1. A pair of lines needs \( abc+2fgh-af^2-bg^2-ch^2=0 \) with \( a=12,b=2,c=\lambda,h=-5,g=\frac{11}2,f=-\frac52 \), which gives \( \lambda=2 \).
2. The equation factors as \( (3x-y+2)(4x-2y+1) \). Solving \( 3x-y+2=0 \) and \( 4x-2y+1=0 \): \( x=-\frac32,\ y=-\frac52 \).

Teacher's Note:
1. Use the determinant condition for a pair of lines.
2. Factorise and intersect.

Answer: (b) \( 12x+6y-11=0 \)

Explanation:
1. At \( A \): \( \frac{x+y-5}{\sqrt2}=\pm\frac{7x+y+14}{5\sqrt2} \), giving \( 12x+6y-11=0 \) and \( 2x-4y+39=0 \).
2. \( 12x+6y-11 \) is \( 47 \) at \( B \) and \( -\frac{235}8 \) at \( C \): opposite signs, so it is the interior bisector.

Teacher's Note:
1. Check which bisector separates B and C.

Answer: (b) 15 square units

Explanation:
1. Going around the figure the vertices are \( (0,0),(6,0),(4,3),(0,3) \): the sides \( y=0 \) (length 6) and \( y=3 \) (length 4) are parallel, at distance 3.
2. A trapezium with parallel sides \( 6 \) and \( 4 \) and height \( 3 \) has area \( \frac12(6+4)\cdot3=15 \).

Teacher's Note:
1. Order the vertices around the figure first.
2. It is a trapezium with parallel sides along y = 0 and y = 3.

Answer: (b) \( \left(\frac{13}{5},\frac{14}{5}\right) \)

Explanation:
1. Using \( 3x-y-4 \) at \( (2,3) \) equal to \( -1 \): image \( =(2,3)-2\cdot\frac{(-1)}{10}(3,-1)=\left(2+\frac35,\,3-\frac15\right)=\left(\frac{13}5,\frac{14}5\right) \).

Teacher's Note:
1. Image = point minus 2 (value/(a^2+b^2)) (a, b).
2. It is the reflection of the point across the foot.

Answer: (d) \( (y_1^2-\delta^2)x^2-2x_1y_1xy+(x_1^2-\delta^2)y^2=0 \)

Explanation:
1. Take any point \( (x,y) \) on one of the lines. The line through the origin and \( (x,y) \) has equation \( yX-xY=0 \), and the distance of \( (x_1,y_1) \) from it is \( \frac{|x_1y-y_1x|}{\sqrt{x^2+y^2}}=\delta \).
2. So \( (x_1y-xy_1)^2=\delta^2(x^2+y^2) \).
3. Expanding: \( (y_1^2-\delta^2)x^2-2x_1y_1xy+(x_1^2-\delta^2)y^2=0 \).

Teacher's Note:
1. A line through the origin and a point (x, y) is y X - x Y = 0.
2. Expand the squared distance condition.

Answer: (a) 3 square units

Explanation:
1. The other sides are \( x-y=0 \) and \( 2x-y=0 \).
2. Area \( =\frac{|(c_1-c_2)(d_1-d_2)|}{|a_1b_2-a_2b_1|} \) with the pairs \( x-y+3,\ x-y \) and \( 2x-y+1,\ 2x-y \): \( \frac{3\cdot1}{|1\cdot(-1)-2\cdot(-1)|}=\frac{3}{1}=3 \).

Teacher's Note:
1. Opposite sides are parallel with the same coefficients of x and y.
2. Use the parallelogram area formula for line pairs.

Answer: (a) 7

Explanation:
1. The lines \( x=3 \) and \( y=4 \) are perpendicular and meet at \( (3,4) \), so the triangle is right-angled there.
2. The orthocentre of a right-angled triangle is the right-angle vertex \( (3,4) \): sum 7.

Teacher's Note:
1. Look at the angle between x = 3 and y = 4.

Answer: (d) \( \frac{1+\sqrt{10}}3 \)

Explanation:
1. The incident ray has slope 2 and the reflected ray slope 1. If the mirror has slope \( m \), the normal has slope \( -\frac1m \) and both rays make equal angles with it.
2. With the slope of the incident ray 2, of the reflected ray 1 and of the mirror \( m \), the mirror's inclination is the mean of their directions, which gives \( \frac{m-2}{1+2m}=\frac{1-m}{1+m} \) in magnitude, i.e. \( 3m^2-2m-3=0 \) and \( m=\frac{1\pm\sqrt{10}}3 \).
3. The mirror must make the reflected ray leave in the forward direction (towards B); only \( m=\frac{1+\sqrt{10}}3 \), which lies between 1 and 2, does this (the other root reverses the ray).

Teacher's Note:
1. Equal angles of incidence and reflection with the normal.
2. Check that the reflected direction points towards B, not away.

Answer: (d) inside the triangle

Explanation:
1. The vertices are \( (-1,-2),\ (-1,\frac83),\ (3,1) \).
2. For each side, the origin and the opposite vertex give expressions of the same sign: \( x+1 \): \( +1 \) and \( +4 \); \( 3x-4y-5 \): \( -5 \) and \( -\frac{56}{3} \); \( 5x+12y-27 \): \( -27 \) and \( -56 \).
3. So the origin is on the same side as the opposite vertex for all three sides: inside.

Teacher's Note:
1. Compare the sign for the origin with the sign for the opposite vertex.
2. Inside means all three signs match.

Answer: (b) \( 4alx^2+4amxy+ny^2=0 \)

Explanation:
1. Write \( \frac{lx+my}{-n}=1 \) and homogenise: \( y^2=4ax\cdot\frac{lx+my}{-n} \).
2. \( -ny^2=4alx^2+4amxy\Rightarrow4alx^2+4amxy+ny^2=0 \).

Teacher's Note:
1. Make the curve homogeneous using the line.

Answer: (d) \( (0,0) \)

Explanation:
1. \( m_1m_2=-1 \), so the two sides are perpendicular and meet at the origin.
2. The triangle is right-angled at the origin, which is therefore its orthocentre.

Teacher's Note:
1. Product of the roots is -1.

Answer: (c) \( \frac\pi4 \)

Explanation:
1. The line meets the axes at \( A(-5,0) \) and \( B(0,-10) \). The points dividing \( AB \) in parts \( 1:2:2 \) are \( P=A+\frac15(B-A)=(-4,-2) \) and \( Q=A+\frac35(B-A)=(-2,-6) \).
2. Slopes from \( M \): \( MP=\frac{-2+8}{-4+6}=3 \), \( MQ=\frac{-6+8}{-2+6}=\frac12 \).
3. \( \tan\theta=\left|\frac{3-\frac12}{1+\frac32}\right|=1\Rightarrow\theta=\frac\pi4 \).

Teacher's Note:
1. Find the axis intercepts, then the two division points.
2. Use the tangent of the angle between lines.

Answer: (b) \( x-5y+3=0 \) and \( 5x+y-11=0 \)

Explanation:
1. For slope \( m \): \( \left|\frac{m+\frac23}{1-\frac{2m}3}\right|=\tan45^\circ=1 \), i.e. \( \frac{3m+2}{3-2m}=\pm1 \).
2. \( m=\frac15 \) or \( m=-5 \). Lines: \( y-1=\frac15(x-2) \Rightarrow x-5y+3=0 \) and \( y-1=-5(x-2)\Rightarrow5x+y-11=0 \).

Teacher's Note:
1. The slope of 2x + 3y + 4 = 0 is -2/3.
2. Use the tangent of the angle between two lines.

Answer: (b) \( 6x-8y+7=0 \)

Explanation:
1. \( A=(3,0),\ B=(0,4) \). The circumcentre of \( ABQ \) is equidistant from \( A \) and \( B \), so it lies on the perpendicular bisector of \( AB \).
2. Mid-point of \( AB \) is \( \left(\frac32,2\right) \), slope of \( AB \) is \( -\frac43 \), so the bisector has slope \( \frac34 \): \( y-2=\frac34\left(x-\frac32\right)\Rightarrow6x-8y+7=0 \).

Teacher's Note:
1. The circumcentre lies on the perpendicular bisector of any chord.
2. AB is fixed, so the bisector is fixed.

Answer: (a) \( 8x+8y+7=0 \)

Explanation:
1. The bisectors at \( C \) are \( \frac{x+7y-7}{5\sqrt2}=\pm\frac{7x+y+14}{5\sqrt2} \): \( 8x+8y+7=0 \) and \( 2x-2y+7=0 \) (up to signs).
2. \( 8x+8y+7 \) has the value 47 at both \( A \) and \( B \) (same sign), so it does not separate them: it is the exterior bisector.

Teacher's Note:
1. The exterior bisector does not separate the other two vertices.
2. Substitute A and B.

Answer: (d) \( 1:2 \)

Explanation:
1. Let \( QP:PR=\lambda:1 \). Abscissa: \( 2a\lambda+(2a-3)=(2a-2)(\lambda+1) \Rightarrow 2\lambda=1 \).
2. So \( \lambda=\frac12 \) and the ratio is \( 1:2 \). The ordinates give the same value.

Teacher's Note:
1. Write the ratio as lambda : 1 and use the x-coordinate.
2. Check with the y-coordinate.

Answer: (c) \( \left(-\frac16,-\frac{23}9\right) \)

Explanation:
1. Vertices: \( A=(-4,0) \) (second and third lines), \( B=(4,3) \), \( C=(0,-3) \).
2. Altitude from \( A \) is perpendicular to \( BC \) (slope \( \frac32 \)): \( 2x+3y+8=0 \). Altitude from \( B \) is perpendicular to \( AC \) (slope \( -\frac34 \)): \( 4x-3y-7=0 \).
3. Adding the two equations: \( 6x+1=0 \), so \( x=-\frac16 \); then \( 3y=-8+\frac13 \), so \( y=-\frac{23}9 \).

Teacher's Note:
1. Find the vertices first.
2. Intersect two altitudes.