JEE Mathematics Straight Lines MCQs Set 20

Answer: (a) \( (2a, 2b) \)

Explanation:
1. If the orthocentre coincides with vertex B, then \( \angle B=90^\circ \), so the circumcentre is the midpoint of the hypotenuse AC.
2. \( \frac{x+0}2=a,\ \frac{y+0}2=b \) gives \( C=(2a,2b) \).

Teacher's Note:
1. A right angle at B puts the orthocentre at B.
2. The circumcentre of a right triangle is the midpoint of the hypotenuse.

Answer: (d) none of these

Explanation:
1. In an orthocentric system A, B, C, H each is the orthocentre of the triangle formed by the other three.
2. So the orthocentre of \( \triangle BCH \) is \( A=(-1,0) \), which is not among the listed points.

Teacher's Note:
1. Use the orthocentric property of four points.
2. Compare the point obtained with the options.

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

Explanation:
1. Let the centroid be \( (h,k) \) and \( A(x_1,y_1) \): \( h=\frac{1+4+x_1}3,\ k=\frac{1+2+y_1}3 \), so \( x_1=3h-5,\ y_1=3k-3 \).
2. A lies on \( y=x+3 \): \( 3k-3=3h-5+3 \), i.e. \( 3h-3k+1=0 \).

Teacher's Note:
1. Express the vertex A through the centroid coordinates.
2. Substitute into the given line.

Answer: (c) \( \frac1{x^2}+\frac1{y^2}=\frac4{p^2} \)

Explanation:
1. The intercepts are \( A\left(\frac p{\cos\alpha},0\right) \) and \( B\left(0,\frac p{\sin\alpha}\right) \). The mid-point is \( (h,k) \) with \( 2h=\frac p{\cos\alpha},\ 2k=\frac p{\sin\alpha} \).
2. \( \cos\alpha=\frac p{2h},\ \sin\alpha=\frac p{2k} \); squaring and adding gives \( \frac{p^2}{4h^2}+\frac{p^2}{4k^2}=1 \), i.e. \( \frac1{h^2}+\frac1{k^2}=\frac4{p^2} \).

Teacher's Note:
1. Find the axis intercepts of the line.
2. Eliminate alpha using sin^2 + cos^2 = 1.

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

Explanation:
1. For the point \( (h,k) \) the distances from the axes are \( |k| \) and \( |h| \).
2. The condition gives \( h^2+k^2=3 \), so the locus is \( x^2+y^2=3 \).

Teacher's Note:
1. Distance from the x-axis is |y| and from the y-axis is |x|.
2. Square and add.

Answer: (b) \( bx+ay-3xy=0 \)

Explanation:
1. Line: \( y-b=m(x-a) \) meets the axes at \( A\left(a-\frac bm,0\right) \) and \( B(0,b-am) \).
2. Centroid \( (h,k) \): \( 3h=a-\frac bm,\ 3k=b-am \). Then \( (a-3h)(b-3k)=\frac bm\cdot am=ab \).
3. Expanding: \( -3ak-3bh+9hk=0 \), so the locus is \( bx+ay-3xy=0 \).

Teacher's Note:
1. Write the intercepts in terms of the slope m.
2. Eliminate m by multiplying the two centroid relations.

Answer: (b) a straight line

Explanation:
1. With base AB fixed and the area constant, the height from C is constant.
2. So C moves on a line parallel to AB (two such lines, one on each side), i.e. the locus is a straight line.

Teacher's Note:
1. Area = half base times height, and the base is fixed.
2. Constant distance from a fixed line gives lines parallel to it.

Answer: (c) \( \frac12\left(a_2^2+b_2^2-a_1^2-b_1^2\right) \)

Explanation:
1. Equate squared distances: \( (x-a_1)^2+(y-b_1)^2=(x-a_2)^2+(y-b_2)^2 \).
2. This gives \( 2(a_1-a_2)x+2(b_1-b_2)y+a_2^2+b_2^2-a_1^2-b_1^2=0 \).
3. Dividing by 2 and comparing: \( c=\tfrac12(a_2^2+b_2^2-a_1^2-b_1^2) \).

Teacher's Note:
1. The locus is the perpendicular bisector of the segment.
2. Expand both squared distances and cancel x^2, y^2.

Answer: (b) \( 2x-16y-5=0 \)

Explanation:
1. Distance of \( (-11,4) \) from \( 2x-16y-5=0 \): \( \frac{|-22-64-5|}{\sqrt{260}}=\frac{91}{\sqrt{260}} \).
2. Distance from \( 64x+8y+35=0 \): \( \frac{|-704+32+35|}{\sqrt{4160}}=\frac{637}{\sqrt{4160}}=\frac{159.25}{\sqrt{260}} \).
3. A point of L1 is nearer to the acute-angle bisector (it makes the smaller angle with L1). Since \( 91<159.25 \), the acute angle bisector is \( 2x-16y-5=0 \).

Teacher's Note:
1. Compare the distances of a point of L1 from the two bisectors.
2. The nearer bisector is the acute angle bisector.

Answer: (b) \( 2x+y-12=0 \)

Explanation:
1. \( AB=\sqrt{10} \) and \( AC=\sqrt{10} \), so the bisector from A meets BC at its mid-point \( D(4,4) \).
2. Slope of AD \( =\frac{4-2}{4-5}=-2 \), so the line is \( y-2=-2(x-5) \), i.e. \( 2x+y-12=0 \).

Teacher's Note:
1. Triangle is isosceles with AB = AC, so the bisector hits the mid-point of BC.
2. Use the two points A and D.

Answer: (a) \( 21x+27y-121=0 \)

Explanation:
1. At \( (-1,4) \): \( 3x-4y+12=-7<0 \) and \( 12x-5y+7=-25<0 \), both negative.
2. The bisector of the angle containing the point has the same signs: \( \frac{3x-4y+12}5=\frac{12x-5y+7}{13} \).
3. Cross-multiplying: \( 13(3x-4y+12)=5(12x-5y+7) \) gives \( 21x+27y-121=0 \).

Teacher's Note:
1. Evaluate both expressions at the given point.
2. Equal signs mean the plus sign is used in the bisector formula.

Answer: (a) 8

Explanation:
1. The first two lines meet at \( (t+a,-16) \). Concurrency requires \( -16=a(t+a) \), i.e. \( a^2+ta+16=0 \).
2. For real a: \( t^2-64\ge0 \), so the least positive t is \( 8 \).

Teacher's Note:
1. Substitute the common point into the third line.
2. Treat it as a quadratic in a with a non-negative discriminant.

Answer: (c) \( 5x-2y-7=0 \)

Explanation:
1. The first family passes through the intersection of \( 5x+3y-2=0 \) and \( 3x-y-4=0 \), which is \( (1,-1) \).
2. The second passes through the intersection of \( x-y+1=0 \) and \( 2x-y-2=0 \), which is \( (3,4) \).
3. The common member is the line through both points: slope \( \frac{4+1}{3-1}=\frac52 \), so \( y-4=\tfrac52(x-3) \), i.e. \( 5x-2y-7=0 \).

Teacher's Note:
1. Find the fixed point of each family.
2. The common line joins the two fixed points.

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

Explanation:
1. Write the line as \( (2+\lambda)x+(1-2\lambda)y+(4-3\lambda)=0 \).
2. Distance from \( (2,-3) \): \( \frac{|5\lambda+5|}{\sqrt{5\lambda^2+5}}=\sqrt{10} \) gives \( 25(\lambda+1)^2=50(\lambda^2+1) \), so \( (\lambda-1)^2=0 \) and \( \lambda=1 \).
3. The line is \( 3x-y+1=0 \).

Teacher's Note:
1. Write the family in general form and apply the point to line distance.
2. The quadratic in lambda is a perfect square.

Answer: (b) \( \left(\frac34,\frac12\right) \)

Explanation:
1. Divide the condition by 4: \( \frac34a+\frac12b+c=0 \).
2. This says \( ax+by+c=0 \) is satisfied by \( (x,y)=\left(\frac34,\frac12\right) \) for every such line.

Teacher's Note:
1. Rewrite the condition so it looks like a x + b y + c = 0.
2. Read off the common point.

Answer: (d) \( 2x-3y+10=0 \)

Explanation:
1. The intercept line is \( \frac x{-2}+\frac y{-3}=1 \), i.e. \( 3x+2y+6=0 \) with slope \( -\frac32 \).
2. Family: \( (x+y-5)+\lambda(x-y+3)=0 \) with slope \( -\frac{1+\lambda}{1-\lambda} \).
3. Perpendicular: \( \frac{3(1+\lambda)}{2(1-\lambda)}=-1 \) gives \( \lambda=-5 \).
4. Line: \( -4x+6y-20=0 \), i.e. \( 2x-3y+10=0 \).

Teacher's Note:
1. Find the slope of the line given by its intercepts.
2. Use the product of slopes equal to -1 on the family member.

Answer: (a) below the x-axis at a distance of 3/2 from it

Explanation:
1. Family: \( (a+b\lambda)x+(2b-2a\lambda)y+3b-3a\lambda=0 \). For a horizontal line, \( a+b\lambda=0\Rightarrow\lambda=-\frac ab \).
2. Then \( y\left(2b+\frac{2a^2}b\right)+3b+\frac{3a^2}b=0 \), i.e. \( y\cdot2(a^2+b^2)=-3(a^2+b^2) \).
3. So \( y=-\frac32 \): below the x-axis at distance 3/2.

Teacher's Note:
1. Make the coefficient of x vanish to get a horizontal line.
2. Cancel a^2 + b^2, which is non-zero.

Answer: (a) 16

Explanation:
1. Slopes satisfy \( m_1+m_2=-10,\ m_1m_2=a \).
2. With \( m_1=4m_2 \): \( 5m_2=-10\Rightarrow m_2=-2,\ m_1=-8 \).
3. So \( a=m_1m_2=16 \).

Teacher's Note:
1. The slopes are roots of m^2 + 10m + a = 0.
2. Use sum and product of roots.

Answer: (a) 2

Explanation:
1. Since \( x^2+2\sqrt2xy+2y^2=(x+\sqrt2y)^2 \), factor as \( (x+\sqrt2y+p)(x+\sqrt2y+q)=0 \) with \( p+q=4,\ pq=1 \).
2. Distance between the parallel lines: \( \frac{|p-q|}{\sqrt{1+2}}=\frac{\sqrt{16-4}}{\sqrt3}=2 \).

Teacher's Note:
1. The quadratic part is a perfect square, so the lines are parallel.
2. Use |p - q| over the norm of the normal vector.

Answer: (b) \( \frac\pi2 \)

Explanation:
1. Homogenise using \( 2x+y=1 \): \( 3x^2+4xy-4x(2x+y)+(2x+y)^2=0 \).
2. Expanding: \( 3x^2+4xy-8x^2-4xy+4x^2+4xy+y^2=0 \), i.e. \( -x^2+4xy+y^2=0 \).
3. Coefficient of \( x^2 \) plus coefficient of \( y^2 \) is \( -1+1=0 \), so the lines are perpendicular: angle \( \frac\pi2 \).

Teacher's Note:
1. Make the curve homogeneous with the line equation.
2. Perpendicular lines have a + b = 0.