JEE Mathematics Coordinates and Straight Lines MCQs Set 15

Answer: (d) \( K\le2a\ \text{or}\ K\ge2b \)

Explanation:
1. Here \( h=\lambda,\ g=f=K,\ c=2K \). The condition \( abc+2fgh-af^2-bg^2-ch^2=0 \) gives \( 2abK+2K^2\lambda-aK^2-bK^2-2K\lambda^2=0 \).
2. For \( K\ne0 \): \( 2\lambda^2-2K\lambda+(a+b)K-2ab=0 \). A real \( \lambda \) needs \( 4K^2-8[(a+b)K-2ab]\ge0 \), i.e. \( K^2-2(a+b)K+4ab\ge0 \).
3. The roots are \( K=2a \) and \( K=2b \), so \( K\le2a \) or \( K\ge2b \).

Teacher's Note:
1. Use the determinant condition for a second-degree equation to represent a pair of lines.
2. Treat the condition as a quadratic in the free parameter and require a non-negative discriminant.

Answer: (a) \( x^2+y^2=13 \)

Explanation:
1. Let the terms be \( a,\ a+2d,\ a+6d \) (that is, \( a,b,c \)). Then \( c-a=3(b-a) \), so \( 2a-3b+c=0 \).
2. This says the point \( (2,-3) \) satisfies \( ax+by+c=0 \) for every such triple, so the fixed point is \( (2,-3) \).
3. \( 2^2+(-3)^2=13 \), so it lies on \( x^2+y^2=13 \).

Teacher's Note:
1. When coefficients satisfy a linear relation, read the fixed point from the relation.
2. Check the fixed point against each option.

Answer: (b) 20

Explanation:
1. The line is \( ax+by+c=1 \), and the given condition is \( 5a+5b+20c=t \).
2. For a fixed point \( (x_0,y_0) \) we need \( (x_0,y_0,1) \) proportional to \( (5,5,20) \), i.e. \( (x_0,y_0,1)=\tfrac1{20}(5,5,20)=\left(\tfrac14,\tfrac14,1\right) \).
3. Substituting \( ax_0+by_0+c=\dfrac{5a+5b+20c}{20}=\dfrac t{20}=1 \) gives \( t=20 \).

Teacher's Note:
1. A line \( ax+by+c=1 \) with a linear condition on \( a,b,c \) passes through a fixed point only for a special value of the constant.
2. Compare the coefficient ratios with \( (x,y,1) \).

Answer: (c) \( \frac53\le\beta\le\frac72 \)

Explanation:
1. The vertices are \( (-1,1) \), \( (-2,4) \) and \( (2,3) \).
2. The line \( x=0 \) cuts the side \( 3y-2x-5=0 \) at \( y=\tfrac53 \) and the side \( 4y+x-14=0 \) at \( y=\tfrac72 \); it does not meet the third side within the triangle.
3. So \( \tfrac53\le\beta\le\tfrac72 \).

Teacher's Note:
1. Find the vertices, then see which two sides the vertical line x = 0 crosses.
2. Sketch the triangle to confirm the entry and exit points.

Answer: (c) \( y=-x \)

Explanation:
1. \( (a-b)^2=c^2 \), so \( c=a-b \) or \( c=b-a \).
2. If \( c=a-b \): \( ax+by+a-b=a(x+1)+b(y-1)=0 \), so the point is \( (-1,1) \). If \( c=b-a \): \( a(x-1)+b(y+1)=0 \), so the point is \( (1,-1) \).
3. Both points satisfy \( y=-x \).

Teacher's Note:
1. Both signs of c give a fixed point.
2. Substitute each fixed point into the options.

Answer: (b) \( l,n,m \) are in G.P.

Explanation:
1. The four lines form a rhombus with diagonals along the axes, with intercepts \( \dfrac nl \) on the \( x \)-axis and \( \dfrac nm \) on the \( y \)-axis (taking positive values).
2. Area \( =\dfrac12(2n/l)(2n/m)=\dfrac{2n^2}{lm} \). Setting this equal to 2 gives \( n^2=lm \).
3. So \( l,n,m \) are in G.P.

Teacher's Note:
1. The diagonals of the rhombus lie along the coordinate axes.
2. Half the product of the diagonals gives the area.

Answer: (a) i, iii and iv only

Explanation:
1. \( \sec^2\ge1 \), so \( a^2-1\le-1 \) forces \( a=0 \) and \( \sec^2(2b)=1 \), i.e. \( 2b=k\pi \): \( b\in\{-\tfrac\pi2,0,\tfrac\pi2\} \).
2. A line of slope \( \tfrac12 \) through \( (0,b) \) is \( x-2y+2b=0 \), giving \( x-2y=0 \), \( x-2y+\pi=0 \) and \( x-2y-\pi=0 \).
3. \( x-2y=1 \) does not occur, so (i), (iii) and (iv) are correct.

Teacher's Note:
1. Use \( \sec^2\theta\ge1 \) to pin down a immediately.
2. The values of b come from \( \tan^2=0 \), i.e. \( \sec^2=1 \).

Answer: (b) ii and iv only

Explanation:
1. The pair-of-lines condition with \( h=2\lambda,\ g=f=2a,\ c=4a \) gives \( 4a\,pq+16a^2\lambda-4a^2(p+q)-16a\lambda^2=0 \).
2. Dividing by \( 4a \): \( 4\lambda^2-4a\lambda+a(p+q)-pq=0 \). A real \( \lambda \) needs \( 16a^2-16[a(p+q)-pq]\ge0 \), i.e. \( (a-p)(a-q)\ge0 \).
3. So \( a\le p \) or \( a\ge q \): intervals (ii) and (iv) lie within this set.

Teacher's Note:
1. The existence of a real parameter value is a discriminant condition.
2. Check each interval against the solution set \( (-\infty,p]\cup[q,\infty) \).

Answer: (d) i, ii and iv only

Explanation:
1. Substituting the point for parameter \( t \): \( lt^3+m(t^2-3)+n(t-1)=0 \), i.e. \( lt^3+mt^2+nt-(3m+n)=0 \).
2. \( a,b,c \) are its three roots: \( a+b+c=-\dfrac ml \), \( ab+bc+ca=\dfrac nl \), \( abc=\dfrac{3m+n}{l} \).
3. So (i) and (ii) hold and (iii) is false. For (iv): \( \dfrac{3m+n}{l}-\dfrac nl-\dfrac{3m}{l}=0 \), which is true.

Teacher's Note:
1. Turn the collinearity condition into a cubic whose roots are a, b, c.
2. Use Vieta's relations.

Answer: (b) i, ii and iii only

Explanation:
1. The vertices are \( O \), \( \left(\dfrac a{m_1},a\right) \) and \( \left(\dfrac a{m_2},a\right) \). Area \( =\dfrac12\cdot|a|\cdot a\left|\dfrac1{m_1}-\dfrac1{m_2}\right|=\dfrac{a^2}{2}\dfrac{|m_1-m_2|}{|m_1m_2|} \).
2. \( m_1+m_2=a \), \( m_1m_2=-(a+1) \), so \( |m_1-m_2|=\sqrt{a^2+4a+4}=|a+2| \).
3. Area \( =\dfrac{a^2|a+2|}{2|a+1|} \). Removing the modulus signs gives the expressions in (i), (ii), (iii) in the stated ranges; the area is not zero.

Teacher's Note:
1. Remove modulus signs separately in each interval of a.
2. The base is the horizontal segment on y = a.

Answer: (d) 5

Explanation:
1. For a pair of lines through no common point with origin condition, lines are equidistant from the origin iff \( f^4-g^4=c(bf^2-ag^2) \).
2. So \( \dfrac{f^4-g^4}{bf^2-ag^2}=c=10 \).
3. \( 2s=10 \), so \( s=5 \).

Teacher's Note:
1. This is a standard result for a pair of lines equidistant from the origin.
2. The constant term c is the value of the ratio.

Answer: (b) 3

Explanation:
1. At the origin the expression equals \( 1>0 \).
2. The point \( (1,1) \) gives \( a^2+ab+1 \), which must be positive for all \( a\ne0 \): the discriminant \( b^2-4<0 \) is needed (at \( b=\pm2 \) it vanishes for some \( a \), putting the point on the line).
3. \( |b|<2 \) gives the integers \( -1,0,1 \): three values.

Teacher's Note:
1. A quadratic in a is positive for all a if its discriminant is negative.
2. Check the boundary cases b = 2 and b = -2 separately.

Answer: (b) 2

Explanation:
1. Let the intercepts be \( p \) and \( q \): \( \dfrac8p+\dfrac2q=1 \). Then \( p+q=(p+q)\left(\dfrac8p+\dfrac2q\right)=10+\dfrac{8q}p+\dfrac{2p}q \).
2. By AM-GM, \( \dfrac{8q}p+\dfrac{2p}q\ge2\sqrt{16}=8 \), so \( p+q\ge18 \).
3. So \( 9t=18 \) and \( t=2 \).

Teacher's Note:
1. Use the intercept form and multiply by the sum to apply AM-GM.
2. Equality holds at p = 12, q = 6.

Answer: (a) 0

Explanation:
1. At \( (1,4) \): \( \sqrt3-4+1=\sqrt3-3<0 \). So we need \( \sqrt3\sin\theta-\cos\theta+1\le0 \).
2. \( 2\sin\left(\theta-\dfrac\pi6\right)\le-1 \), so \( \dfrac{7\pi}6\le\theta-\dfrac\pi6\le\dfrac{11\pi}6 \) and \( \theta\in\left[\dfrac{4\pi}3,2\pi\right] \).
3. On this interval \( \sin\theta\le0 \), with maximum \( 0 \) at \( \theta=2\pi \).

Teacher's Note:
1. Rewrite \( \sqrt3\sin\theta-\cos\theta \) as \( 2\sin(\theta-\pi/6) \).
2. The sine is negative throughout the allowed interval except at its end.

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

Explanation:
1. \( a,b,c \) in A.P. means \( a-2b+c=0 \), so every line of the family passes through \( (1,-2) \). Both \( AB \) and \( BC \) pass through it, so \( B=(1,-2) \).
2. Check: \( AB \) is the line \( x=1 \) through \( A(1,1) \) and \( B(1,-2) \).

Teacher's Note:
1. A family of lines whose coefficients are in A.P. is concurrent at \( (1,-2) \).
2. Both sides contain the common point, which must be the vertex B.

Answer: (a) obtuse angled triangle

Explanation:
1. \( a,b,c \) in A.P. means \( a-2b+c=0 \), so every line of the family passes through \( (1,-2) \). Both \( AB \) and \( BC \) pass through it, so \( B=(1,-2) \).
2. The altitude from \( A \) passes through \( (2,4) \) with slope 3, so \( BC \) has slope \( -\tfrac13 \): \( y+2=-\tfrac13(x-1) \). The altitude from \( C \) is perpendicular to \( AB \) (the line \( x=1 \)), so it is \( y=4 \), giving \( C=(-17,4) \).
3. \( \overrightarrow{AB}\cdot\overrightarrow{AC}=(0,-3)\cdot(-18,3)=-9<0 \), so angle \( A \) is obtuse.

Teacher's Note:
1. The orthocentre is the intersection of the three altitudes.
2. An obtuse angle has a negative dot product of the two side vectors.

Answer: (c) \( \frac{20}3 \)

Explanation:
1. The angle \( \varphi \) between the diagonals: slopes \( -\tfrac12 \) and \( -2 \) give \( \tan\varphi=\left|\dfrac{-\frac12+2}{1+1}\right|=\dfrac34 \), so \( \sin\varphi=\dfrac35 \).
2. Area \( =\tfrac12\cdot AC\cdot BD\cdot\sin\varphi=\tfrac12\cdot4\cdot BD\cdot\dfrac35=8 \).
3. \( BD=\dfrac{20}{3} \).

Teacher's Note:
1. Area of a quadrilateral with diagonals d1, d2 at angle φ is \( \tfrac12d_1d_2\sin\varphi \).
2. Use the slopes to find the angle between the diagonals.