JEE Mathematics Straight Lines MCQs Set 28

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

Explanation:
1. \(a+b+c=3,\ ab+bc+ca=6,\ abc=-1\).
2. \(\dfrac1{ab}=\dfrac c{abc}=-c\), so the y-coordinate is \(-\dfrac{a+b+c}3=-1\); the x-coordinate is \(\dfrac{ab+bc+ca}3=2\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (d) \(\left(\dfrac{31}7,\dfrac{31}7\right)\)

Explanation:
1. \(A,B\) are on the same side of \(y=x\). Reflect \(B\) to \(B'(13,7)\).
2. Line \(AB'\): \(y-4=\dfrac3{10}(x-3)\). Setting \(y=x\): \(0.7x=3.1\), \(x=\dfrac{31}7\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (c) \(\left(\dfrac{2b}3,\dfrac{2b}3\right)\)

Explanation:
1. \(C=(c,c)\) on \(AB\): \(c\left(\dfrac1a+\dfrac1b\right)=1\), so \(c=\dfrac{ab}{a+b}\).
2. \(\dfrac{[AOC]}{[BOC]}=\dfrac{a}{b}=2\) (same height \(c\)), so \(a=2b\) and \(c=\dfrac{2b^2}{3b}=\dfrac{2b}3\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (b) \(\dfrac53\le\beta\le\dfrac72\)

Explanation:
1. Vertices: \((-1,1),\ (2,3),\ (-2,4)\).
2. The y-axis meets the sides at \(y=-2\), \(y=\tfrac53\), \(y=\tfrac72\); the y-axis cuts the triangle in the segment from the side \(3y-2x-5=0\) to the side \(4y+x-14=0\).
3. So \(\dfrac53\le\beta\le\dfrac72\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (d) \((-\infty,-\tfrac43)\cup(\tfrac43,\infty)\)

Explanation:
1. The points are \(\left(\dfrac2m,2\right)\) and \(\left(\dfrac6m,6\right)\), distance \(\dfrac{4\sqrt{1+m^2}}{|m|}\).
2. \(16(1+m^2)<25m^2\) gives \(m^2>\dfrac{16}9\), i.e. \(|m|>\dfrac43\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (b) \(3x+4y+3=0\)

Explanation:
1. For \(C(x,y)\), the centroid is \(\left(\dfrac{x-5}3,\dfrac{y+3}3\right)\).
2. \(3(x-5)+4(y+3)+6=0\) gives \(3x+4y+3=0\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (c) \((a-1)^2x-(a+1)^2y=0\)

Explanation:
1. \(G=\left(\dfrac{(a+1)^2}2,\dfrac{(a-1)^2}2\right)\).
2. With circumcentre \(O=(0,0)\), \(H=3G-2O=3G\), so \(H=\left(\tfrac32(a+1)^2,\tfrac32(a-1)^2\right)\) satisfies \((a-1)^2x=(a+1)^2y\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (a) \(ay-bx+2b=0\)

Explanation:
1. It cuts the x-axis at \((2,0)\).
2. A line perpendicular to \(ax+by+c=0\) is \(bx-ay+k=0\); through \((2,0)\): \(k=-2b\), giving \(bx-ay-2b=0\), i.e. \(ay-bx+2b=0\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (d) \(\left(-\dfrac12,3\right)\)

Explanation:
1. The two expressions at \((a,2)\) are \(a-3\) and \(2a+1\).
2. A point between the lines gives opposite signs: \((a-3)(2a+1)<0\), so \(-\tfrac12<a<3\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (c) \((-4,-7)\)

Explanation:
1. Let \(C=(x,y)\). \(OA\perp BC\): \(5(x+2)-(y-3)=0\), i.e. \(5x-y+13=0\).
2. \(OB\perp AC\): \(-2(x-5)+3(y+1)=0\), i.e. \(2x-3y-13=0\). Solving gives \(C=(-4,-7)\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (b) \(x(\sqrt{13}+2\sqrt2)+y(\sqrt{13}-3\sqrt2)+(\sqrt{13}-5\sqrt2)=0\)

Explanation:
1. At \((10,-20)\): \(x+y+1=-9\) and \(2x-3y-5=75\) have opposite signs.
2. For an angle containing such a point the bisector is \(\dfrac{x+y+1}{\sqrt2}=-\dfrac{2x-3y-5}{\sqrt{13}}\), i.e. \(\sqrt{13}(x+y+1)+\sqrt2(2x-3y-5)=0\), which is the first option.

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

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

Explanation:
1. Bisectors: \(\dfrac{4x-3y+7}5=\pm\dfrac{3x-4y+14}5\) give \(x+y-7=0\) and \(x-y+3=0\).
2. Both slopes are positive (\(\tfrac43,\tfrac34\)), so the acute angle lies around the line of slope 1: \(x-y+3=0\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (c) 7

Explanation:
1. Distances of \(O\) from the lines: \(\dfrac9{\sqrt{20}}\) and \(\dfrac6{\sqrt5}=\dfrac{12}{\sqrt{20}}\).
2. The lines lie on opposite sides of \(O\), so \(OP:OQ=9:12=3:4\) and \(m+n=7\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (a) 0

Explanation:
1. At \((1,4)\): \(\sqrt3-4+1<0\). So \(\sqrt3\sin\theta-\cos\theta+1\le0\), i.e. \(\sin\left(\theta-\dfrac\pi6\right)\le-\dfrac12\).
2. This gives \(\theta=0\) or \(\theta\in\left[\dfrac{4\pi}3,2\pi\right]\), where \(\sin\theta\le0\); the maximum is 0.

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (c) 1

Explanation:
1. \(L_2\) is the reflection of \(L_1\) in \(L\). Slope of \(L_1\) is \(-\tfrac12\), and \(L\) has inclination \(135^\circ\).
2. Inclination of \(L_2=270^\circ-\theta_1\), so \(m=\cot\theta_1=-2\).
3. \(812\cdot4+3=3251\): unit digit 1.

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (d) 5

Explanation:
1. \(A=AB\cap CA=(1,-2)\); \(AP=(1,5)\perp BC\) so the normal \((1,p)\parallel(1,5)\) and \(p=5\).
2. \(B=(b,-2b)\) on \(AB\) with \(BP\perp CA\): \((2-b)+(3+2b)=0\), \(b=-5\), \(B=(-5,10)\).
3. \(B\) on \(x+5y=q\): \(q=45\); \((5+45)/10=5\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

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

Explanation:
1. Condition \(abc+2fgh-af^2-bg^2-ch^2=0\) with \(g=f=K,\ c=2K,\ h=\lambda\): \(2abK+2K^2\lambda-aK^2-bK^2-2K\lambda^2=0\).
2. For \(K\ne0\): \(2\lambda^2-2K\lambda+(a+b)K-2ab=0\). Real \(\lambda\) needs \(K^2-2(a+b)K+4ab\ge0\), i.e. \((K-2a)(K-2b)\ge0\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

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

Explanation:
1. \(a=A,\ b=A+2d,\ c=A+6d\): \(A(x+y+1)+d(2y+6)=0\).
2. Fixed point: \(y=-3,\ x=2\); \(4+9=13\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

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

Explanation:
1. \((a-b)^2=c^2\) so \(c=\pm(a-b)\).
2. \(c=a-b\): \(a(x+1)+b(y-1)=0\) passes through \((-1,1)\); \(c=b-a\): through \((1,-1)\). Both lie on \(y=-x\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (a) \(l,n,m\text{ are in G.P.}\)

Explanation:
1. The vertices are \((\pm\dfrac nl,0)\) and \((0,\pm\dfrac nm)\), so area \(=\dfrac12\cdot\dfrac{2n}l\cdot\dfrac{2n}m=\dfrac{2n^2}{lm}\).
2. \(=2\) gives \(n^2=lm\): \(l,n,m\) in G.P.

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (c) \(\dfrac{-a^2(a+2)}{2(a+1)}\)

Explanation:
1. Vertices \((0,0),\ \left(\dfrac a{m_1},a\right),\ \left(\dfrac a{m_2},a\right)\); area \(=\dfrac{a^2|m_2-m_1|}{2|m_1m_2|}\).
2. \(|m_2-m_1|=|a+2|\) and \(|m_1m_2|=|a+1|\). For \(-2<a<-1\): \(|a+2|=a+2,\ |a+1|=-(a+1)\), so area \(=\dfrac{-a^2(a+2)}{2(a+1)}\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (a) 5

Explanation:
1. For lines equidistant from the origin, \(f^4-g^4=c(bf^2-ag^2)\).
2. So the ratio is \(c=10=2s\) and \(s=5\). (Check with lines \(10(3x+4y+1)(5x+1)\): ratio 10.)

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (b) 3

Explanation:
1. At the origin the value is \(1>0\); at \((1,1)\): \(a^2+ab+1>0\) for all \(a\ne0\).
2. Discriminant \(b^2-4<0\); for \(b=\pm2\) the value is \((a\pm1)^2\), zero at \(a=\mp1\), so excluded. Hence \(|b|<2\): \(b=-1,0,1\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

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

Explanation:
1. \(a+c=2b\) means \(a-2b+c=0\): every line of the family passes through \((1,-2)\).
2. \(AB\) and \(BC\) meet at \(B\) and both pass through \((1,-2)\), so \(B=(1,-2)\).

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (d) obtuse angled

Explanation:
1. \(AB\) is \(x=1\); altitude from \(C\) is horizontal: \(C_y=4\). \(BH\) has slope 6, so \(AC\) has slope \(-\tfrac16\): \(C=(-17,4)\).
2. \(\vec{AB}\cdot\vec{AC}=(0,-3)\cdot(-18,3)=-9<0\), so angle \(A\) is obtuse.

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.

Answer: (a) \(\dfrac{20}3\)

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

Teacher's Note:
1. Write the condition in terms of intercepts, slopes or distances and reduce it to an algebraic relation.
2. For families of lines, rewrite the equation as L1 + k L2 = 0 to find the fixed point.