JEE Mathematics Straight Lines MCQs Set 16

Answer: (a) (B) and (C) only

Explanation:
1. A point on \(x+y=4\) is \((t,4-t)\). Its distance from \(4x+3y-10=0\) is \(\dfrac{|4t+12-3t-10|}5=\dfrac{|t+2|}5\).
2. Setting this equal to \(1\): \(t+2=\pm5\Rightarrow t=3\) or \(t=-7\), giving \((3,1)\) and \((-7,11)\).
3. So (B) and (C) are correct; (A) and (D) do not even lie on x + y = 4.

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Answer: (a) \(4\sqrt2\)

Explanation:
1. Along \(x-y+1=0\) (direction \(\tfrac1{\sqrt2}(1,1)\)) the points are \(\left(2+\tfrac r{\sqrt2},\,3+\tfrac r{\sqrt2}\right)\).
2. Substituting in \(2x-3y+9=0\): \(4-9+9+\tfrac{2r-3r}{\sqrt2}=0\Rightarrow r=4\sqrt2\).

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Answer: (c) \(a+b+c=0\)

Explanation:
1. Concurrency needs \(\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}=-(a^3+b^3+c^3-3abc)=0\).
2. \(a^3+b^3+c^3-3abc=(a+b+c)\cdot\tfrac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\). As \(a,b,c\) are distinct, the second factor is non-zero, so \(a+b+c=0\).

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Answer: (a) \(210^\circ,\ 1\)

Explanation:
1. Rewrite as \(-\sqrt3x-y=2\) so that the constant on the right is positive.
2. Divide by \(\sqrt{(\sqrt3)^2+1^2}=2\): \(-\dfrac{\sqrt3}2x-\dfrac12y=1\).
3. So \(\cos\theta=-\dfrac{\sqrt3}2,\ \sin\theta=-\dfrac12\) gives \(\theta=210^\circ\), and \(p=1\).

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Answer: (c) 25 units

Explanation:
1. The side is \(5\), so the distance between the parallel lines is \(\dfrac{|k_1-k_2|}{\sqrt{9+16}}=5\).
2. Hence \(|k_1-k_2|=25\).

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Answer: (b) \((-a,0)\)

Explanation:
1. The points lie on \(y^2=4ax\). The chord through parameters \(t_1,t_2\) is \((t_1+t_2)y=2x+2at_1t_2\).
2. With \(t_1+t_2=-\lambda\) and \(t_1t_2=1\): \(-\lambda y=2x+2a\), i.e. \(2(x+a)+\lambda y=0\) for all \(\lambda\).
3. This passes through \((-a,0)\) for every \(\lambda\).

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Answer: (c) \([1,\infty)\)

Explanation:
1. Adding the equations: \(x(1+a)=|a|+1\Rightarrow x=\dfrac{|a|+1}{a+1}\), which needs \(a>-1\) for \(x\ge0\).
2. For \(a\ge0\): \(x=1\) and \(y=a-1\ge0\Rightarrow a\ge1\).
3. For \(-1 \lt a \lt 0\): \(y=-\dfrac{a^2+1}{1+a} \lt 0\), not allowed. So \(a\ge1\).

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Answer: (a) \(\left(\tfrac12,3\right)\)

Explanation:
1. Inside the angle means \(x>0\) and \(\dfrac x2 \lt y \lt 3x\).
2. With \((x,y)=(a,a^2)\): \(a>0\) and \(\dfrac a2 \lt a^2 \lt 3a\Rightarrow\dfrac12 \lt a \lt 3\).

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Answer: (b) \(2\alpha\beta\)

Explanation:
1. Since \(P\) lies on the line, \(\dfrac\alpha a+\dfrac\beta b=1\). By AM-GM, \(1\ge2\sqrt{\dfrac{\alpha\beta}{ab}}\Rightarrow ab\ge4\alpha\beta\).
2. \(S=\dfrac12ab\ge2\alpha\beta\), with equality when \(a=2\alpha,\ b=2\beta\).

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Answer: (c) \((a,a)\)

Explanation:
1. Take \(A\) as origin, \(B(a,0)\), \(C(0,a)\). Let \(BP=p,\ CQ=q\) with \(pq=a^2\): \(P(a+p,0),\ Q(0,a+q)\).
2. Line \(PQ\): \(\dfrac x{a+p}+\dfrac y{a+q}=1\). At \((a,a)\): \(\dfrac a{a+p}+\dfrac a{a+q}=\dfrac{a(2a+p+q)}{a^2+a(p+q)+pq}=\dfrac{a(2a+p+q)}{2a^2+a(p+q)}=1\).
3. So \((a,a)\) lies on \(PQ\) for all such \(p,q\).

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Answer: (a) \(x^2+y^2=\dfrac{9+\tan^2\alpha}4\)

Explanation:
1. Equal angles with \(AB\) make triangle \(PAB\) isosceles with \(PA=PB\), so \(P\) lies on the perpendicular bisector of \(AB\), at distance \(\tfrac32\) from \(O\) along \(AB\).
2. The height of \(P\) above \(AB\) is \(\tfrac12AB\tan\alpha=\tfrac12\tan\alpha\).
3. \(OP^2=\tfrac94+\tfrac14\tan^2\alpha\), so the locus is \(x^2+y^2=\dfrac{9+\tan^2\alpha}4\).

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Answer: (c) \(\left(-\dfrac1{\sqrt2},\dfrac7{\sqrt2}\right)\)

Explanation:
1. I: \((4,1)\to(1,4)\). II: \((1,4)\to(3,4)\).
2. III: rotation by \(\tfrac\pi4\): \(x'=\dfrac{3-4}{\sqrt2}=-\dfrac1{\sqrt2},\ y'=\dfrac{3+4}{\sqrt2}=\dfrac7{\sqrt2}\).

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Answer: (a) \(\pm1\)

Explanation:
1. Put \(y=mx\) (\(m=\tan\theta\)): \((1+m^2)x^3+2(1+m^2)x^2+3x+1=0\). The distances \(OA,OB,OC\) are proportional to \(|x|\), so \(\dfrac1x\) are in A.P.
2. With \(u=\dfrac1x\): \(u^3+3u^2+2(1+m^2)u+(1+m^2)=0\). The A.P. roots sum to \(-3\), so the middle root is \(u=-1\).
3. Substituting \(u=-1\): \(-1+3-2(1+m^2)+(1+m^2)=0\Rightarrow m^2=1\), so \(\tan\theta=\pm1\).

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Answer: (a) 0

Explanation:
1. The lines \(x+y=4\) and \(x+y=\tfrac52\) are parallel.
2. The distance between them is \(\dfrac{4-\frac52}{\sqrt2}=\dfrac{3}{2\sqrt2}\approx1.06\ne1\).
3. Every point of x + y = 4 is at this same distance from the other line, which is not 1, so no such point exists.

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Answer: (b) \(x-4y+5=0\)

Explanation:
1. The sides \(x=0,x=2,y=0,y=3\) form a rectangle with centre \((1,\tfrac32)\).
2. A line through the centre halves the area. The line \(x-4y+c=0\) through \((1,\tfrac32)\): \(1-6+c=0\Rightarrow c=5\).

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Answer: (b) \(12x+33y-7=0\)

Explanation:
1. Write it as \((x+y)+k(2x-y+1)=0\): all lines pass through \(\left(-\tfrac13,\tfrac13\right)\).
2. The farthest line from \((1,4)\) is perpendicular to the segment joining \((1,4)\) and \(\left(-\tfrac13,\tfrac13\right)\), direction \(\left(\tfrac43,\tfrac{11}3\right)\propto(4,11)\).
3. Line: \(4x+11y+c=0\) through \(\left(-\tfrac13,\tfrac13\right)\Rightarrow c=-\tfrac73\), i.e. \(12x+33y-7=0\).

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Answer: (c) \(6x-5y+14=0\)

Explanation:
1. \(x=1,6\) and \(y=4,10\): vertices \((1,4),(6,4),(6,10),(1,10)\).
2. The diagonal from \((1,4)\) to \((6,10)\) has slope \(\tfrac65\): \(y-4=\tfrac65(x-1)\Rightarrow6x-5y+14=0\).

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Answer: (b) \(\left(\tfrac43\pm\sqrt2,\ -\tfrac43\pm\sqrt2\right)\)

Explanation:
1. The roads cross at \(\left(\tfrac43,-\tfrac43\right)\). Their directions are \((2,1)\) and \((1,2)\), with \(\cos=\tfrac45\), so the angle between them is acute and its bisector has direction \((1,1)\).
2. Walking \(2\) km in either direction along the unit vector \(\tfrac1{\sqrt2}(\pm1,\pm1)\): displacement \((\pm\sqrt2,\pm\sqrt2)\).
3. Points: \(\left(\tfrac43\pm\sqrt2,\ -\tfrac43\pm\sqrt2\right)\) (same sign in both coordinates).

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Answer: (c) \(\tfrac13f(6)\) square units

Explanation:
1. The functional equation with \(f(1)=2\) gives \(f(x)=2^x\) for integer arguments (so \(f(4)=16,\ f(5)=32,\ f(6)=64\)).
2. \(3|x|+2|y|\le8\) is a rhombus with semi-diagonals \(\tfrac83\) and \(4\); area \(=2\cdot\tfrac83\cdot4=\tfrac{64}3\).
3. \(\tfrac13f(6)=\tfrac{64}3\).

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Answer: (c) \(\left(2-\sqrt3\right)x+y=2\left(\sqrt3-1\right)\)

Explanation:
1. After rotation \(B'=(2\cos30^\circ,2\sin30^\circ)=(\sqrt3,1)\) and \(D'=(-2\sin30^\circ,2\cos30^\circ)=(-1,\sqrt3)\).
2. Slope \(=\dfrac{\sqrt3-1}{-1-\sqrt3}=-(2-\sqrt3)\).
3. \(y-1=-(2-\sqrt3)(x-\sqrt3)\Rightarrow(2-\sqrt3)x+y=2(\sqrt3-1)\).

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Answer: (a) \(abc+3(a+b+c)=ab+bc+ca\)

Explanation:
1. A line \(px+qy+r=0\) meets the curve where \(pt^3+q(t^2-3)+r(t-1)=0\), i.e. \(pt^3+qt^2+rt-(3q+r)=0\).
2. If the roots are \(a,b,c\): \(a+b+c=-\dfrac qp\), \(ab+bc+ca=\dfrac rp\), \(abc=\dfrac{3q+r}p\).
3. Hence \(abc=-3(a+b+c)+(ab+bc+ca)\), i.e. \(abc+3(a+b+c)=ab+bc+ca\).

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