JEE Mathematics Straight Lines MCQs Set 17

Answer: (b) G.P.

Explanation:
1. \(p_1=\dfrac{m^2\cos^2\alpha+2m\sin\alpha\cos\alpha+\sin^2\alpha}{|\cos\alpha|}=\dfrac{(m\cos\alpha+\sin\alpha)^2}{|\cos\alpha|}\).
2. Similarly \(p_3=\dfrac{(m'\cos\alpha+\sin\alpha)^2}{|\cos\alpha|}\) and \(p_2=\dfrac{(m\cos\alpha+\sin\alpha)(m'\cos\alpha+\sin\alpha)}{|\cos\alpha|}\) (up to sign).
3. So \(p_2^2=p_1p_3\): G.P.

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Answer: (d) \(\dfrac{\cos\alpha+\cos\beta+\cos\gamma}{\sin\alpha+\sin\beta+\sin\gamma}\)

Explanation:
1. With the circumcentre at the origin, the orthocentre is \(H=A+B+C\) (position vectors).
2. If the circumradius is \(R\), \(x_i=R\cos\theta_i\) and \(x_i\tan\theta_i=R\sin\theta_i\) for \(\theta_i=\alpha,\beta,\gamma\) (as \(|OA|=|OB|=|OC|=R\)).
3. So \(a=R\sum\cos\), \(b=R\sum\sin\) and \(\dfrac ab=\dfrac{\cos\alpha+\cos\beta+\cos\gamma}{\sin\alpha+\sin\beta+\sin\gamma}\).

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Answer: (b) a line segment

Explanation:
1. The left side is the sum of distances of \((x,y)\) from \((2,0)\) and \((-2,0)\).
2. The distance between these two points is \(4\), equal to the given sum, so the point must lie on the segment joining them (a degenerate ellipse): a line segment.

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Answer: (a) -1

Explanation:
1. The origin gives \(+1\) in \(3x-4y+1\) and \(-5\) in \(x+2y-5\). So \(3\alpha^2-4(\alpha+1)+1>0\) and \(\alpha^2+2(\alpha+1)-5 \lt 0\).
2. First: \(3\alpha^2-4\alpha-3>0\Rightarrow\alpha \lt \dfrac{2-\sqrt{13}}3\approx-0.54\) or \(\alpha>1.87\). Second: \(-3 \lt \alpha \lt 1\).
3. Together \(-3 \lt \alpha \lt -0.54\), so \([\alpha]\in\{-3,-2,-1\}\); among the options only \(-1\) is possible.

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Answer: (b) 588

Explanation:
1. The lines \(4x+3y=12\) and \(3x+4y=12\) meet at \(\left(\tfrac{12}7,\tfrac{12}7\right)\).
2. Let the variable line have intercepts \(2h\) and \(2k'\) (midpoint \((h,k')\)): \(\dfrac{12}{7\cdot2h}+\dfrac{12}{7\cdot2k'}=1\Rightarrow 6(h+k')=7hk'\).
3. So \(2hk'=\tfrac{12}7(h+k')\), \(k=\tfrac{12}7\), and \(343k=343\cdot\tfrac{12}7=588\).

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Answer: (b) Incentre

Explanation:
1. The orthocentre is the right-angle vertex (integral) and the circumcentre is the midpoint of the hypotenuse (rational); the centroid is the average of integral points (rational).
2. With an integer hypotenuse \(c\) and equal legs of length \(c/\sqrt2\), the incentre is \(\dfrac{aA+bB+cC}{a+b+c}\), a weighted average with weights \(\tfrac c{\sqrt2},\tfrac c{\sqrt2},c\), so its coordinates involve \(\sqrt2\) (e.g. triangle \((0,0),(2,0),(1,1)\) has inradius \(\sqrt2-1\)): never rational.

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Answer: (c) \(x+y=3-2\sqrt6\)

Explanation:
1. Unit vector along \(L\): \(\tfrac1{\sqrt2}(1,1)\). So \(Q=(2,1)\pm2\sqrt3\cdot\tfrac1{\sqrt2}(1,1)=(2\pm\sqrt6,\ 1\pm\sqrt6)\).
2. Third quadrant: \(Q=(2-\sqrt6,\ 1-\sqrt6)\).
3. The perpendicular to \(x-y=4\) is \(x+y=c\): \(c=3-2\sqrt6\).

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Answer: (a) \(41x-38y+38=0\)

Explanation:
1. The mirror has direction \(u=(1,7)\). The reflected ray has direction \(v=(1,-2)\).
2. The incident ray's direction is the mirror image of \(v\) in the mirror (reversed): \(v'=2\dfrac{v\cdot u}{u\cdot u}u-v=2\cdot\dfrac{-13}{50}(1,7)-(1,-2)=\left(-\tfrac{38}{25},-\tfrac{41}{25}\right)\propto(38,41)\).
3. Line through \((0,1)\) with direction \((38,41)\): \(41x-38(y-1)=0\Rightarrow41x-38y+38=0\).

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Answer: (d) lie on a straight line

Explanation:
1. Slope between the first two: \(\dfrac{3-\frac83}{1}=\dfrac13\).
2. Slope between the last two: \(\dfrac{30-3}{82-1}=\dfrac{27}{81}=\dfrac13\). Equal slopes: collinear.

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Answer: (a) \(\left(\tfrac45,\tfrac{12}5\right)\)

Explanation:
1. The intercepts are \(2\) and \(4\) (as \(P\) is their midpoint): \(L:\ 2x+y=4\).
2. \(L_1\) is perpendicular through \((-2,1)\): \(x-2y=-4\).
3. Solving: \(x=2y-4\Rightarrow5y=12\), \(y=\tfrac{12}5,\ x=\tfrac45\).

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Answer: (b) \(y-\sqrt3x+2+3\sqrt3=0\)

Explanation:
1. The given line has slope \(-\sqrt3\) (inclination \(120^\circ\)). A line at \(60^\circ\) to it has inclination \(60^\circ\) or \(180^\circ\) (slope \(\sqrt3\) or \(0\)).
2. Slope \(0\) gives \(y=-2\), which is parallel to the \(x\)-axis and does not meet it. So the slope is \(\sqrt3\).
3. \(y+2=\sqrt3(x-3)\Rightarrow y-\sqrt3x+2+3\sqrt3=0\).

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Answer: (b) \(x^2+y^2-2x-2y-14=0\)

Explanation:
1. Inradius \(r=\dfrac{|3+4+3|}5=2\). In an equilateral triangle the circumradius is \(R=2r=4\) and the circumcentre coincides with the incentre \((1,1)\).
2. Circle: \((x-1)^2+(y-1)^2=16\Rightarrow x^2+y^2-2x-2y-14=0\).

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Answer: (b) \(\dfrac{4\sqrt3}{15}\)

Explanation:
1. The height is the distance from \((1,2)\) to the base: \(h=\dfrac{|3+8-9|}5=\dfrac25\).
2. Side \(=\dfrac{2h}{\sqrt3}=\dfrac4{5\sqrt3}=\dfrac{4\sqrt3}{15}\).

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

Explanation:
1. Centroid \(G=\left(\dfrac{(a+1)^2}2,\dfrac{(a-1)^2}2\right)\).
2. With circumcentre \(O\) at the origin, \(H=3G\), so \(H\) lies on the line \(OG\): \(\dfrac yx=\dfrac{(a-1)^2}{(a+1)^2}\).
3. That is \((a-1)^2x-(a+1)^2y=0\).

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Answer: (b) \(\dfrac5{\sqrt{13}}\)

Explanation:
1. \(L\) is \(x+5y=k\), with intercepts \(k\) and \(\tfrac k5\). Area \(=\tfrac12\cdot|k|\cdot\left|\tfrac k5\right|=\tfrac{k^2}{10}=5\Rightarrow|k|=5\sqrt2\).
2. \(L\) is parallel to \(x+5y=0\), so the distance is \(\dfrac{|k|}{\sqrt{26}}=\dfrac{5\sqrt2}{\sqrt{26}}=\dfrac5{\sqrt{13}}\).

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Answer: (d) \(5/2\)

Explanation:
1. Reflect \(R\) in the \(x\)-axis: \(R'(6,-7)\). The path \(P\to Q\to R\) is the straight line \(PR'\).
2. Slope of \(PR'=\dfrac{-7-3}{6-1}=-2\). The line \(y-3=-2(x-1)\) meets \(y=0\) at \(x=1+\tfrac32=\tfrac52\).

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Answer: (a) \(a^2-9a+18=0\)

Explanation:
1. The slopes are \(\tfrac13,\ -\tfrac a2,\ -a\). A right angle needs a product of two slopes equal to \(-1\).
2. \(\tfrac13\cdot\left(-\tfrac a2\right)=-1\Rightarrow a=6\); \(\tfrac13(-a)=-1\Rightarrow a=3\); \(\left(-\tfrac a2\right)(-a)=-1\) has no real solution.
3. So \(a=3\) or \(6\): \((a-3)(a-6)=a^2-9a+18=0\).

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Answer: (d) -3/16

Explanation:
1. \(3x+4y=12\) has intercepts \(4\) and \(3\). So \(L\) has \(x\)-intercept \(8\) and \(y\)-intercept \(\tfrac32\).
2. Slope \(=-\dfrac{3/2}{8}=-\dfrac3{16}\).

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35. t
36. )
37. .

Answer: (a) Statement 1 is true; Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Explanation:
1. The angle between two lines depends only on their slopes, not on the constants \(c_i\).
2. The lines \(-x+5y+c_2=0\) and \(-x+5y+c_3=0\) are parallel, so \(\theta_1=\theta_2\) for all \(c_2,c_3\): Statement 2 is true.
3. Statement 1 is a special case of Statement 2, which explains it: true and correctly explained.

Teacher's Note:
1. P
2. a
3. r
4. a
5. l
6. l
7. e
8. l
9.
10. l
11. i
12. n
13. e
14. s
15.
16. m
17. a
18. k
19. e
20.
21. e
22. q
23. u
24. a
25. l
26.
27. a
28. n
29. g
30. l
31. e
32. s
33.
34. w
35. i
36. t
37. h
38.
39. a
40. n
41. y
42.
43. g
44. i
45. v
46. e
47. n
48.
49. l
50. i
51. n
52. e
53. .

Answer: (c) 7

Explanation:
1. The vertex opposite the side of slope \(m\) lies on the altitude through the origin of slope \(-\tfrac1m\): \(y=x\), \(y=\tfrac x2\), \(y=-\tfrac x3\) respectively.
2. Write the vertices as \((p,p),(2q,q),(3r,-r)\). Requiring the two sides through each pair to have slopes \(3\) and \(-1\) gives \(p=\tfrac52q\), \(r=\tfrac32q\) (and then the third side has slope \(-2\), as required).
3. Centroid \(=\left(\dfrac{p+2q+3r}3,\dfrac{p+q-r}3\right)=\left(3q,\tfrac23q\right)\), so \(y=\tfrac29x\): \(a=2,\ b=9\), \(b-a=7\).

Teacher's Note:
1. U
2. s
3. e
4.
5. t
6. h
7. e
8.
9. f
10. a
11. c
12. t
13.
14. t
15. h
16. a
17. t
18.
19. e
20. a
21. c
22. h
23.
24. v
25. e
26. r
27. t
28. e
29. x
30.
31. l
32. i
33. e
34. s
35.
36. o
37. n
38.
39. a
40. n
41.
42. a
43. l
44. t
45. i
46. t
47. u
48. d
49. e
50.
51. t
52. h
53. r
54. o
55. u
56. g
57. h
58.
59. t
60. h
61. e
62.
63. o
64. r
65. t
66. h
67. o
68. c
69. e
70. n
71. t
72. r
73. e
74. .

Answer: (c) 6

Explanation:
1. \(d_1+d_2=\dfrac{|x-y|+|x+y|}{\sqrt2}=\dfrac{2\max(|x|,|y|)}{\sqrt2}=\sqrt2\max(|x|,|y|)\).
2. The condition is \(\sqrt2\le\max(|x|,|y|)\le2\sqrt2\): the region between two concentric squares of side \(4\sqrt2\) and \(2\sqrt2\).
3. \(A=32-8=24\), so \(\dfrac A4=6\).

Teacher's Note:
1. U
2. s
3. e
4.
5. |
6. a
7. -
8. b
9. |
10.
11. +
12.
13. |
14. a
15. +
16. b
17. |
18.
19. =
20.
21. 2
22.
23. m
24. a
25. x
26. (
27. |
28. a
29. |
30. ,
31. |
32. b
33. |
34. )
35. .