JEE Mathematics Straight Lines MCQs Set 18

Answer: (b) 3

Explanation:
1. Let \(L_1:\ y=mx+c\). Then \(x_1=\dfrac{1-c}{m-3}\) and \(x_2=\dfrac{3-c}{m+2}\).
2. The midpoint's abscissa \(\tfrac12(x_1+x_2)\) must not depend on \(c\): \(\dfrac1{m-3}+\dfrac1{m+2}=0\Rightarrow2m-1=0\Rightarrow m=\tfrac12\).
3. So \(\dfrac pq=\dfrac12\) and \(p+q=3\).

Teacher's Note:
1. A
2.
3. v
4. e
5. r
6. t
7. i
8. c
9. a
10. l
11.
12. l
13. o
14. c
15. u
16. s
17.
18. m
19. e
20. a
21. n
22. s
23.
24. t
25. h
26. e
27.
28. x
29. -
30. c
31. o
32. o
33. r
34. d
35. i
36. n
37. a
38. t
39. e
40.
41. o
42. f
43.
44. t
45. h
46. e
47.
48. m
49. i
50. d
51. p
52. o
53. i
54. n
55. t
56.
57. i
58. s
59.
60. t
61. h
62. e
63.
64. s
65. a
66. m
67. e
68.
69. f
70. o
71. r
72.
73. e
74. v
75. e
76. r
77. y
78.
79. c
80. .

Answer: (b) 0

Explanation:
1. Let \(A=(a,a),\ B=(b,2b),\ C=(c,3c)\). Collinearity of \(A,(1,0),B\) gives \(a=\dfrac{2b}{b+1}\); collinearity of \(B,(0,-1),C\) gives \(c=\dfrac b{1-b}\).
2. Writing the line \(AC\) for different values of \(b\) (e.g. \(b=\tfrac13,\tfrac25\)) and intersecting gives the common point \(\left(\tfrac12,-\tfrac12\right)\), which also lies on \(AC\) for \(b=3\).
3. Hence \(h+k=0\).

Teacher's Note:
1. E
2. x
3. p
4. r
5. e
6. s
7. s
8.
9. A
10.
11. a
12. n
13. d
14.
15. C
16.
17. i
18. n
19.
20. t
21. e
22. r
23. m
24. s
25.
26. o
27. f
28.
29. o
30. n
31. e
32.
33. p
34. a
35. r
36. a
37. m
38. e
39. t
40. e
41. r
42.
43. a
44. n
45. d
46.
47. f
48. i
49. n
50. d
51.
52. t
53. h
54. e
55.
56. p
57. o
58. i
59. n
60. t
61.
62. c
63. o
64. m
65. m
66. o
67. n
68.
69. t
70. o
71.
72. a
73. l
74. l
75.
76. t
77. h
78. e
79.
80. l
81. i
82. n
83. e
84. s
85.
86. A
87. C
88. .

Answer: (a) 14

Explanation:
1. \(a,b,c\) in A.P. means \(a-2b+c=0\), so every line of the family passes through \((1,-2)\). Hence \(B=(1,-2)\).
2. \(AB\) is the vertical line \(x=1\). The altitude from \(C\) is perpendicular to it through \(H(2,4)\): \(y=4\), so \(k=4\).
3. Slope of \(AH=3\), so \(BC\) has slope \(-\tfrac13\): \(y+2=-\tfrac13(x-1)\). With \(y=4\): \(x=-17\).
4. \(C=(-17,4)\) and \(2h+12k=-34+48=14\).

Teacher's Note:
1. T
2. h
3. e
4.
5. A
6. .
7. P
8. .
9.
10. c
11. o
12. n
13. d
14. i
15. t
16. i
17. o
18. n
19.
20. g
21. i
22. v
23. e
24. s
25.
26. a
27.
28. f
29. i
30. x
31. e
32. d
33.
34. p
35. o
36. i
37. n
38. t
39.
40. c
41. o
42. m
43. m
44. o
45. n
46.
47. t
48. o
49.
50. b
51. o
52. t
53. h
54.
55. s
56. i
57. d
58. e
59. s
60. .

Answer: (c) 5

Explanation:
1. Group by \(\sec\theta\) and \(\csc\theta\): \(\sec\theta(3x+7y+11)+\csc\theta(5x-3y-11)=0\). So \(B\) is the intersection of \(3x+7y+11=0\) and \(5x-3y-11=0\): \(B=(1,-2)\).
2. \(A\) and \(B\) lie on the same side of \(x-y+3=0\) (values \(2\) and \(6\)), so the maximum of \(|PA-PB|\) is \(AB=\sqrt{4+36}=2\sqrt{10}\).
3. \(2\sqrt{2n}=2\sqrt{10}\Rightarrow n=5\).

Teacher's Note:
1. F
2. o
3. r
4.
5. p
6. o
7. i
8. n
9. t
10. s
11.
12. o
13. n
14.
15. t
16. h
17. e
18.
19. s
20. a
21. m
22. e
23.
24. s
25. i
26. d
27. e
28. ,
29.
30. t
31. h
32. e
33.
34. m
35. a
36. x
37. i
38. m
39. u
40. m
41.
42. o
43. f
44.
45. |
46. P
47. A
48.
49. -
50.
51. P
52. B
53. |
54.
55. e
56. q
57. u
58. a
59. l
60. s
61.
62. A
63. B
64. .

Answer: (c) 8

Explanation:
1. The base joins \(B(4,0)\) and \(C(0,2)\). The side \(x=4\) passes through \(B\), so the apex is \(A(4,t)\) with \(AB=AC\).
2. \(t^2=16+(t-2)^2\Rightarrow4t=20\Rightarrow t=5\), so \(A=(4,5)\).
3. The other equal side \(AC\): through \((4,5)\) and \((0,2)\), slope \(m=\tfrac34\), \(c=2\). \(8m+c=6+2=8\).

Teacher's Note:
1. F
2. i
3. n
4. d
5.
6. t
7. h
8. e
9.
10. a
11. p
12. e
13. x
14.
15. f
16. r
17. o
18. m
19.
20. t
21. h
22. e
23.
24. e
25. q
26. u
27. a
28. l
29. -
30. s
31. i
32. d
33. e
34. s
35.
36. c
37. o
38. n
39. d
40. i
41. t
42. i
43. o
44. n
45. .

Answer: (b) 1

Explanation:
1. \(A\) does not lie on \(3x-4y+5=0\) (value \(10\)), so this is diagonal \(BD\).
2. The line through the midpoints of \(AB\) and \(AD\) is parallel to \(BD\), midway between \(A\) and \(BD\). The expression \(3x-4y\) equals \(5\) at \(A\) and \(-5\) on \(BD\), so on the midline it equals \(0\).
3. The line is \(3x-4y=0\): \(a+b+c=3-4+0=-1\), so \(|a+b+c|=1\).

Teacher's Note:
1. T
2. h
3. e
4.
5. l
6. i
7. n
8. e
9.
10. j
11. o
12. i
13. n
14. i
15. n
16. g
17.
18. m
19. i
20. d
21. p
22. o
23. i
24. n
25. t
26. s
27.
28. o
29. f
30.
31. t
32. w
33. o
34.
35. s
36. i
37. d
38. e
39. s
40.
41. o
42. f
43.
44. a
45.
46. t
47. r
48. i
49. a
50. n
51. g
52. l
53. e
54.
55. i
56. s
57.
58. p
59. a
60. r
61. a
62. l
63. l
64. e
65. l
66.
67. t
68. o
69.
70. t
71. h
72. e
73.
74. t
75. h
76. i
77. r
78. d
79.
80. s
81. i
82. d
83. e
84. ,
85.
86. a
87. t
88.
89. h
90. a
91. l
92. f
93.
94. t
95. h
96. e
97.
98. d
99. i
100. s
101. t
102. a
103. n
104. c
105. e
106. .

Answer: (c) 6

Explanation:
1. \(\beta=2\alpha+3\). Area \(=\tfrac12\left|\alpha(2-3)+1(3-\beta)+2(\beta-2)\right|=\tfrac12|\beta-\alpha-1|=\tfrac12|\alpha+2|\).
2. \(2\le\tfrac12|\alpha+2|\le3\Rightarrow4\le|\alpha+2|\le6\).
3. \(\alpha+2\in\{4,5,6,-4,-5,-6\}\): \(6\) integer values of \(\alpha\), each giving an integer \(\beta\). So \(6\) points.

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

Answer: (b) 2

Explanation:
1. \(x^2-(y-1)^2=0\) gives the lines \(y=x+1\) and \(y=1-x\), meeting at \((0,1)\).
2. The angle bisectors are the vertical line \(x=0\) and the horizontal line \(y=1\).
3. With \(x+y=3\) they form a triangle with vertices \((0,1),(0,3),(2,1)\): area \(=\tfrac12\cdot2\cdot2=2\).

Teacher's Note:
1. T
2. h
3. e
4.
5. b
6. i
7. s
8. e
9. c
10. t
11. o
12. r
13. s
14.
15. o
16. f
17.
18. t
19. h
20. e
21.
22. l
23. i
24. n
25. e
26. s
27.
28. (
29. x
30. -
31. 0
32. )
33. ^
34. 2
35.
36. =
37.
38. (
39. y
40. -
41. 1
42. )
43. ^
44. 2
45.
46. a
47. r
48. e
49.
50. t
51. h
52. e
53.
54. a
55. x
56. e
57. s
58. -
59. p
60. a
61. r
62. a
63. l
64. l
65. e
66. l
67.
68. l
69. i
70. n
71. e
72. s
73.
74. t
75. h
76. r
77. o
78. u
79. g
80. h
81.
82. (
83. 0
84. ,
85.
86. 1
87. )
88. .

Answer: (c) 8

Explanation:
1. The graphs are the lines \(x+y=2\), \(x+y=-2\), \(x=1\), \(x=-1\), enclosing a parallelogram.
2. For \(-1\le x\le1\), \(y\) runs from \(-2-x\) to \(2-x\), a height of \(4\).
3. Area \(=4\times2=8\).

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

Answer: (b) 10

Explanation:
1. The mouse is closest to the cheese at the foot of the perpendicular from \((12,10)\) to \(5x+y-18=0\).
2. Foot: parameter \(\dfrac{5\cdot12+10-18}{26}=2\), so \((a,b)=(12-10,\ 10-2)=(2,8)\).
3. \(a+b=10\).

Teacher's Note:
1. D
2. i
3. s
4. t
5. a
6. n
7. c
8. e
9.
10. s
11. t
12. o
13. p
14. s
15.
16. d
17. e
18. c
19. r
20. e
21. a
22. s
23. i
24. n
25. g
26.
27. a
28. t
29.
30. t
31. h
32. e
33.
34. f
35. o
36. o
37. t
38.
39. o
40. f
41.
42. t
43. h
44. e
45.
46. p
47. e
48. r
49. p
50. e
51. n
52. d
53. i
54. c
55. u
56. l
57. a
58. r
59. .

Answer: (b) 3

Explanation:
1. The triangle's area is \(\tfrac12\cdot8\cdot1=4\), so the left part must have area \(2\).
2. For \(c\le1\) the left part has area \(\tfrac{4c^2}9\le\tfrac49 \lt 2\), so \(c>1\). For \(1\le c\le9\): area \(=\tfrac49+\left[(c-1)-\tfrac{c^2-1}{18}\right]=2\).
3. This gives \(c^2-18c+45=0\Rightarrow c=3\) (the root \(15\) is outside the triangle).

Teacher's Note:
1. C
2. o
3. m
4. p
5. u
6. t
7. e
8.
9. t
10. h
11. e
12.
13. a
14. r
15. e
16. a
17.
18. t
19. o
20.
21. t
22. h
23. e
24.
25. l
26. e
27. f
28. t
29.
30. o
31. f
32.
33. x
34.
35. =
36.
37. c
38.
39. p
40. i
41. e
42. c
43. e
44. w
45. i
46. s
47. e
48. .