Line and Lines
Line and Lines
Line and Lines
1. The equation of line equidistant from the points A(1, –2) and B(3,4) and making congruent
angles with the coordinate axes is . . .
(a) x +y = 1 (b) y – x + l = 0 (c)y – x – 1 = 0 (d) y – x = 2
2. The equation of line passing through the point (–5,4) and making the intercept of length
2 between the lines x + 2y – 1 = 0 and x + 2y + 1 = 0 is . . .
5
(a) 2x – y + 4 = 0 (b) 2x – y –14 = 0 (c) 2x – y + 14 = 0 (d) None of these
3. The equation of line containing the angle bisector of the lines 3x – 4y – 2 = 0 and 5x –
12y + 2 = 0 is . . .
(a) 7x + 4y – 18 = 0 (b) 4x – 7y – 1 = 0 (c) 4x – 7y + 1 = 0 (d) None of these
4. The equation of line passing through the point of intersection of the lines
3x – 2y = 0 and 5x + y – 2 = 0 and making the angle of measure tan–1(–5) with the positive
direction of x – axis is . . .
(a) 3x – 2y = 0 (b) 5x + y – 2 = 0 (c) 5x + y = 0 (d) 3x + 2y + 1 = 0
5. If for a + b + c 0, the lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are
concurrent, then . . .
a b c
(a) ab + be + ca = 0 (b) + + 1 (c) a = b (d) a = b = c
b c a
6. The equation of line passing through the point (1,2) and making the intercept of length 3
units between the lines 3x + 4y = 24 and 3x + 4y = 12, is . . .
(a) 7x – 24y + 41 = 0 (b) 7x + 24y = 55 (c) 24x – 7y = 10 (d) 24x + 7y – 38 = 0
x
7. If (a, a2) lies inside the angle between the lines y = , x > 0 and y = 3x, x > 0,
2
then a . . .
3 3
(a) 2
xy0 (b) x 2
y0 (c) 3x y 0 (d) x 3y 0
37
x y 1
9. If the non zero numbers a, b,c are in harmonic progression, then the line 0
a b c
passes through the point . . .
10. A line passing through 0(0,0) intersect the parallel lines 4x + 2y = 0 and 2x + y + 6 = 0 at
P and Q respectively, then in what ratio does 0 divide PQ from P ?
(a) 1 : 2 (b) 3 : 4 (c) 2 : 1 (d) 4 : 3
11. The points on the line 3x – 2y – 2 = 0, which are 3 units away from the line
3x + 4y – 8 = 0 are . . .
(a) (3, 3), 3, 13 (b) 3, 27 , 13 , 32 (c) 72 ,3 , 13 ,3 (d) (3,1),(1,3)
12. If A(1, –2), 5(–8,3), A–P–B and 3 AP = 7AB, then P = . . .
(a) 22, 41
3 (b) 22, 41
3 (c) not possible (d) None of these
13. For the collinear points P – A – B, AP = 4AB, then P divides AB from A in the ratio.....
(a) 4 : 5 (b) – 4 : 5 (c) –5 : 4 (d) –1 : 4
14. If the length of perpendicular drawn from (5,0) on kx + 4y = 20 is 1, then k = . . .
16
(a) 3, 3
(b) 3, 163 (c) 3, 16
3
(d) 3, 163
15. If the lengths of perpendicular drawn from the origin to the lines xcos – ysin =
sin2a and xsin + ycos = cos2 are p and q respectively, then p2 + q 2 = . . .
(a) 4 (b) 3 (c) 2 (d) 1
16. The points onY – axis at a distance 4 units from the line x + 4y = 12 are . . .
38
19. A(1,0) and B(–1,0), then the locus of points satisfying AQ – BQ = ±1 is . . .
(a) 12x2 + 4y2 = 3 (b) 12x2 – 4y2 = 3 (c) 12x2 – 4y2 = –3 (d) 12x2 + 4y2 = –3
20. A rod having length 2c moves along two perpendicular lines, then the locus of the mid point
of the rod is . . .
(a) x2 – y2 = c2 (b) x2 + y2 = c2 (c) x2 + y2 = 2c2 (d) None of these
21. Consider a square PQR having the length of side a, where O(0,0). The sides OP and
OR are along the positive X – axis and Y – axis respectively. If A and B are the mid points
of PQ and QR respectively, then the angle between OA and OB would be... . .
22. 3x y 2 is the equation of line containing one of the sides of an equilateral triangle
and if (0,–1) is one of the vertices, then the length of the side of the triangle is . . .
3 2
(a) 3 (b) 2 3 (c) 2
(d) 3
24. If two perpendicular lines passing through origin intersect the line x
a
yb 1, a 0, b 0 at
1 1 ..........
A and B, then OA 2 OB2
1 1 ab a 2 b2
(a) 2
2 (b) (c) (d) None of these
a b a 2 b2 a 2b2
25. The equation of a line at a distance 5 units from the origin and the ratio of the intercepts
on the axes is 1 : 2, is . . .
(a) 2x + y + 5 = 0 (b) 2x + y + 5 = 0 (c) x – 2y + 5 = 0 (d) None of these
26. For any values of p and q, the line (p + 2q)x + (p – 3q)y – p – q passes through which
fixed point ?
39
27. If A(x 1,y 1 ), B(x2,y2) and P(tx 2 + (1 – t)x 1 , ty 2 + (1 – t)y 1) where t < 0, then P divides
AB from A in the ratio . . .
t 1 t
(a) 1 – t (b) (c) (d) t – 1
t 1 t
28. A(1,2), B(5,7) and P(x,y) AB , then y – x – 1 is . . .
(a) < 0 (b) > 0 (c) = 0 (d) –3
29. A(2,3), B(4,7) and P(x,y) AB , then the maximum value of 3x + y is . . .
(a) 19 (b) 9 (c) –19 (d) –9
30. A(– 2,5), 5(6,2), then AB AB ..........
(a) {(8t – 2, 5 – 3t / t < 0) (b) {(8t – 2, 5 – 3t) / 0 < t < 1}
(c) {(8t – 2, 5 – 3t) / t R – [0, 1]} (d) {(8t – 2, 5 – 3t) / t > 1}
π π π π
(a) xcos ysin 2 (b) xcos ysin 2
6 6 3 3
π π π π
(c) xcos ysin 2 (d) xcos ysin 2
3 3 6 6
32. The length of side of an equilateral triangle is a. There is circle inscribed in a triangle.
What is the area of a square inscribed in a circle ?
33. If the lines x + 2ay + a – 0, x + 3by + 3 = 0 and x + 4cy + c = 0 are concurrent, then a,
b, c are in . . .
(a) A.P. (b) H.P. (c) G.P. (d) A.G.P
34. The foot of perpendicular drawn from (2,3) to the line 4x – 5y – 34 = 0 is . . .
(a) x
2
3y 1, x
2
1y 1 (b) x
2
3y 1, x
2
1y 1
(c) x
2
3y 1, x
2
1y 1 (d) x
2
3y 1, x
2
1y 1
40
36. A line intersects X – axis and Y – axis at A and B respectively. If AB = 15 and AB makes a
triangle of area 54 units with coordinate axes, then the equation of AB is . . .
(a) 4x ± 3y = 36 or 3x ± 4y = 36 (b) 4x ± 3y = 24 or 3x ± 4y = 24
(c) –4x ± 3y = 24 or – 3x ± 4y – 24 (d) –4x ± 3y = 12 or – 3x ± 4y – 12
37. The angle between the lines xcos85° + ysin85° = 1 and xcos40° + ysin40° = 2 is :
(a) 90° (b) 80° (c) 125° (d) 45°
38. If a1 , a2, a 3 and b 1, b 2, b3 are in geometric progression and their common ratios are equal,
then the poi nts A(a1, b1), B(a2,b 2) and C(a3,b 3) . . .
(a) lie on the same line (b) lie on a circle (c) lie on an ellipse (d) None of these
39. The image of the point (4, –13) in the line 5x + y + 6 = 0 is . . .
(a) (1,2) (b) (3,4) (c) (–4,13) (d) (–1, –14)
40. If the lines x + (a – l)y + 1 = 0 and 2x + a2y –1 = 0 are perpendicular then . . .
(a) | a | = 2 (b) 0 < a < 1 (c) –1 < a < 1 (d) a = –1
41. If x + 3y – 4 = 0 and 6x – 2y – 7 = 0 are the lines containing the diagonals of a
parallelogram PQRS, then parallelogram PQRS is . . .
(a) rectangle (b) square (c) cyclic quadrilateral (d) rhombus
42. For a + b + c = 0, the line 3ax + 4by + c = 0 passes through the fixed point . . .
(a) 13 , 14 (b) 13 , 1
4 (c) 13 , 14 (d) 13 , 14
43. If 3l + 2m + 6n = 0, then the family of lines lx + my + n = 0 passes through the fixed
point . . .
41
48. The locus of mid points of the segment intercepted between the axes by the line
xseca + ytana = p is . . .
p2 p2 x2
2
p2 p2 p2 p2
(a) 1 (b) y2 4 (c) 1 (d) 1
4x2 4y2 p2 p x2 y2 4x 2 4y 2
49. If the y – intercept of the perpendicular bisector of the segment obtained by joining
P(l,4) and Q(k, 3) is –4 then k = . . .
(a) 1 (b) 2 (c) –2 (d) –4
50. The y – intercept of the line passing through the point (2,2) and perpendicular to the line
3x + y – 3 = 0 is . . .
3
(a) 4
(b) 4
3
(c) 43 (d) 34
51. The line parallel to the X – axis and passing through the intersection of the lines
ax + 2by + 3b = 0 and bx – 2ay – 3a = 0 where (a, b) (0,0) is :
(A) above the X – axis at a distance of 2 from it
3
3
(B) above the X – axis at a distance of 2
from it
(C) below the X – axis at a distance of 2 from it
3
3
(D) below the X – axis at a distance of 2
from it
52. A square of side a lies above the x – axis and has one vertex at the origin. The side passing
through the origin makes an angle a with the positive direction of x – axis.
The eq. of its diagonal not passing through the origin is :
(A) y(cos + sin ) + x(sin – cos ) = a (B) y(cos + sin ) + x(sin + cos ) = a
(C) y(cos + sin ) + x(cos – sin ) = a (D) y(cos – sin ) – x(sin – cos ) = a
53. If P and Q divides AB from A in the ratios and – , then A divides PQ from p in the
ratio . . . . . . .
(a) 1 (b) 1
(c) 2 (d) 2
+1 1 2 2
42
57. P(–1,0), Q(0,0) and R(3,3 3) are given points, then the equation of the bisector of
If
PQR is . . .
(a) 3 (b) x 3 (c) (d) x 3y 0
2
xy 2
y0 3x y 0
58. For the line y – y 1 = m(x – x t ), m and x 1 are fixed values, if different lines are drawn
according to the different value of y1; then all such lines would be . . .
(a) all lines intersect the line x = x 1 (b) all lines pass through one fixed point
(c) all lines are parallel to the line y = x 1 (d) all lines will be the set of perpendicular lines
59. If the length of perpendicular drawn from origin to a line is 10 and 56 then the
equation of line would be . . .
(a) 3x y (b) 3x y (c) 3x y 20 0 (d) 3x y 20 0
50
60. Find the equation of line making a triangle of area 3
units with two axes and on which a
62. Find the equation of line passing through the point ( 3, 1) and at a distance 2 units
from the origin.
43
63. If (3,–2) and (–2,3) are two vertices and (6,–1) is the orthocentre of a triangle, then the
third vertex would be . . .
(a) (1,6) (b) (–1,6) (c) (1, –6) (d) none of these
64. The circumcentre of the triangle formed by the lines x + y = 0, x – y = 0 and x – 7 = 0 is . .
(a) (7,0) (b) (3.5,0) (c) (0,7) (d) (3.5,3.5)
65. If 1, 1, 1
a b c
are in arithmetic sequence, then the line x y 1 0 passes through the fixed
a b c
point . . .
44
73. The equation of a straight line passing through the point (–5, 4) and which cut off an
intercept of 2 unit between the lines x + y + 1 = 0 and x + y – 1 = 0 is
(a) x – 2y + 13 = 0 (b) 2x – y + 14 = 0 (c) x – y + 9 = 0 (d) x – y + 10 = 0
74. If P(1, 2), Q(4, 6), R(5, 7) and S(a, b) are the vertices of a parallelogram PQRS then
(a) a = 2, b = 4 (b) a = 3, b = 4 (c) a = 2, b = 3 (d) a = 2, b = 5
75. The sum of squares of intercepts on the axes cut off by the tangents to the curve
2 2 2
x 3 y 3 a 3 (a > 0) at a8 , a8 is 2. Thus a has the value.
(a) 1 (b) 2 (c) 4 (d) 8
76. If two vertices of a trinangle eare (5, –1) and (–2, 3) and if its orthocentre lies at the origin
then the cooridnates of the third vertex are
(a) (4, 7) (b) (–4, –7) (c) (2, –3) (d) (5, –1)
77. Line ax + by + p = 0 makes angle which x cos ysin p, p R + . If these lines and
the line x sin y cos 0 are concurrent then
(a) a2 + b2 = 1 (b) a2 + b2 = 2 (c) 2(a2 + b2) = 1 (d) a2 – b2 = 2
78. A straight line passess through a point A(1, 2) and makes an angle 600 with the x–axis. This
line intersect the line x + y = 6 at P. Then AP will be
(a) 3( 3 1) (b) 3( 3 1) (c) ( 3 1) (d) 3 3
79. The image of origin in the line x + 4y = 1 is
(a) 17
2 , 8
17 2 , 8
(b) 17 17 2 , 8
(c) 17 17 2 , 8
(d) 17 17
80. Orthocentre of triangle with vertices (0, 0), (3, 4) and (4, 0) is
(a) 3, 54 (b) (3, 12)
(c) 3, 34 (d) (3, 9)
81. The equation of three sides of triangle are x = 2, y + 1 = 0 and x + 2y = 4. The coordinates
of the circumcentre of the triangle is
(a) (4, 0) (b) (2, –1) (c) (0, 4) (d) (–1, 2)
82. If a, b, c are in A.P. then ax + by + c = 0 represents
(a) a single line (b) a family of concurrent lienes
(c) a family of parallel lines (d) a family of circle
83. A(4, 0), B(0, 3), C(6, 1) be vertices of triangle ABC. Slope of bisector of angle C will be
(A) 3 2 7 (b) 5 2 7 (c) 6 2 7 (d) none
45
84. The locus of the variable point whose distance from (–2, 0) is 23 times its distance from
the line x 92 is
(a) ellipse (b) parabola (c) circle (d) hyperbola
85. The line 3x – 4y + 7 = 0 is rotated through an angle in the clockwise direction about
the point (–1, 1). The equation of the line in its new position is
(a) 7y + x – 6 = 0 (b) 7y – x – 6 = 0 (c) 7y + x + 6 = 0 (d) 7y – x + 6 = 0
86. The area of the triangle formed by the point (a, a2), (b, b2), (c, c2) is ..... (a, b, c are three
consecutive odd integers)
(a) 15
7 sq unit (b) 10
7 sq unit (c) 18
7 sq unit (d) 10
13 sq unit
88. In triangle ABC, equation of right bisectors of the sides AB and AC are x + y = 0 and
y – x = 0 respectively. If A = (5, 7) then equation of side BC is
(a) 7y = 5x (b) 5x = y (c) 5y = 7x (d) 5y = x
89. The equations of the two lines each passing through (5, 6) and each making an acute angle
of 450 with the line 2x – y + 1 = 0 is
(a) 3x + y – 21 = 0, x – 3y + 13 = 0 (b) 3x + y + 21 =0, x + 3y + 13 = 0
(c) y = 2x, y = 3x (d) 3x + y – 21 = 0, x – 3y – 13 = 0
90. If the equation of base of an equilateral triangle is 2x – y = 1 and the vertex is (–1, 2),
then the length of the side of the triangle is
20 2 8 15
(a) 3
(b) 15
(c) 15
(d) 2
4
91. Four points (x1, y1), (x2, y2), (x3, y3) and (x4, y4) are such that xi 2 yi 2
i 1
46
92. A variable straight line passess through a fixed point (a, b) intersecting the coordinate axes
at A and B. If ‘O’ is the origin, then the locus of the centroid of the triangle OAB is
(a) bx + ay = 3xy (b) bx + ay = 2xy (c) ax + by = 3xy (d) ax + by = 2xy
93. If the poitns (k, 2–2k), (1–k, 2k) and (–k–4, 6 – 2k) are collinear, the possible value of k
are
(a) 12 , 43 (b) (1, 3) (c) (3, 1)
(d) 34 , 12
98. The area of parallelogram whose two sides are y = x + 3, 2x – y + 1 = 0 and remaining
two sides are passing through (0, 0) is
47
103. The equations of two striaght lines which are parallel to x + 7y + 2 = 0 and at unit distance
from the point (1, –1) are
(a) x 7 y 6 4 2 0 (b) x 7 y 6 5 2 0
(c) 2 x 7 y 6 5 2 0 (d) x y 6 5 2 0
104. The points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10 are
(a) (3, 1), (–7, 11) (b) (7, 11, (2, 2) (c) (7, –11), (–3, 7) (d) (1, 3), (–5, 9)
105. One side of the rectangle lies along the line 4x + 7y + 5 = 0. Two of its vertices are (–3, 1
and (1, 1). Then the equations of other side is
(a) 7x – 4y + 25 = 0 (b) 4x + 7y = 11 (c) 7x – 4y – 3= 0 (d) All of these
106. Equation of a straight line passing through the point (4, 5) and equally inclined to the lines
3x = 4y + 7 and 5y = 12x + 6 is (angle bisector)
(a) 9x – 7y = 1 (b) 9x + 7y = 71 (c) 7x – y = 73 (d) 7x – 9y + 17 = 0
107. The nearest point on the line 3x + 4y = 1 from origin is
7 , 4
(a) 25 25 (b) 25
7 , 2
25
3 , 4
(c) 25 25
1 , 3
(d) 25 25
108. The locus of the mid point of the intercept of the variable line x cos a + y sin a = p
between the coordinate axes is
(a) x 2 y 2 p2 (b) x 2 y 2 2p 2 (c) x 2 y 2 4p 2 (d) non of these
109. Three straight lines 2x + 11y – 5 = 0, 4x – 3y – 2 = 0 and 24x + 7y – 20 = 0
(a) form a triangle (b) are only concurrent
(c) are concurrent with one line bisecting the angle between the other two.
(d) none of these
110. A straight line through the point (2, 2) intersects the line 3x y 0 and 3x y 0 at
the points A and B. The equation to the line AB so that the triangle OAB is equilateral is
(a) x = 2 (b) y = 2 (c) x + y = 4 (d) none
111. A triangle with vertices (4, 0), (–1, –1), (3, 5) is
(a) isosceles and right angled (b) isosceles but not right angled
(c) right angled but not isosceles (d) neither right angled nor isosceles
112. Equation of a line at a distance 5 unit from origin with intercepts 1:2 on axes is ....
(a) 2x – y + 5 = 0 (b) 2x + y + 5 = 0 (c) x – 2y + 5 = 0 (d) x + 2y + 5 = 0
48
113. The equation of the lines with slope –2 and intersecting x–axis at points distance 3 unit
from the origin is ........
(a) 2x + y +6 = 0 (b) x + 2y + 6 = 0 (c) 2x + y + 3 = 0 (d) x + 2y + 3 = 0
114. The equation of a line containing a side of an equilateral triangle is 3 x 4 . If (0, –1)
is one of the vertices then the length of its side is.....
3 2
(a) 3 (b) 2 3 (c) 2
(d) 3
115. If the equation of the locus of a point equidistant from the points (a1, b1) and (a2, b2) is
(a1 – a2)x + (b1, b2)y + c = 0, then the value of C will be
(c) 2 a1 a2 b1 b 2
1 2 2 2 2
(d) a12 b12 a22 b 22
116. Locus of the centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, – b cost)
and (1, 0), where t is a parameter is
(a) (3x – 1)2 + (3y)2 = a2 – b2 (b) (3x – 1)2 + (3y)2 = a2 + b2
(c) (3x + 1)2 + (3y)2 = a2 + b2 (d) (3x + 1)2 + (3y)2 = a2 – b2
117. A square of side ‘a’ lies above the x–axis and has one vertex at the origin. The side passing
through the origin makes an angle with the positive direction of x–axis. The
equation of the diagonal not passing through the origin is
(a) y(cos – sin ) – x (sin – cos ) = 0 (b) y(cos + sin ) + x (sin – cos ) = 0
(c) y(cos – sin ) – x (sin + cos ) = 0 (d) y(cos – sin ) – x (cos – sin ) = 0
118. If x1, x2, x3 and y1, y2, y3 both are in GP with the same common ratio, then the points (x1, y1)
(x2, y2) and (x3, y3)
(a) lie on a striaght line (b) lie on a ellipse
(c) lie on a circle (d) are vertices of a triangle
119. The length of a side of a square OPQR is a, O is the origin OP and OR are along positive
direction of the X and Y axes respectively. If A and B are mid points of PQ and QR
respectively then measure of angle between OA and OB is....
49
120. The incentre of a triangle whose vertices A(2, 4), B(2, 6) and C(2+ 3, 5) is....
1
(a) 2 3 , 5 1 5
(b) 1 2 3 , 2 (c) (2, 5) (d) None of these
121. If a line 3x + 4y = 24 intersects the axes at A and B, then inradius of OAB is .....
(a) 1 (b) 2 (c) 3 (d) 4
122. The equation of straight line passing through (1, 2) and having intercept of length 3
between the straight lines 3x + 4y = 24 and 3x + 4y = 12 is
(a) 7x – 24y + 41 = 0 (b) 7x + 24y – 55 = 0 (c) 24x – 7y – 10 = 0 (d) 24x + 7y – 38 = 0
123. Let A(2, –3) and B(–2, 1) be vertices of a triangle ABC. If the centroid of this triangle
moves on the line 2x + 3y = 1, then locus of the vertex C is the line
(a) 2x + 3y = 0 (b) 2x – 3y = 7 (c) 3x + 2y = 5 (d) 3x – 2y = 3
124. The line parallel to the x–axis and passing through the intersection of lines ax + 2by + 3b
= 0 and bx – 2ay – 3a = 0 where (a, b) (0, 0) is
(a) 1, 12 (b) (1, –2) (c) (–1, –2) (d) (–1, 2)
126. If a vertex of a triangle is (1, 1) and the mid–points of two sides through this vertex are
(–1, 2) and (3, 2), then centroid of the triangle is
(a) 13 , 73
(b) 1, 73
(c) 13 , 73
(d) 1, 73
127. The reflection of the point (4, –13) in the line 5x + y + 6 = 0 is.....
(a) (1, 2) (b) (3, 4) (c) (–4, 13) (d) (–1, –14)
128. If P 1 and P 2 denote the lengths of the perpendiculars from the origin on the lines
2
x sec + y cosec = 2a and x cos + y sin = a cos 2a respectively then p1 p2
2 1
p p
is
equal to .....
(a) 4 sin24 (b) 4 cos24 (c) 4 cosec24 (d) 4 sec24
50
129. Locus of mid point of rod having length 2c begins to slide on two perpendicular lines is...
(a) x2 – y2 = c2 (b) x2 + y2 = c2 (c) x2 + y2 = 2c2 (d) x2 – y2 = 2c2
2
130. A(3t , 6t), B 32 , 6t and S(3, 0). Then value of SA
t
1 1 is
SB
(a) 22
3
, 13
3 (b) 22 3
, 13
3 (c) 223, 13
3
(d) 223
, 13
3
132. A straight line through the point A(3, 4) is such that its intercept between the axes is
bisected at A. It’s equation is
(a) 3x – 4y + 7 = 0 (b) 4x + 3y = 24 (c) 3x + 4y = 25 (d) x + y = 7
133. If (a, a2) falls inside the angle made by the lines y 2x , x 0 and y = 3x, x > 0 the ‘a’
belongs to
(a) (3, )
(b) 12 , 3
(c) 3, 12
(d) 0, 12
134. Let A(h, k), B(1, 1) and C(2, 1) be the vertices of right angled triangle with AC as its
hypotenuse. If the area of a triangle is 1, then the set of vvalues which ‘k’ can take is given
by
(a) (1, 3) (b) (0, 2) (c) (–1, 3) (d) (–3, –2)
135. Let P(–1, 0), Q(0, 0) and R (3, 3 3) be three points. The equation of the bisector of the
PQR is
3 3
(a) 3x y 0 (b) x 2
y0 (c) 2
xy0 (d) x 3y 0
136. The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y–intercept
–4. Then a possible value of k is
(a) –4 (b) 1 (c) 2 (d) –2
137. If A(1, 2) and B(6, 2), 3AB = 2BC and A – B – C athe value of C can be
(a) 32 , 33 (b) 272
,2
(c) 272
,2 (d) 27
2
, 2
138. The equation of a striaght line passing through the point (4, 3) and making intercepts on
the coordinate axes whose sum is –1 is given by
(a) 3x – 2y = 6 and x – 2y = –2 (b) 3x – 2y = –6 and x – 2y = 2
(c) 3x – 2y = 6 and x + 2y = 2 (d) 3x – 2y = –6 and x – 2y = –2
139. The obtuse angle bisector of the lines x – 2y + 4 = 0 and 4x – 3y + 2 = 0 is
(a) x(4 5) + y(2 5 3) + (2 4 5) 0 (b) x(4 5) + y(2 5 3) + (2 4 5) 0
(c) x(4 5) + y(2 5 3) + (2 4 5) 0 (d) x(4 5) + y(2 5 3) + (2 4 5) 0
51
140. Equation of line which is equally inclined to the axis and passes through a common points
of family of lines 4acx + y(ab + bc + ca – abc) + abc = 0
(a) y x (b) y x (c) y x 1 (d) y x 1
141. The equation of a line passing through the point of intersection of 3x – 2y = 0 and
5x + y – 2 = 0 and making an angle of measure tan–1(–5) with positive direction of x–axis is
(a) 3x – 2y = 0 (b) 5x + y – 2 = 0 (c) 5x + y = 0 (d) 3x + 2y + 1 = 0
142. The straight line perpendicular to the straight line 3 x y makes which of the
following angles with the positive direction of y–axis
(a) 300 (b) 600 (c) 450 (d) none
2 2 2 2
143. The lines p(p + 1)x – y + q = 0 and (p + 1) x + (p + 1)y + 2q = 0 are perpendicular to a
common line in 2D geometry for
(a) exactly one value of p (b) exactly two value of p
(c) more than two value of p (d) no value of p
y
144. The line L given by 5x + b passes through the point (13, 32). The line K is parallel to L
y
and has the equation cx 3 . Then distance between L and K is
17 23 23
(a) 17 (b) 15 (c) 17 (d) 15
145. The lines x + y = | a | and ax – y = 1 intersect each other in the first quadrant. Then the set
of all possible values of ‘a’ is the interval
(a) (0, ) (b) [1, ) (c) (1, ) (d) (–1, 1]
146. Consider three points P = (–sin ( – ), –cos ), Q = (cos ( – ), sin ) and
R = (cos ( – + ), sin ( – )) where 0 < < π4 then
(a) P lies on the RQ (b) Q lie on the PR (c) R lie on the QP (d) P, Q, R are non collinear
147. Triangle is formed by the coordinates (0, 0), (0, 21) and (21, 0). Find the number of
intergral co–ordinates strictly inside the triangle (intergral coordinates has both x and y)
(a) 190 (b) 105 (c) 231 (d) 205
148. A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0
at points P and Q respectively, then the point O divide the segment PQ in the ratio
(a) 1:2 (b) 3:4 (c) 2:1 (d) 4:3
149. A triangle is formed by the tangents to the curve f(x) = x2 + bx – b at the point (1, 1) and
the coordinate axes, lies in the first Quadrant. If the area is 2, then value of b is :
(a) –1 (b) 3 (c) –3 (d) 1
52
150. Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nx and
y = nx + 1 equals
m + n 2 1 1
(a) (m n)2 (b) m + n (c) (m n) (d) m + n
151. Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). The
equation of the line passing through (1, –1) and parallel to PS is
(a) 2x – 9y – 7 = 0 (b) 2x – 9y – 11 = 0 (c) 2x + 9y – 11 = 0 (d) 2x + 9y + 7 = 0
152. P(3, 1) and Q(6, 5) and R(x, y) are three points such that the angle PRQ is a right angle and
the area of RQP = 7, then the number of such points R is
(a) 0 (b) 1 (c) 2 (d) 4
153. If one of the diagonal of a square is along the line x = 2y and one of its vertices is (3, 0)
then its sides through this vertex are given by the equaions.
(a) y – 3x + 9 = 0, 3y + x – 3 = 0 (b) y + 3x + 9 = 0, 3y + x – 3 = 0
(c) y – 3x + 9 = 0, 3y – x + 3 = 0 (d) y – 3x + 3 = 0, 3y + x + 9 = 0
(a) 3,
2
3 3
6
(b) 2, 12
(c) 4 ,
3 3 2
4
(d) 12 , 12
155. If the extremities of the base of an isoscelese triangle are the points (2a, 0) and (0, a) and
the equation of one of the sides is x = 2a, then area of the triangle is
53
159. Comrehensive type : A straight line L with negative slope passes through the point (9, 4)
and cuts the positive corodinate axes at the points P and Q respectively. Now answer the
following
(A) Minimum value of OP + OQ, as L varies, where O is the origin is
(a) 18 (b) 25 (c) 36 (d) 49
(B) Area of OPQ, when OP + OQ becomes minimum is _____ sq units
(a) 75 (b) 225 (c) 125 (d) 200
(C) Let R be a moving point on the x – y plane such that OPRQ becomes a ractangle then
locu of R as L varies is
x 4 x 4 9 4 4 1
(a) 9 y (b) 9 y 1 (c) x y 1 (d) x y 1
160. If the lines x = a + m, y = –2 and y = mx are concurrent, the least value of | a | is.....
(a) 0 (b) 2 (c) 2 2 (d) none
161. A(–3, 4), B(5, 4) and C and D form a rectangle, x – 4y + 7 = 0 is a diameter of the
circumcircle of rectangle ABCD, the area of ABCD is
(a) 8 (b) 16 (c) 32 (d) 64
162. The line 3x + 2y = 24 meets y–axis at A and x–axis at B. The perpendicular bisector of AB
meets the x–axis at c, then area of ABC is
54
UNIT- 11 line-lines
Hints
1. ANS : B
y y1 m( x x1 ) where m = 1
equidistant from (1, -2) and (3, 4)
2 1 a 4 3 a
a 1
2 2
RL y-x+1=0
2. ANS : C
2
distance between 1 and 2 =
5
3 is to both 1 and 2 3 : 2x y 14 0
also A(-5, 4) 3 k=14
3. ANS : C
eqn of angle bisector
3x 4 y 2 5x 12 y 2
9 16 25 144
7x 4y 18 0 or 4x 7y 1 0
4. ANS : B
4 6
point of intersection of the lines is ,
13 13
m=-5 ´Ü³ÜÜ y y1 m( x x1 )
RL : 5x+y-2=0
5. ANS : D
a b c
lines are concurrent b c a 0
c a b
(a b c)(ab bc ca a 2 b 2 c 2 ) 0
(a b ) 2 ( b c ) 2 (c a ) 2 0 [ a b c 0]
a bc
6. ANS : A
y y1 m( x x1 ) and eqn of line passes through (1, 2)
mx-y+2-m=0—(A)
55
4 4m 6 9m 16 4m 6 21m
This line intersect to the given line at point A , and B ,
3 4m 3 4m 3 4m 3 4m
7
also AB 3 m
24
RL : 7x-24y+41=0
7. ANS : C
there for x 0 a 0 (a , a 2 )
y x2 a 2 a2 0 a 12 ____(1) 1
a 3
y 3x a 2 3a 0 a 3 __( 2) 2
8. ANS : C Figure
slop of QR tan 3
3
QS is bisector of PQR m 3
which passes through (0, 0)
from, y y1 m( x x1 ) , y 0 3(x 0)
3x y 0
1 1 1 1a b2 1c 0
9. ANS : A , , H.P and x y 1
a b c 4 b c 0
by comparing x = 1, y=-2
line passes thorugh (1, -2)
10. ANS : B
6
perpendicular distance between (0, 0) and 2x+y+6=0 = OQ
5
9
perpendicular distance between (0, 0) and 4x+2y-9=0 = OP =
2 5
OP 3
required ratio
OQ 4
11. ANS : B The point lies on the line 3x-2y-2=0
3a 2
X - co-ordinate : a then y - co-ordinate :
2
1
then the perpendicular distance formula : 9a 12 15 a 3,
3
1 1 3
a 3 x 3, y 7 or a x , y
2 3 3 2
56
1 3
7
required points are 3, 2 , 2 , 2
7
12. ANS : C AP AB AP AB P AB
3
A - P - B is not possible
AP PA 4 PA 4
13. ANS : B 0, also 4 : 5
PB AB 1 PB 5
5k 0 20
14. ANS : A p 1 (3k 16)(k 3) 0
k 2 16
16
k , or k 3
3
15. ANS : D p sin 2, q cos 2
p 2 q 2 sin 2 2 cos2 2 1
4b 12
16. ANS : C (0, b) be the point on the y-axis then 4
17
17 4 or b 17 3
p(0, 3 17 ) or p(0, 17 3)
1
17. ANS : B a .b
2
1
(2b)(p b) b2 †Ü‘äÝ´Ü
2
p 0 or p 2b
vertex of triangle lies on line x=0
12x 2 4y 2 3
a b
20. ANS : B (h, k ) ,
2 2
OA 2 OB 2 AB 2 a 2 b 2 4c 2
57
h 2 k 2 c2
locus of the mid point : x 2 y 2 c 2
1
21. ANS : D slope of OA , slope of OB = 2
2
1
2
tan 1 2
1
tan 1 3 4
1 2 .2
sin 1 3 5 †Ü‘äÝ´Ü
22. ANS : A AM BC AM 3 2
2
a
2 2
from right AMB , AM a
2
a 3 †Ü‘äÝ´Ü
23. ANS : C
4 2
if A lies on the line x+2y=1 then, t=
3
5 2
if A lies on the line 2x +4y=15 then, t =
6
4 2 5 2
t
3 6
24. ANS : C A(r1 cos , r1 sin ), B(r2 sin , r2 cos ) are on line
r2 sin r2 cos r cos r2 sin
1 and 1 1
a b a b
1 1 1 1
Now OA 2 OB2 r 2 r 2
1 2
1 1 a 2 b2
OA 2 OB2 a 2b2
25. ANS : B
b x y
take a in 1
2 a b
also take distance between 2x+y-b=0 and (0, 0) is 5
2x y 5 0 which is RL.
26. ANS : D
2 3
x+y-1=0 and 2x-3y+1=0 (sloving the eqn) ,
5 5
58
27. ANS : C
t t
t<0 0
1 t 1 t
28. ANS : B x=4t+1, y=5t+2 y x 1 t 0
y x 1 Positive
29. ANS : A x=2t+2, y=4t+3 3x y 10 t 9
(x, y) AB 0 t 1 9 10t 9 19
3x y maximum value = 19
30. ANS : C x=8t-2, y=5-3t and t R [0, 1]
AB AB {(8t 2,5 3t) / t R [0,1]}
a 1 sin b 3
31. ANS : B cos ,
a 2 b2 2 a 2 b2 2 3
4
p 2 , Now from x cos y sin p
1 3
x cos 3 y sin 3 2
32. ANS : B circumcentre = controid
AD AB sin 60 [from ABD ]
a
r
2 3
2 13 AD †Ü‘äÝ´Ü
a2
one side of PQRS = x x 2 x 2 (2r ) 2 ,
6
a2
area of square
6
2 1 1
33. ANS : B = 0 [concurrent] 2ac ab bc
b c a
a, b, c H.P
34. ANS : A 5x+4y+k=0 which passes thorugh (2, 3)
required vertex : (6, -2) “å.
x y
35. ANS : D take b=-a-1 in 1 , also (4, 3) on given line
a b
x y
a 2, RL , 1, and x y 1
2 3 2 1
36. ANS : A 12 2 9 2 15 2 †Ü‘äÝ´Ü
x y
1
9 12
59
3x 4y 36 or 4x 3y 36
m1 m 2
37. ANS : D m1 cot 85 , m 2 cot 40 tan
1 m1m 2 ,
fixed point is 1 2 , 13
1 1
44. ANS : A indentical : 5, 1
5 5
4
5
45. ANS : A by sloving x ,
3 4m
1 1
3 4m 1, 5. m , -1 , , -2
2 2
integer no. of m = 2
46. ANS : C
60
M(4, 5) is foot of perpendicular from 0(0, 0) slope of OM = 5 4 slope of
k 1 7
49. ANS : D mid point of PQ ,
2 2
7 k 1
eqn of perpendicular bisector : y (k 1) x whose y-intercept = - 4
2 2
k 4
50. ANS : B the eqn of line perpendicular to the given line and passing through (2, 2) is : x-3y+4=0
y - intercept 4 3
61
x x1 1
k
x2 x 1
54. ANS : D
none of the point outof A, B, C is not on the line or point of intersection 5 2 , 15 2
dy
55. ANS : A slope of the curve is constant m.
dx
y mx c
5 12 5 12
56. ANS : C x y 1 cos , sin
13 13 13 13
12
tan 1 [ is in the third quadrant]
5
57. ANS : C slope tan 3 mPQS 60
slope of QS 3
using y y1 m( x x1 ) †Ü‘äÝ´Ü
3x y 0
58. ANS : A y is not fixed so all the lines are not parallel to x x1
they intersect to the line x x1
59. ANS : C from the eqn x cos y sin p
we get x cos y sin 10
6 6
3x y 20 0
60. ANS : D x cos y sin P, where 30 †Ü‘äÝ´Ü
2p
A , 0 , B(0, 2p)
3
50 1 50
BOA (OA)(OB) p 2 25, p 5
3 2 3
3x y 10
61. ANS : C
slope of BC m1 1, slope of AB m 2
m1 m 2 m 2 2 3 OR
tan 1 m1m 2
m2 2 3
, which passes thorugh
3, 1 .
y (2 3)x 3
or y (2 3)x 3
62
62. ANS : A
x cos y sin p where p 2 which passes through
3, 1
4 sin 2 2 2 sin 1 0
3 1 3 1
sin , cos
2 2 2 2
( 3 1)x ( 3 1)y 4
OR sin
3 1
, cos
3 1
,
2 2 2 2
( 3 1)x ( 3 1)y 4
63. ANS : B slope of BC slope of AM = -1
3a b 9 0 — (1)
slope of AC slope of BH = -1
2a b 4 0 — (2) solve (1) and (2)
c(a, b) c( 1, 6)
64. ANS : A x+y=0 and x-y=0 are perpendicular
the circumcenter of is on the line x-7=0
circumcenter is (7, 0)
1 2 1 x y 2 1
65. ANS : D from we get 0
c b b a b b a
1 1
( x 1) ( y ( 2)) 0
a b
which passes through (1, -2)
66. ANS : B slope of line x+y+3=0 =-1 †Ü‘äÝ´Ü
slope of the line perpendicular to it =1
67. ANS : A X=3 is a vertical line and slope of other line tan 3
3
2 2 3 6
1
68. ANS : C slope of y=e is m1 = 0 slope of other line m 2
3
1
0
m1 m 2 3 1 ,
tan 1 0 3 6
1 m1m 2
73. (C) x – y + 9 = 0 distance between two liners is 2 . eq of RL passes through (–5, 4) any line
to given line is x – y + k = 0 –5 –4 + K = 0
K=9
74. (C) a = 2 b = 3
1 5 7 2 a 4 b 6
diagonals bisect each other choose the 4th vertecx as (a, b) , ,
2 2 2 2
a = 2 and b = 3
2 2 2
75 (C) x 3 y 3 a 3
differentiating
2 13 2 13 dy
x y 0
3 3 dx
1
dy y 3
1 at a , a , dy 1
dx 8 8 dx
x 3
eq of tanget at a 8 , a 8 is
64
a
8 8
ya xa x y 0
4
a a a
sum of intercepts = 2
4 4 2
a=4
k 0 4
1 2h 4k
h 0 7
OB AC
k 3 1
1 5h k 13 0
h 2 5
h = – 4 k = –7
x 1 y 2
r
1 3
2 2
r
x= 1 x + y = 6
2
3 r 3
y= r2 1 r 2 = 6
2 2 2
65
6
r 3 3 1 = AP
3 1
2 8
79. (D) ,
17 17
image of (x1 y1) is (x2 y2) in line ax + by + c = 0 then
x 2 x1 y 2 y1 ax by c
2 1 2 12
a b a b
x 2 0 y 2 0 2 0 0 1
1 4 17
2 8
x2 y2
17 17
3
80. (C) 3,
4
66
82. (B) family of concurrent lines
2b = a + c
a – 2b + c = 0
ax + by + c = 0 passesthrough (1, –2)
83. (B) 5 2 7
BC = 40
AC = 5
BC BD 40 8
AC AD 5 1
67
1 a a2
1
86. Area of = 1 b b2
2
1 c c2
1
= (a – b) (b – c) (c – a)
2
1
= (–2) (–2) (4) = 8 sq unit
2
6 6
87. (B) Vertices of are , 2 , 2 , 0,5
7 7
18
its area = sq unit
7
88. (A) 7y = 5x
eq of AB eq of AC
y – 7 = 1 (x – 5) y – 7 = – (x – 5)
y – x = 2 ......(1) x + y = 12 .....(3)
Also y + x = 0 ......(2) x–y=0 .....(4)
P (–1,1) Q (6, 6)
P is midpoint of AB Q is midpoint AC
B = (–7, –5) C = (7, 5)
68
x7 y 5
eq of BC = 10x + 70 = 14y + 70
7 7 5 5
x7 y5
\ 5x = 7y
14 10
m2
89. (A) 1
1 2m
1
m = –3 and m =
3
20
90. (A)
3
2 1 2 1
AD = 5
2 2
2 1
5
tan 60° = a
2
20
a=
3
x1 x 3 2 x 2 x 4 2 y1 y2 2 y3 y4 2 0
x1 = x3 y1 = y2
x2 = x4 y3 = y4
69
92. (A) bx + ay = 3xy
eq of AB = y – b = m (x – a)
b
a m b am
G= ,
3 3
b
3h = a – , 3k = b – am
m
eliminating ‘m’ we will get bh + ak – 3hk = 0 ie bx + ay – 3xy = 0
1
93. (B) , 1 Slope of AB = Slope of BC
2
2 2k 2k 2k 6 2k
k 1 k 1 k k 4
(4k – 6) (2k – 1) + 10(2k – 1) = 0
1
k= or k = –1
2
94. (B) (7, –2) (4, 3)
x1 + y1 = 5
x2 = 4
G = (4, 1)
1 x1 x 2 y y2 2
4& 1 1
3 3
x1 + x2 = 11 y1 + y2 = 1
x1 = 7 x2 = 4
y2 = 3 y1 = –2
95. (A) 4 : 1
2 1 2
P= ,
1 1
2 1 2
3 4 70
1 1
= 4
70
96. (A) 11x – 3y + 9 = 0
eq of lines
3x – 4y + 7 = 0
–12x – 5y + 2 = 0
a1a2 + b1b2 = –36 + 20 < 0 eq of acute
angle bisector is
3x 4y 7 12x 5y 2
5 13
11x – 3y + 9 = 0
3 1
97. (D) ,
4 2
3 1
ax + by + c = 0 ,
4 2
3 0 1 0 3
=
2 1
11
71
100. (A) 8
1 1 1
3 m 3 7
m = –1
m 1
1 1
3 21
7 1
REOL y 1 x 5x + 5y = 3
10 10
3x 4y 12
102. (A) 21x + 27y – 121 = 0 ; at (–1, 4), 0
12x 5y 12
we have to take +ve sign
3x 4y 12 12x 5y 7
5 13
21x + 27y – 121 = 0
103. (B) x + 7y + 6 + 5 2 = 0
iet line is x + 7y + = 0 distance of this line from (1, –1) is
1 7 1 7
But as per Que 0
50 50
=6+ 5 2
72
104. (A) (3, 1) and (–7, 11), any pt on line x + y = 4 can be taken as (t, 4 – t) the distance of this pt
from the line 4x + 3y – 10 = 0 is 1
4t 3 4 t 10
1
5
t2
1
5
t = 3 or t = –7
= 25 or = – 3
106. (A) 9x – 7y = 1
3x 4y 7 12x 5y 6
5 13
ie 21x + 27y + 121 = 0 &
99x – 77y – 61 = 0
7 9
there slopes = and
9 7
eq of lines passing through (4, 5)
7
y – 5 = x 4 7x + 9y = 73
9
9
y–5= x 4 9x – 7y = 1
7
107. (C)
We know that foot of from (x1 y1) on the line ax + by + c = 0 is
73
x x1 y y1
ax1 by1 c
a b a 2 b2
0 0 1 3 4
ie \ ,
3 4 25 25 25
24x 7y 20 4x 3y 2
109. (C)
25 5
27x + 7y – 20 = 20x – 15y – 10 (by +ve sign)
4x + 22y – 10 = 0
2x + 11y – 5 = 0
x y a 1
112. (B) 2x + y + 5 = 0, 1,
a b b 2
2x y
1 2x + y – b = 0
b b
b 5
5 b=+5 a=+
5 2
2x y
REOL 1
5 5
113. (A) 2x + y + 6 = 0
Line intersect x axis at pt (3, 0), (–3, 0) with slope – 2
y – 0 = –2(x – 3) y – 0 = –2 (x + 3)
y + 2x – 6 = 0 y + 2x + 6 = 0
74
114. (A) 3
0 1 2 3
AM =
3 1 2
a2 9
a2
4 4
a2 = 3
a= 3
a cos t bsin t 1
116. (B)
3
, centriod
a sin t b cos t
3
a cos t bsin t 3 1
a sin t bcos t 3
sq sadd a2 + b2 = (3 – 1)2 + (3)2
117. (D) eq of AB :
a cos a sin
y a sin x a cos
a sin a cos
cos sin
y a sin x a cos
cos sin
75
118. (A) Lie on a straight line
1 3
119. (D) sin
5
1
slope of OA = m1
2
slope of OB 2 m 2
1 2
^ 2 3 3
OA OB tan
1
1
tan1 sin1
1 2 4 5
2
1
120. (A) 2 ,5
3
Incentre = centroid
AB = BC = CA = 2
121. (B) 2
1
8 6
2
Inradius = =2
S 1 8 6 10
2
76
123. (A) 2x + 3y = 9
Let C is ()
2
controdi is ,
3 3
2
2 3 1
3 3
2 3 9
124. (D)
7
126. (B) 1,
3
y 1 6 1 4
y=5 =3
+ 1 = –2 +1=4
=–3 =3
1 3 5 1 3 3
centroid ,
3 3
7
= 1,
3
77
127. (D) (–1, –14)
L et B (x 1, y1) is reflection of A(4, –12)
x 4 y1 13
c 1 , it lie on line 5x + y + 6 = 0
2 2
x y y1 13
5 1 60 5x1 y1 7 0
2 2
Slope of AB Slope of (5x + y + 6) = –1
y1 13
x 4 5 1 5y1 x1 69 0
1
x1 = –1 y1 = –14
a 2 4 tan 2 a 2 cos 2 2
=
1 tan 2 2 cos 2 sin 2
2
22 tan
=a 2
a 2 cos 2 2
1 tan
= a 2 sin 2 2 cos 2 2 a 2
1 4 2
p12 p 22 a sin 4
4
p1 p 2 p12 p22 2
2cosec 4
p2 p1 p1p2 sin 4
129. (B) x2 + y2 = c2
a = 2h
b = 2k
OA2 + OB2 = AB2
a2 + b2 = 4c2
4h2 + 4k2 = 4c2
h2 + k2 = c2
78
1
130. (C)
3
SA2 = (3 – 3t2)2 + (6t)2
= 9[1 – 2t2 + t4 + 4t2]
= 9 (1 + t2)2
2 2
3 6
SB2 = 3 2 0
t t
2 1 4
= 9 1 2 4 2
t t t
2
1
= 9 1 2
t
1 1 1
SA SB 3
22 13
131. (B) ,
3 3
AB = 64 16 80 4 5
BC = 12 4 125 5 5
BC 5
BA 4
5 5
4 69 4
7 1
coordinate of D = 5 ,
5
1 1
4 4
22 13
= ,
3 3
132. (B) 4x + 3y = 24
x y
1
a b
79
3 a b
, 4 , a= 6, b = 8
a 2 2
6y + 8x = 24
1
33. (B) , 3
2
x
y ,x0
2
y 3x, x 0
a
a 2 3a 0, a2 0
2
1
– a 3
2
134. (C) (–1, 3)
135. (A) 3x y 0
y – 0 = tan120° (x – 0)
Slope of QR = 3
136. (A) (–4)
1
Slope of PQ = –
k 1
Slope of AB = k –1
k 1 7
R is mid point of PQ , R
2 2
7 k 1
eq of y k 1 x
AB 2 2
k2 = 16 k=+4
27
137. (B) , 2
2
80
BC 3
and A B C
AB 2
B divide AC from C in ratio 3 : 2
3 3
2 1 x 2 y
,2
(6, 2) = 3 3
1 1
2 2
27
x & y2
2
x y 4 3
138. (A) 1 1 Also a + b = -1
a b a b
4 3
1 a=+2
a 1 a
a = 2 b = –3
a = –2 b = 1
81
4 6
POI of 3x – 2y = 0 and 5x + y – 2 = 0 is ,
13 13
–1
The line makes an angle of measure tan (–5) with x-axis
tan 1 5 tan 5
6 4
REOL y 5x
13 13
5x y 2 0
3x y 20 0
1
0
3 1
tan 6
1 0 3
angle with the +ve direction of y-axis is
2 2 6 3
23
144. (C)
17
x y
since L : 1 Passess through (13, 22)
5 b
82
13 32
1 b = –20
5 b
line L becomes
x y
1 4x – y – 20 = 0 .....(1)
5 20
x y
L is // to K : 1
c 3
1
4
e c =
3
1 1 4
3
K becomes 4x – y + 3 = 0 ........(2)
23
distance between K & L = 17
83
148. (B) 3 : 4 Let r1 cos , r1 sin is on
9
4x + 2y = 9 r1
4cos 2sin
Let r2 cos , r2 sin lie on
2x y 6 0
6
r2
2cos sin
OP r1 3
OQ r2 4
1 b
Ininter sept form OA and OB 1 b
2b
1
Area of OAB OA OB 2 given
2
1 b2 4 2 b
b = –3
1
150. (D)
mn
1 n
coordinate of pare ,
nm nm
Area of // gm OPQR = 2 area of OPQ
84
0 0 1
1
Desired area = 2 0 1 1
2
1 n
1
nm n m
1
=
n m
151. (D) 2x + 9y + 7 = 0
6 7 1 3 13
Mid point of Q (6, –1) and R(7, 3) is , ,1
2 2 2
1 2 2
P
13
Slope of median through 2 9
2
Equation of the required line is
2
y + 1 = (x – 1) or 2x + 9y + 7 = 0
9
85
m 1 1
Now tan 2 m = 3,
4 1 m 3
2
1 1
154. (B) 2, Given is right angled at vertex 2,
2 2
AC = BC = t
2
= 4a 2 a t
5a
t=
2
5a
coordinats of third vertex C = 2a,
2
155. (B) 3x y 10 0
Let p is length of from the original on the given line. Then its equation in normal from
x cos 30° + y sin 30° = p or
3x y 2p
This meets the coordinats arces at
A 2p ,0 and B 0, 2p
3
1 2p 2p 2 50
Area of OAP = 2p .
2 3 3 3
– p = + 5, REq 3x y 10
156. (C) 1
lines are concurrent
1 a a
b 1 b 0
c c 1
86
1 1 1
a
abc 1 1 1 0
b
1 1 1
c
1
1 1
a
1 1
abc 1 1 0 0
a b
1 1
1 0 1
a c
R 3 R 3 R1
R 2 R 2 R1
1 1 1 1 1 1 1
abc 1 1 1 1 1 1 0
a b c a b c a
(1 – b) (1 – c) + c(1 – a) (1 – b) + b(1 – c) (1 – c) (1 – a) = 0
1 c b
0
1 a 1 c 1 b
a b c
1 0
1 a 1 a 1 c
a b c
1
1 a 1 b 1 c
87
fhe ykÃkðk{kt ykðþ
2
6
eq of BC : y – 6 = 5 x 7
11
7
5
ie 14x + 23y – 40 = 0
9 4
158. (A) (b) 25 (B) (a) 75 C (c) 1
x y
(A) Let eq of line is y – 4 = m(x – 9)
9m 4
P= ,0 Q = (0, 4 – 9m)
m
4 4
OP + OQ = 9 4 9m 13 2 9m 25
m m
(B) OP + OQ is minimum when
4 4 2
9m m2 m
m 9 3
P = (15, 0) & Q = (0, 10)
1
Area of OPQ = 15 10 75
2
9m 4 9 4
(C) h 1
m h k
k = 4 – 9m
9 4
1
x y
88
þu.
1 3 24
3 = 5, 2 =0
3 3
D = (5, 0)
x1 x 2 x x3 x x
1 2 5 & 3 1 3
2 2 2
y1 y 2 y y3 y y3
2 2 0 & 1 4
2 2 2
A = (–1, 6) B = (3, –2) C = (7, 2)
eq of AB = 2x + y = 4
2 Ar ABC
Height of altitude from A = 6 2
BC
99
160. (B)
19
AP = CQ = x
45 x 153 x
–
10 28
135
– x
19
135
45
19 99
slope of PQ =
10 19
161. (C) 2 2
lines are concurrent
1 0 a m
0 1 2 0 m2 + am + 2 = 0
m 1 0
89
Answer Key
1 B 35 D 69 A
2 C 36 A 70 C
3 C 37 D 71 A
4 B 38 A 72 B
5 D 39 D
6 A 40 D
7 C 41 D
8 C 42 C
9 A 43 C
10 B 44 A
11 B 45 A
12 C 46 C
13 B 47 A
14 A 48 A
15 D 49 D
16 C 50 B
17 B 51 B
18 A 52 C
19 B 53 B
20 B 54 D
21 D 55 A
22 A 56 C
23 C 57 C
24 C 58 A
25 B 59 C
26 D 60 D
27 C 61 C
28 B 62 A
29 A 63 B
30 C 64 A
31 B 65 D
32 B 66 B
33 B 67 A
34 A 68 C
73 c 101 c 129 b
74 c 102 a 130 c
75 c 103 b 131 b
76 b 104 a 132 b
77 b 105 d 133 b
78 b 106 a 134 c
79 d 107 c 135 a
80 c 108 c 136 a
81 a 109 c 137 b 157 c
82 b 110 b 138 a 158 c
90
83 b 111 a 139 b 159 a&b&c
84 a 112 b 140 a&d 160 b
85 a 113 a 141 b 161 c
86 b 114 a 142 b 162 a
87 c 115 a 143 b
88 a 116 b 144 a
89 a 117 d 145 c
90 a 118 a 146 b
91 b 119 d 147 d
92 a 120 a 148 b
93 b 121 b 149 c
94 d 122 a 150 b
95 a 123 a 151 d
96 a 124 b 152 c
97 d 125 b 153 a
98 b 126 b 154 b
99 a 127 d 155 b
100 a 128 c 156 b
91