Straight Lines (Package Solutions)
-
Upload
devarshwali -
Category
Documents
-
view
44 -
download
6
description
Transcript of Straight Lines (Package Solutions)
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 67 -
Section-A
Q.No. Solution
1. Answer (1)
Distance = 2 2sin cos 1
2. Answer (2)
Centroid = 1 2 6 1 6 1
, (3, 2)3 3
3. Answer (3)
0 0 11
0 1 22
0 4 1
h
4h = ± 4
h = ± 1
Hence sum = 1 – 1 = 0
4. Answer (4)
1
1 1 1 0
2 2 1
a b
(1 2) (1 2) 1(2 2) 0a b
0a b a b
5. Answer (2)
Ratio = 4 6
2 :15
c ab
6. Answer (1)
Ratio = 11
4
b ca
externally
10 Chapter Straight Lines
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 68 -
Q.No. Solution
7. Answer (1)
Ratio = tan tan 3 2 3
2 2 3tan 1
B CA
8. Answer (3)
Orthocentre is not always inside the triangle
9. Answer (4)
Let 1 1 2 2 3 3( , ), ( , ), ( , )A x y B x y C x y
1 1
2 2
3 3
11
12
1
x yA x y
x y = Rational number
But the area of equilateral triangle is also calculated by
23 (side)
4A = Irrational
Hence triangle cannot be equilateral.
10. Answer (3)
Slope = ±1
11. Answer (1)
2tan45
1 2
mm
1
3,3
m
Product = –1
12. Answer (2)
In this case the line will be parallel to y-axis. Hence the angle = 90°.
13. Answer (3)
0 = mx + c
cxm
14. Answer (3)
y – 4 = 3(x – 3)
y – 3x + 5 = 0
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 69 -
Q.No. Solution
15. Answer (4)
16. Answer (3)
In a parallelogram, mid-points coincide
Mid-point of PR = Mid-point of SQ
2
6,
2
4
2
72,
2
15 ba
a = 2, b = 3
17. Answer (3)
A (a, a)
B (a + 1, a + 1)
C(a + 2, a)
AC = 2
BM = 1
Area of 122
1.
2
1 BMAC = 1 square unit
18. Answer (4)
Lines are 3x – 4y + 7 = 0
–12x – 5y + 2 = 0
a1 a2 + b1 b2 = –36 + 20 < 0
positive sign gives acute angle bisector, 25144
25–12–
169
74–3
yxyx
11x – 3y + 9 = 0
19. Answer (3)
Slopes of diagonals are 3
1– and 3.
1–33
1–21
mm Diagonals are perpendicular.
Parallelogram is a Rhombus.
20. Answer (3)
Let point on line x + y = 4 be (x, 4 – x)
916
10––4341
xx
| x + 2 | = 5 x = 3 & –7
Points are (3, 1) & (–7, 11)
S( , )a b R(5,7)
Q(4,6)P(1,2)
M
a + 1
aA
B
CM
a a +1X
Y
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 70 -
Q.No. Solution
21. Answer (1)
Equation of a lines which passes through the intersection of two lines
L1 + L2 = 0
Let L1 x + 2y – 10 = 0
L2 2x + y + 5 = 0
L1 + 2L2 = 0
(x + 2y – 10) + 2 (2x + y + 5) = 0
5x + 4y = 0
Answer is option (1).
22. Answer (2)
1
2
5
10
BCAB
CMAM
By section formula the coordinates of m
3
1,
3
1
12
7–8,
12
1–2
Equation of BM 1–
31
1–
5–31
5– yx
x – 7y + 2 = 0
23. Answer (1)
Reflection of y = log10x
About y = x is x = log10y
y = 10x
24. Answer (1)
Equation of BC
x + y – 2 = 0
2
1
11
2–1–2
AM [Altitude of equatorial
2
3 side]
2
1side
2
3
Side = 3
2
A
B
C
(5, 1)
(–1, –7) (1, 4)M
B
A
C
(2,–1)
M
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 71 -
Q.No. Solution
25. Answer (4)
Solve these equations
1–23
1–2
CCCx
&
1–235–
1–
CCCy
Here C 1 at C = 1 lines are coincident.
23
1
1–23
1–lim
2
1
C
CCC
CxC
5
2x
25
1–
235–
1lim
1–235–
1–lim
11
CCCCy
CC
Point of intersection is
25
1–,
5
2
26. Answer (2)
Lines xcos + ysin = p and xsin – ycos = 0 are perpendicular to each other. Thus ax + by + p = 0 is equally inclined to these lines and it will be the angle bisector of these lines. Now equations of angle bisectors is
xsin – ycos = ± (xcos + ysin p)
x(cos – sin) + y(sin + cos) = p
or x(sin + cos) y(cos sin) = p
Comparing these lines with ax + by + p = 0, we get
1cossinsincos
ba
a2 + b2 = 2 or 1cossincossin
ba
a2 + b2 = 2
27. Answer (3)
We observe that the sum of coefficients in all equations is zero.
Hence the lines are concurrent at (1, 1).
28. Answer (2)
Equation of CN be x = 4
Let coordinate of N be (4, b)
N is mid point of AB
Coordinate of B (7,2 b – 2)
B lie on the line x + y = 5
7 + 2b – 2 = 5
b = 0
B (7, –2)
A
B C
G
(1, 2)
(4, 1)
(x y )1 1
N M
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 72 -
Q.No. Solution
29. Answer (3)
Slope of line = ± 1
Let equation of straight line be y = x + C
x – y + C = 0 …(1)
Line (1) is equidistance from points (1, –2) & (3, 4)
2
4–3
2
21 CC
| C + 3 | = | C – 1|
C + 3 = –C + 1 C = –1
From equation (1) line is x – y – 1 = 0
30. Answer (4)
Perpendicular distances of the lines from origin are
5
6OM &
20
9–ON =
52
9
O divides MN in the ratio = 2
3:2
52
9:
5
6 = 4 : 3
31. Answer (3)
By complete squaring method 2(x – 2)2 + 3(y – 3)2 = k
If k = 0
2(x – 2)2 + 3(y – 3)2 = 0
Then necessarily (x – 2)2 = 0 & (y – 3)2 = 0 x = 2 & y = 3
Equation represents a point if k = 0
32. Answer (2)
Equation of angle bisectors of x2 – 2pxy – y2 = 0
be pxyyx–1––1
– 22
px2 + 2xy – py2 = 0 …(1)
x2 – 2qxy – y2 = 0 …(2)
Represents coincident lines
1–
–
2–
2
1
pq
p pq = – 1
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 73 -
Q.No. Solution
33. Answer (1)
2
cab
02
cycaax
012
1
2
1
ycyxa
012
1
2
1
yacyx
This line will always pass through the intersection point of two lines
02
1 yx & 01
2
1y
Solve these equations y = –2, x = 1
Fixed point (1, –2)
34. Answer (1)
44–4 22 xyyx = 1 + 2y – x
Squaring both sides
x2 + 4y2 – 4xy + 4 = 1 + 4y2 + x2 + 4y – 4xy – 2x
2x – 4y + 3 = 0
Equation represents a straight line
35. Answer (4)
Equation of pair of lines which passes through the intersection of given curves be L1 + L2 = 0
(x2 + y2 + 2xy – 4) + (3x2 + 5y2 – xy – 7) = 0
(1 + 3)x2 + (1 + 5)y2 + (2 – ) xy – (4 + 7) = 0
Lines passes through origin
Equation should be homogeneous
Put 074
7
4–
Equation of lines is 5x2 + 13y2 – 18xy = 0
36. Answer (3)
px2 + 2axy + qy2 = r(1)2
px2 + 2axy + qy2 = r [ax + by]2
(p – ra2) x2 + (q – rb2)y2 + (a – rab) 2xy = 0
These lines are perpendicular
p – ra2 + q – rb2 = 0
p + q = r(a2 + b2)
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 74 -
Q.No. Solution
37. Answer (4)
Since the product of the slopes of the four lines represented by the given equation is 1 and a pair of lines represents the bisector of the angles between the other two. The product of the slopes of each pair is 1. So let equation of one pair be ax2 + 2hxy ay2 = 0
The equation of its bisector is hxy
ayx
2
22
By hypothesis,
x4 + x3y + cx2y2 xy3 + y4
= (ax2 + 2hxy ay2) × (hx2 – 2axy hy2)
= ah(x4 + y4) + 2(h2 a2)(x3y xy3) 6ahx2y2
Comparing the respective coefficients, we get
ah = 1, c = 6ah = 6
38. Answer (1)
Here the lines x + y = 0 and x – y + 1 = 0 are perpendicular to each other
So take 1
2
x yx and
2
x yy
1 2x y x …(i)
2x y y …(ii)
Solving (i) & (ii), we get
1 1and
2 22 2
x y y xx y
Putting these value in the given locus we get
2 22 2 1 0x y x
Changing (x, y) into (x, y) we get
2 22 2 1 0x y x
39. Answer (2)
Here the origin remains fixed and axes are rotated through angle , in anticlockwise sense
Let new co-ordinates of the point (x, y) becomes (x, y)
Then equation of transformation will be
x = xcos – ysin
y = xsin + ycos
Changed equation will be
a(xcos – ysin)2 + 2h.(xcos – ysin) (xsin + ycos) + b(xsin + ycos) = 0
(Since this expression is free from xy)
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 75 -
Q.No. Solution
–2a sincos+ 2h(cos2 – sin2) + 2bsincos = 0
– a sin2 + 2h cos2+ 3 sin2= 0
sin2 (a – b) = 2h cos2
2
tan2h
a b
1 22 tan
ha b
11 2tan
2
ha b
40. Answer (3)
The given equation is
ax2 + 2hxhy + by2 + 2gx + 2fy + c (a1 x + b1 y + c1) (a2 x + b2 y + c2) = 0
Where
a1a2 = a a1b2 + b1a2 = 2h
b1b2 = b a1c2 + c1a2 = 2g
c1c2 = c b1c2 + c1b2 = 2f
Lines are
a1x + b1y + c1 =0 …(i)
And a2x + b2y + c2 =0 …(ii)
Product of distances of lines (i) and (ii) from origin is given by
= 1 2 1 2
2 2 2 2 2 2 2 2 2 2 2 21 1 2 2 1 2 1 2 1 2 1 2
c c c c
a b a b a a a b b a b b
= 2 2 24 2
ca b h ab
= 2 2( ) 4
c
a b h
41. Answer (1)
Any curve passing through the intersection of the given curves is
ax2 + 2hxy + by2 +2gx + (ax2 + 2hxy + by2 + 2gx) = 0 …(i)
This will be pair of straight lines passing through origin if it is IInd degree homogeneous in x and y. For this the condition on (i) is
Coefficient of x = 2g + 2g = 0 g
g
Also the lines are perpendicular
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 76 -
Q.No. Solution
i.e. coefficient of x2 + coefficient of y2 = 0
a + a + b + b = 0
a + b + (a + b) = 0
a + b = – (a + b)
g(a + b) = g(a + b)
42. Answer (2)
The given equation of the line is x – y = 2
A = (2, 0) and B = (4, 2)
1 2 0tan 45
4 2BAX
B
45°45°
A(2, 0)
B(4, 2)
(0, 0)x
2
B AX , where B is the new position of B so
2
Where AB makes an angle with +ve direction of x-axis
Equation of AB= x – 2 = 0
43. Answer (2)
Let A be the point of incidence
A is intersection of
x – 2y – 3 = 0 …(i)
and 3x – 2y – 5 = 0 …(ii)
A = (1, –1) A
P
Q
3x y – 2 – 5 = 0
xy
– 2 – 3 = 0
Let P be any point on the line of incidence x – 2y – 3 = 0. So we take P = (3, 0)
Let Q(, ) be angle of P in the line 3x – 2y – 5 = 0
PQ the line 3x – 2y – 5
13 2
B
…(iii)
And 3
3 2 5 02 2
…(iv)
Equation (iii) 3 + 2 = 6
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 77 -
Q.No. Solution
Equation (iv) 3 – 2 – 1 = 0
Solving these we get 15 16
,13 13
15 16
,13 13
Q
Line containing the reflected ray is the line joining the points A(1, –1) and 15 16
,13 13
Q
Required equation is
161
131 ( 1)16
113
y x
29x – 2y – 31 = 0
44. Answer (3)
Let P be the middle point of the line segment joining A(3, –1) and B(1, 1)
Q
A(3, –1) P(2, 0) B(1, 1)
Then P = (2, 0)
Let P be shifted to Q where PQ = 2 and y–coordinate of Q is greater than that of P (from graph)
Now, Slope of AB = 1 1
11 3
Slope of PQ = 1
Coordinates in Q by distance formula
= (2 ± 2cos, 0 ± 2sin), where tan= 1
= ,(2 2 2)
As y-coordinate of Q is greater than that of P
(2 2, 2)Q , which is the required point.
45. Answer (1)
The given lines are concurrent if
2
1 1 –1
2 3 0
4 9
Solving we get
2 + 13 – 30 = 0
Which gives two values of whose sum is –13
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 78 -
Q.No. Solution
46. Answer (3)
Let BAO = , then
OA = c cos
OB = c sin
Let m (h, k) be foot of the perpendicular from P on AB
Let MN OX
O
B P
AN
m h k( , )
ON = h = OA – NA
= c cos – MA.cos
= c cos – PA.cos2
. cos
= c cos – c sinsin cos
= c cos (1 – sin2)
h = c cos3 …(i)
k = MN = MA sin
k = c.sin3 …(ii)
h2/3 + k2/3 = c2/3(sin2 + cos2) = c2/3
Replacing (h, k) by (x, y) we get
x2/3 + y2/3 = c2/3
47. Answer (2)
The given line L1 : ax + by + c = 0
,0 , 0,c cP Q
a b
Any line L2 is perpendicular to L1 is
bx – ay + = 0
,0 , 0,R Sb a
Equation of line PS is
/
/
a cy xc a a
cy xc a
…(i)
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 79 -
Q.No. Solution
Equation of line QR is
/
/
c by xb b
cy x
b
…(ii)
Locus of the point of intersection of (i) and (ii) is obtained by eliminating from (i) and (ii)
From (ii) c cy xb
… (iii)
Multiplying (i) and (iii) we get
c cy y x xb a
2 2 0c cx y x ya b
48. Answer (1)
Here circumcentre O = (0, 0)
A x x( , tan )1 11
C x( , tan )3 3x3B x( , tan )2 2x2
So, OA = OB = OC
2 2 2 2 2 21 1 1 2 2 2tan tanx x x x
= 2 2 23 3 3tanx x
2 2 2 2 2 21 1 2 2 3 3sec sec secx x x
31 2
1 2 3cos cos cos
xx xk
(suppose)
Vertices of the triangle become
A = (k.cos1, k.sin1)
B = (k.cos2, k.sin2)
C = (k.cos3, k.sin3)
Centroid 1 2 3 1 2 3(cos cos cos ) (sin sin sin ),
3 3G k k
We know that orthocenter H, centroid G and circumcentre O are collinear
So, Slope of HO = slope of GO
1 2 3
1 2 3
sin sin sin
cos cos cos
yx
1 2 3 1 2 3(cos cos cos ) (sin sin sin )y x
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 80 -
Q.No. Solution
49. Answer (3)
The given equation is
x2 + 6xy + 9y2 + 4x + 12y – 5 = 0 …(i)
Here abc + 2gfh – af2 – bg2 – ch2 = 0
And h2 – ab = 0
Equation (i) Represents the parallel straight lines
From (i) we know
9y2 + 6(x + 2)y + (x2 + 4x – 5) = 0
2 26( 2) 86( 2) 36( 4 5)
2 9
x x x xy
Or 3y + x = 1 and 3y + x + 5 = 0
There are two parallel lines and distance between these two lines is
2 2
5 ( 1) 6
103 1
50. Answer (3)
The given point is the centroid of the triangle.
51. Answer (2)
Let ABC = 2, and r the radius of the inscribed circle then AB and CD can be expressed in terms of r and . Area of quadrilateral that is trapeziun in our case, can be get in term of r and and then we can solve the equation for r.
52. Answer (4)
For collinear points
sin( ) cos 1
cos( ) sin 1
cos( ) sin( ) 1
Clearly 0 for any value of , , , hence points are non-collinear.
IInd method : (by observation)
P R Qsincos
(–sin( – ), –cos ) (cos( – ), sin )
In this case cos ·cos( ) sin sin( ) cos sin sin cos
,sin cos sin cos
R
cos( ) sin( )
, (cos( ), sin( ))sin cos sin cos )
R
, if sin + cos = 1
Which is not possible if 0 < < 4
Hence points are non-collinear.
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 81 -
Q.No. Solution
53. Answer (4)
We have
(1 + p)x – py + p(1 + p) = 0
1– 1
x yp p
Also
(1 + q)x – qy + q(1 + q) = 0
1– 1
x yq q
Equation of AM is
–
( )1
py x ap
and equation of BM is –
( )1
qy x pa
( ) ( )1 1
p qx a x pp q
x = pq …(i)
Also
–
( )1
py pq qp
y = –pq …(ii)
from (i) and (ii)
x = –y
x + y = 0
Which is the required locus representing a straight line.
54. Answer (2)
The question is too simple from the diagram, the given line
60°120°
0,
3
1
(0, 1)
3 1x y makes an angle 120° with x-axis and intersects at 1
, 03
. A line making an angle 60°
with the given line is either x-axis or different from x-axis. By observation it is clear that the straight line
3 2 3 3 0y x is the required line.
x
y
C
pcM
N
(– , 0)q(–
, 0)
p
A B
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 82 -
Q.No. Solution
Second Solution
The equation of the line through (–3, – 2) may be written as
y + 2 = m(x – 3)
which will make 60° with 3 1x y if
3
tan601 3
mm
3
31 3
mm
3 or 0m m
Since the line intersects x-axis also, hence m 0 consequently 3m and the required line is
2 3( 3)y x
3 2 3 3 0y x
55. Answer (1)
0ax by c ... (1)
0bx ay c ... (2)
Solving, cx
a b
Also from (1) & (2)
y = x
Point of intersection lies on y = x
cy
a b
Given, 2 2
1 1 2 2c c
a b a b
2 1 2 2c
a b
2a b c
a b
2 2a b c a b
0a b c
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 83 -
Section-B
Q.No. Solution
1. Answer (1, 2, 3, 4)
Area = 1
4 3 6 362
Hence all options are correct.
2. Answer (1, 2, 3, 4)
AB = 3
BC = 3 2
AC = 3
AB2 +AC2 = BC2
Hence A = 90°
Orthocentre = A = (4, 4)
Circumcentre = 11 11
,2 2
Centroid = 4 4 7 4 4 7
, (5,5)3 3
Incentre = 3 7 3 4 3 2 4 3 4 3 2 4 4 7
,3 3 3 2 3 3 3 2
= 11 4 2 11 4 2
,2 2 2 2
3. Answer (1, 2, 3)
Let the slope of the line is m.
tan 45° = 2 1
3,1 2 3
m m mm
Lines y – 3 = –3 (x – 2)
3 9y x
and y – 3 = 1
( 2)3
x
3y – 9 = x – 2
3 7y x
A B
3
3
(4, 7)
(7, 4)
3 2
(4, 4)
x
y
(–3, 0)
(0, 6)
(3, 0)
(0, –6)
x3
y6
– + = 1 x3
y6+ = 1
x3
y6– = 1x
3y6
– – = 1
m
(2, 3)
45°
2 – + 3 = 0x y
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 84 -
Q.No. Solution
4. Answer (2, 3, 4)
(A) Let the image is ( , ) then
3 4 2(3 3 4 5) 40 20
41 3 1 9 10 5
1, 8
(B) Distance = 3 4 3 5 20
2 101 9 10
(C) True
(D) Equation is 3x – y + k = 0
At x = 3, y = 4
9 – 4 + k = 0
5k
Hence equation is 3x – y – 5 = 0
5. Answer (1, 3, 4)
6. Answer (1, 2)
(a + a 2) (a + a + 2) < 0 a (1, 1)
7. Answer (1, 2, 3)
Vertices are rational parts
Centroid
3
,3
11 yx is rational.
Vertices are rational
Coefficient of equations of lines perpendicular to the sides are also rational
Orthocenter is intersection point of equations of altitudes.
Orthocenter is rational. Orthocenter, centroid and circumcentre are collinear and centroid divides the line segment in the ratio 2 : 1
Circumcentre are also rational
Incentre =
rqpryqypy
rqprxqxpx 321321 ,
Here p, q, r may be irrational
Hence incentre is not always rational.
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 85 -
Q.No. Solution
8. Answer (1, 3)
0
62
31–2
213
2
aa
02
1–22
6
321–
62
31–3 22
aaaa
a2 – 2a – 6 = 0
71 a
9. Answer (3, 4)
A line parallel to given line
3x – 4y + = 0 …(1)
3x – 4y – 2 = 0 …(2)
169
2––4
| + 2 | = 20 = 18, –22
Lines are 3x – 4y + 18 = 0 & 3x – 4y – 22 = 0
10. Answer (2, 4)
PA2 = BP2
[p – (a + b)]2 + [q – (b – a)]2 = [p – (a – b)] 2 + [q – (a + b)]2
aq = bp
P(p, q) can be (a, b)
11. Answer (1, 3)
Let line (1) makes angle 1, with positive x axis
21
–1
2tan
aa
aaa 1–
21–
1 tan2–1
2tan
Let line (ii) makes angle 2 with x-axis
22
–1
2tan
bb
bbb 1–
21–
2 tan2–1
2tan
Angle bisector between these two lines makes angle 2
21 with positive axis.
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 86 -
Q.No. Solution
ba 1–1–21 tantan
2
1
2
1 =
abba
–1tan 1–
abba
–1tan
Equation of bisector is
pxabbaqy –
–1–
(a + b) (x – p) – (1 – ab) (y – q) = 0
Two bisector are perpendicular
Second bisector is px
baabqy –
–1––
(1 – ab) (x – p) + (a + b) (y – q) = 0
12. Answer (1, 2, 3, 4)
Point of intersection of lines is
qp
pqqp
pq, which will satisfy all the four lines
13. Answer (2, 4)
B (a + b, b – a), C(a – b, a + b)
Let M be mid point of BC
M (a, b), Slope of BC = ba
ba
–2–
2
Slope of AM = ab
Equation of AM, y – b = axab
–
1–by
= 1–ax
y = xab
, 1ab
, (a, b) and
ab
,1 will satisfy this equation, but (a, b) is the mid-point of BC. Therefore only
1,ba
and , 1ab
can be the required vertex
14. Answer (2, 3)
Third point of equilateral be
2
–3,
2
–3 21212121 yyxxyyxx
3,32
230,
2
230
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 87 -
Q.No. Solution
15. Answer (2, 3)
x + 2y + 3 = 0 & x + 2y – 7 = 0 are parallel lines
Equation of a line parallel to 2x – y – 4 = 0 is 2x – y + = 0
In a square distance between two parallel lines are equal
14
4
41
73
+ 4 = | 10 |
= 6, –14
Lines are 2x – y + 6 = 0 & 2x – y – 14 = 0
16. Answer (2, 3)
Centre of circle be
2
3–,
2
1
Let line L1 be y = mx
Intercepts are equal
Lines are at equal distance from the centre
2
1–2
3–
2
1
1
2
3
2
1
2
m
m
|(m + 3)| = 122 2 m
Squaring both sides
7m2 – 6m – 1 = 0
m = 1, 7
1– Lines are y = x &
7
1–y x
x – y = 0 & x + 7y = 0
17. Answer (1, 3)
Slope of line m = tan30° =3
1
Equation of line y = Cx 3
1
Intersection parts on axis are 0,3– CA & CB ,0
Given AB = 10
103 22 CC C = ± 5 Lines are y = 53
1x
0353– yx
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 88 -
Q.No. Solution
18. Answer (1, 4)
Slope of AC, m = 3
1
1–7
3–5
Sides through A(1, 3) makes angle 4
with line AC.
Equation of sides are
y – 3 = 1–
4tan1
4tan
xm
m
y – 3 = 1–
3
11
13
1
x
y – 3 = 1–
13
31 x
Lines are 2x – y + 1 = 0 & x + 2y – 7 = 0
19. Answer (1, 4)
Case-I : When {0}a R
a2 is positive
x2 + y2 + a2 = 0
So no real locus
Case-II : When a = 0
x2 + y2 = 0
x = 0; y = 0
Which is a point (0, 0)
20. Answer (1, 2, 3, 4)
If at all the equation,
x3 + y3 – kx2y + axy2 = 0 represents three straight lines then they all must pass through origin. So irrespective of k we choose the area with always remain zero.
21. Answer (3, 4)
Let the co-ordinates of C be (h, k)
a
b
AO
B
C h k( , )
Now since AOB = 90° = ACB
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 89 -
Q.No. Solution
So, OACB is a cyclic quadrilateral
AOC = ABC
tan ABC = ab
tana k ay xb h b
The required locus
Similarly the other locus may be ay xb
22. Answer (2, 3)
D C(5, 1)
BA(1, 3)
Section-C
Q.No. Solution
Comprehension-I
1. Answer (2)
A(a, 1); B(1, b) ;C (0, 0)
(CA)2 = (CB)2 = (AB)2
a2 + 1 = b2 + 1 = (a – 1)2 + (b – 1)2
a = b
b2 + 1 = a2 + b2 – 2a – 2b + 2
0 = a2 – 2a – 2b + 1
a2 – 4a + 1 = 0
a = 2 – 3
Side CA = 34–812 a
Area of equilateral,
= 4
3 (Side)2
= 34–84
3
= 3 2 – 3
2 3 – 3
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 90 -
Q.No. Solution
2. Answer (1)
AC2 = BC2
a = b
5 AB2 = 2 AC2
5[(a – 1)2 + (b – 1)2] = 2(a2 + 1)
5.2 (a – 1)2 = 2 (a2 + 1)
2a2 – 5a + 2 = 0
a = 2 & 2
1
a < 1
a (0, 1) and b (0, 1)
a = 2
1 = b
ab = 4
1
3. Answer (1)
AB2 = AC2 + BC2
(a – 1)2 + (b – 1)2 = a2 + 1 + b2 + 1
–2a – 2b = 0
a + b = 0
Comprehension-II
1. Answer (3)
Slope of MP = – 1
Slope of AB = 1
Equation of AB, y = x
Slope of NP = 1
Slope of AC = – 1
Equation of AC, y = – x
AB AC
orthocenter is A(0, 0)
2. Answer (2)
Circumcentre is intersection point of x + y = 3 and x – y = 1 is P(2, 1)
3. Answer (4)
BC = Diameter of circumcircle of ABC
= 2 (radius of circle) = 2 AP = 2 14 = 52 = 20
A B
C
( , 1)a(1, )b
(0,0)
A
B C
M
NP
(0, 0)
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 91 -
Q.No. Solution
Comprehension-III
1. Answer (1)
Equation of OA 4x + 5y = 0 …(i)
Equation of OC 7x + 2y = 0 …(ii)
Equation of AC 11x + 7y = 9 …(iii)
Solve equation (i) & (iii) A
3
4–,
3
5
Solve equation (ii) & (iii) C
3
7,
3
2–
Mid point of AC,
2
1,
2
1
Point B (1, 1) Equation of OB, y = x
2. Answer (4)
Vertices are (0, 0), (1, 1),
3
4,–
3
5
3. Answer (1)
Area of parallelogram = 2 [Area of OAB] = 2.
111
13
4–
3
5100
2
1= 3
Comprehension-IV
1. Answer (1)
Let L ax + by + c = 0
Then,
RBAR
QACQ
PCBP
3 32 2 1 1
3 3 1 1 2 2
| || | |1
| | | | | |
ax by cax by c ax by cax by c ax by c ax by c
2. Answer (3)
3
321 xxx = 0 x1 + x2 + x3 = 0
Similarly y1 + y2 + y3 = 0, let the line is ax + by + c = 0
Then 3 31 1 2 2
2 2 2 2 2 21
ax by cax by c ax by ca b a b a b
a(x1 + x2 + x3) + b(y1 + y2 + y3) + 3c = 2 2a b
3c = 2 2a b
9c2 = a2 + b2 2 2
2 29
a bc c
C B
AO
M
(0, 0)
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 92 -
Q.No. Solution
3. Answer (4)
From Q. (1)
RBAR
QACQ
PCBP
= −1
13
1
1
2
RBAR
2
3
RBAR
R divides AB externally in the ratio 3 : 2
Comprehension-V
1.
2.
3.
Answer (2)
Answer (1)
Answer (2)
Solution of questions no. 1 to 3 As , are roots of x2 – 6p1,x + 2 = 0
+ = 6p1, = 2 …(i)
Also ,, are roots of x2 – 6p2x + 3 = 0
+ = 6p2, = 3 …(ii)
And , are roots of x2 – 6p3x + 6 = 0
+ = 6p3, = 6 …(iii)
() () () = 2.3.6
= 6 …(iv)
From (i), (ii), (iii), (iv) we get = 3, = 2, = 1
And 1 2 3
1 2 5, ,
2 3 6p p p
Now, centroid of ABC, is
1 1 1 1 6 1 1 1
, , 13 3 3 3 2 3
11
2,18
Comprehension-VI
1.
2.
3.
Answer (1)
Answer (2)
Answer (1)
Solution of Question no.-1 to 3 Let ‘O’ be origin and let a1x + b1y = 1 and a2x + b2y = 1 be two given straight lines equation of straight line passing through ‘O’
cos sin
x y
…(i)
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 93 -
Q.No. Solution
This line cuts a1x + b1y = 1 and a2x + b2y = 1 at L and M respectively. Let OL = r1 and OM = r2. Then
L (r1cos, r1sin) and M (r2cos, r2sin)
Let N(h, k) be a variable point in equation (i) such that
ON = r3 h = r3cos
k = r3sin
Since L, M lie on a1x + b1y = 1 and a2x + b2y = 1
r1(a1cos + b1sin) = 1 and r2(a2cos + b2sin) = 1
1 2
3 1 1 3 2 2
1 1and
r rr a h b k r a h b k
…(ii)
Now,
When ON is AM of OL and OM
1 2 1 23
3 3
22
r r r rr
r r
1 1 2 2
1 12
a h b k a h b k
(a1 + a2)h + (b1 + b2)k – (a1 + a2)h – (b1 + b2)k + 2(a1b2 + a2b1)hk = 0 is required locus
When ON is geometric mean of OL and OM
2 1 23 1 2
3 3
1r r
r r rr r
or 3 31 1 2 2
1 2
1 ( ) ( ) 1r r
a h b k a h b kr r
a1a2x2 + b1b2y2 + {a1b2 + a2b1}xy = 1 is required locus
When ON is harmonic mean of OL and OM;
3 1 2
2 1 1
r r r
3 3
1 2
2r rr r
(a1h + b1k) + (a2h + b2k) = 2
(a1 + a2)h + (b1 + b2)k = 2 is required locus
Section-D
Q.No. Solution
1. Answer (1)
Statement-1 Product of slopes = –1
Statement-2 m1, m2 R for perpendicular lines m1m2 = –1
2. Answer (2)
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 94 -
Q.No. Solution
3. Answer (1)
a (x + y – 1) + b(x – 2y) = 0
Intersection point of x + y – 1 = 0 and x – 2y = 0 is
3
2,
3
1
4. Answer (3)
P(x, x – 12) lies on line (1). Distance of equidistant lines from P.
PM =
5
60–7
5
12–12–43 xxx
PN =
5
24–12–34 xx =
5
60–7x
PM = PN
5. Answer (4)
m is variable
Radius of circumcircle are also variable
6. Answer (1)
p, x1, x2 …… and q, y1, y2, y3 …… are in AP with common difference ‘a’ and ‘b’ respectively
andi ix p ai y q ib
1 2 nx x xh
n
and 1 2 ny y yk
n
1 1
andn n
i ii i
nh x nk y
1 1
( ) and ( )n n
i inh p ia nk q ib
( 1) ( 1)
and2 2
n n n n bnh np a nk nq
1
2
h p n k qa b
h p k q
a b
Hence locus of (h, k) is
b{x – p} = a{k – q}
Hence, statement-2 is true and
For statement-1, n = 3
Statement-1 is true and statement-2 is correct explanation for statement-1
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 95 -
Q.No. Solution
7. Answer (3)
As ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents the general equation of second degree.
But it represents a pair of straight line, if
abc + 2fgh – af2 – bg2 – ch2 = 0
Also, (2x – y – 5) (x + 2y – 3) = 2x2 + 3xy – 2y2 – 11x – 7y + 15 = 0
Hence option (3) correct.
8. Answer (4)
The bisectors of angles between lines in new position are same as bisectors of angles between their old positions, is always true.
The equation of angle bisectors of angle between lines in new position is
2 2
2 22 01 ( 1)
x y xy px xy pyp
Hence option (3) correct
9. Answer (2)
10. Answer (1)
11. Answer (3)
Angle bisector of a triangle does not divide the triangle into two similar triangles and hence statement-2 is wrong.
Section-E
Q.No. Solution
1. Answer : A(q), B(p), C(s), D(r)
2. Answer : A(s), B(p), C(q), D(r)
Centroid
3
2,
3
2 ba
Circumcentre is middle point of AB =
2,
2
ba
Orthocenter = (a, b)
Let foot of altitude from C is D (h, k)
AD = AC. sin = b sin
DM = k = AD sin = b sin2 ( sin = 22 ba
b
, cos =
22 ba
a
=
22
3
22
2.
bab
babb
AM = a – h = AD cos
A – h = b sin cos = 2222
..
ba
a
ba
bb
a – h = 22
2.
baba
h = a .
22
2
–1ba
b =
22
3
baa
BC ( , )a b
( , )h kD
k
h a h–M
(0, )b
A
( , 0)a
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 96 -
Q.No. Solution
3. Answer : A(q), B(r), C(s), D(p)
Mid point of AC, M(3, 2) M lie on y = 2x + C
C = –4
Equation of BD
y = 2x – 4
BM = r = AM = 514
Slope of Line BD is tan = 2
sin = 5
2
cos = 5
1
Let B (h, k)
h = 3 ± 5 cos = 3 ± 1 = 4, 2
k = 2 ± 5 sin = 2 ± 2 = 4, 0
B (4, 4) & D (2, 0)
Mid point of AB =
2
7,
2
5
Middle point of BC =
2
5,
2
9
4. Answer : A(q), B(s), C(p), D(r)
5. Answer : A(p, q, r, s, t), B(p), C(q), D(r)
(A) Let the line is ax + by + c = 0 …(i)
and the points are (xi, yi) where i = 1, 2, 3,…..n. Algebraic length of perpendicular from (xi, yi)
2 2
i ii
ax by cp
a b
But 2 2
0 0i ii
a x b y ncp
a b
0i ia x b yc
n n
…(ii)
by (i) and (ii)
,i ix yx y
n n
, which are the coordinates of fixed point
According to the problem
2 1 32
3
3 3 64
3
x
y
Hence fixed point = (2, 4)
A B
C
D
(1, 3)
(5, 1)
rm
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 97 -
Q.No. Solution
(B) (3, 3) 12 (x, y) (1, 1)
(Orthocenter) (centroid) (circumcentre)
As we know that centroid divides the distance between orthocenter and circumcentre in the ratio 2 : 1.
Hence, 2 1 1 3 5
2 1 3x
2 1 1 3 5
2 1 3y
Hence coordinates are 5
,3 3
(C) Using the formula that coordinates of incentre are 1 2 3 1 2 3,ax bx cx ay by cy
a b c a b c
(0, 4)
(0, 0) (3, 0)3
45
According to the problem the incentre is given by
5 0 4 3 3 0 5 0 4 0 3 4 12 12, 1, 1
4 3 5 4 3 5 12 12
(D) Lines are x = 0, y = 0, x + y = 2. As we know that the lines x = 0 and y = 0 are perpendicular hence orthocenter is (0, 0).
6. Answer : A(q, t), B(p), C(r), D(s)
(A) As we know that image of point A through y = x and 2x + 3y + 13 = 0 will lie on BC.
A(2, 3)
B CB1 C1
2 +3 +13=0
xy
y x =
(x1,y1) (x2,y2)
I
Clearly B1 = (x1 , y1) = (3, 2)
C1 = (x2, y2) and the line mirror 2x + 3y + 13 = 0. To find the (x2 , y2) we use the formula for image
2 22 3 2 2 2 3 3 13
2 3 4 9
x y
2 22 3 2 264
2 3 13
x y
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 98 -
Q.No. Solution
x2 = –8 + 2, y2 = –12 + 3
x2 = –6, y2 = –9
C1 = (–6, –9)
Hence equation of BC is 9 22 3
6 3y x
112 3
9y x
9y – 18 = 11x – 33 11x – 9y – 15 = 0
(B) B is the point of intersection of y = x and 11x – 9y – 15 = 0
15 15
,2 2
B
(C) Point ‘C’ is intersection of
2x + 3y + 13 = 0 …(i)
and 11x – 9y – 15 = 0 …(ii)
24 173
,17 51
C
(D) Incentre is the point of intersection of any two internal angle bisectors
y = x ...(i)
2x + 3y + 13 = 0 …(ii)
I = incentre = 13 13
,5 5
7. Answer : A(s), B(s), C(q), D(q)
(A) Reflection about x-axis
Replace x x, y –y
ax2 + 2h(x) (–y) + b(–y)2 = 0
ax2 – 2hxy + by2 = 0
(B) Replace x –x, y y
ax2 – 2hxy + 3y2 = 0
(C) Replace x y and y x
bx2 + 2hxy + ay2 = 0
(D) Replace x –y and y –x
bx2 + 2hxy + ay2 = 0
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 99 -
Q.No. Solution
8. Answer A(s); B(p, q); C(r), D(p, q, s)
(A) Solving L1 and L3
1
36 10 12 25 2 15
x y
x = 2, y = 1
L1, L2, L3 are concurrent, if (2, 1) lies on L2
6 – k – 1 = 0 k = 5
(B) Either L1 is parallel to L2, or L3 is parallel to L2,
then 1 3 3
or3 5 2
kk
k = –9 or 6
5k
(C) L1, L2, L3 form a triangle, if they are not concurrent, or not parallel
6 5
5, 9,5 6
k k
(D) L1, L2, L3 do not form a triangle, if
6
5, 9,5
k
Section-F
Q.No. Solution
1. Answer (3)
Let us consider a triangle ABC where PQR are middle points of sides.
A
B R C
QP
Here PQ || BC and PQ = 1
2BC
Similarly QR || AB and QR = 1
2AB
And PR || AC and PR = 1
2AC
Hence quadrilaterals PQCR, PQRB, PRQA are parallelogram.
So if three points P,Q,R are given then fourth point may have three positions A,B,C to form a parallelogram
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 100 -
Q.No. Solution
2. Answer (1)
5x + 7y = 12 …(i)
y = kx + 2 …(ii)
By (i) and (ii)
5x + 7(kx + 2) = 12
2
5 7x
k
Now for x to be integer
5 + 7k = 1, –1, 2, –2
4 6 3
, , , 17 7 7
k
Hence only one value of k exists.
3. Answer (2)
The equation of the given lines
x + y = 1, x + y = 3
Distance between the lines 2
22
The line passes through (1, 1) intersects an intercepts length 2 unit which can be shown as following.
AC y + x = 3
y + x = 1required lineB
(1, 1) 2
As AB = 2
2 1cos 45
2 2 o
Let the slope of required line is m.
1
tan451
o mm
1
11
mm
, 1
11
mm
If 1
11
mm
m + 1 = m – 1 m = 0
If 1
11
mm
m + 1 = –1 + m, m
Hence the equation of the lines is y = 1 and x = 1
Hence two lines are there.
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 101 -
Q.No. Solution
4. Answer (5)
If origin is fixed then the perpendicular distance of (0, 0) from the line is always constant.
2 22 2
1 1
1 11 13 4
a b
2
2 2
1 1 1 1 25 5
9 16 144 12 a b
5
12 512
k k
5. Answer (0)
The distance of the line x + y = 6 from origin (0, 0) is
6
3 22
So, the minimum distance of the line x + y = 6 from (0, 0) is 3 2 , which is greater than the given distance, hence no such line is possible.
6. Answer (1)
P
xa
yb+ = 1
B
B
O A A
Let ‘O’ be the origin clearly OA = a; OB = b
Let the other line be AB
Let (say)OA OBOA OB
So, the equation to AB is
1x ya b
And that to AB is
1x ya b
Let (h, k) be their point of intersection
1h ka b
…(A)
1h ka b
…(B)
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 102 -
Q.No. Solution
Subtracting (A) and (B) we get
1 1
1 1 0h ka b
h ka b
So, the locus is
x ya b
Hence c = 1
7. Answer (1)
Making the equation
(x – h)2 + (y – k)2 = c2 homogeneous
with kx + hy = 2hk, we get the required condition that
12
kx hyhk
h2 + k2 = c2
Hence t = 1
8. Answer (3)
Q(3, 0)
P(1, 2)
Let the point Q be (h + 3, h)
Now since after reaching point Q it starts moving farther
So, PQ must be perpendicular to the line
Slope of PQ 2
2
hh
Slope of the line 1
h – 2 = – (h + 2)
2h = 0 h = 0
So, Q = (3, 0)
a + b is 3
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 103 -
Section-G
Q.No. Solution
1. Answer (2)
All the statements are clearly true.
2. Answer (2)
STATEMENT-1 is clearly true.
STATEMENT-2 is true because if lines intersect at right angle, then both angles are equal.
STATEMENT-3
Let A = (2, 3), B = (0, 0), the required line will be perpendicular to AB.
Slope of 3
2AB
Hence required line is
23 2
3y x
3y – 9 = –2x + 4
3y + 2x – 13 = 0
3. Answer (1)
A(2, –2)
B(–2, 1) C(5, 2)D
EF
10,
2
7, 0
2
3 3,
2 2
m1 = slope of 2 4
23 3
FD
m2 = slope of
332
2 41
DE
m1m2 = – 1. Hence D is 90° hence A is also 90°. ABC and DEF are both right angled isosceles triangle.
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 104 -
Q.No. Solution
4. Answer (2)
1. Statement-1
X3 + y3 + xy(x + y) = 0
(x + y) {x2 + y2 – xy – xy} = 0
(x + y) (x2 + y2 + ( – 1)xy) = 0
The angle bisector of lines x2 + y2 + ( – 1)xy = 0
2 2 2
( )( ) 01 1 1
x y xy x y x y
Hence x + y = 0 is angle bisector of other two lines
2. Statement-2
Let y = xtan be one of the lines represented by the equation
k(x3 – 3xy2) + y3 – 3x2y = 0
Then 3
2
3 tan tantan3 (say) tan3 tan3
1 3 tank
1
, ;3
n n I angle which the lines makes with x axis are 2
, ,3 3
Angle between lines 1st and 2nd is 3
, between 2nd and 3rd is
3
, so between 1st and 3rd is
2
3
which is 2
3
i.e.,
3
. Hence three lines equally inclined to one another.
3. Statement-3
Distance between two parallel lines is 2
2( )
g aca a b
Hence a = 8, b = 2, h = 4, g = 13
13 169 120 49 7
, 15 2 22 8 10 8 10 2 5
f c
5. Answer (3)
Statement-1
45°
45°
(0, 0) (1, 0)C B
A(0, 2)
D
x
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 105 -
Q.No. Solution
Area of ACD + area of BCD = Area of ABC
1 1 1
1 sin45 2sin45 1 22 2 2
x x
2 2
3 sin45 23
x x
Statement-2: Slope of the line = 3
5
Equation of the line is 3x + 5y = 43
43 3
5 43 35
xy x y
Hence the points are [(1, 8), (6, 5), (11, 2)]
Statement-3: Here
2 1 3
5 3 0
3 1 2
k
. Hence k = – 2
6. Answer (4)
7. Answer (3)
Statement-1
The x intercept of the line = 2
m
2 1 1 2 4
0 02 2 2
mm m m
( 4, 0)m
Statement-2
Let equation of line x + y = a …(i)
2 2 21 1| |
2 2z x y …(ii)
y = a – x putting in equation (ii) is
2 2 2 21 1( ) 2 2 0
2 2x a x x ax a
D = 0 a2 = 1 a = ± 1
Hence the equation of lines is |x + y| = 1
Statement-3
Let the equation of line is
y + 2 = m(x + 4) mx – y = 2 – 4m
or 2 4
1 | 4 2 |2 4 4 2
x y m mm m m
m
1
, 2 0, 4 2 0 or 12
m x y m m
2y = x, x + y + 6 = 0, m = + 1
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 106 -
Section-H
Q.No. Solution
1. Answer (4)
Prime number between 0 & 17; 2, 3, 5, 7, 11, 13
With 2 as x-co-ordinate, total points = 6
With 3 as x-co-ordinate, total points = 6
With 5 as x-co-ordinate, total points = 5
With 7 as x-co-ordinate, total points = 4
With 11 as x-co-ordinate, total points = 3
With 13 as x-co-ordinate, total points = 2
Total points = 26
2. Answer (2)
The slope of the line is
1000 1 111
100 1 11
So, all the points will have the form (1 + 11t, 1 + 111t)
1 1 11 100 1 1 111 1000
0 11 99 0 111 999
0 9 0 9
t tt t
t t
Hence there are 8 such values of t and hence there are 8 such points.
3. Let ( , )P and any line through P be
cos sin
x y r
Any point on this line is ( + rcos, + rsin)
Points where it meets the parabola can be obtained from;
( + rsin)2 = 4a( + rcos)
r2sin2 + 2r(sin – 2acos) + 2 – 4a = 0
2
2(2 cos sin )
sin
aPQ PR
and
2
2
4
sin
aPQ PR
Since PQ, PS and PR are in H.P.
PS = 2( 4 )
(2 cos sin )
aa
Let S (h, k)h = + PS.cos, k = + PS.sin.
2 – 4a = 2a(h –) – (k –)
Hence locus of ‘S’ is 2ax – y + 2a = 0
Which is clearly a straight line whose slope is 2a/, that does not depends upon the absciss of point P.
(0, 17) (17, 17)
(17, 0)(0, 0)
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 107 -
Q.No. Solution
4. Answer (1, 2, 3, 4)
A(, ) lies on y = 2x + 3
= 2 + 3
A (, 2 + 3)
Area of
2 3 11
1 2 12
2 3 1
= 1
[ (2 3) (2 3)(2 1) 1(3 4)]2
= 1
| 2 | S2
4 | + 2| < 6
4 + 2 < 6 or –6 < + 2 –4
2 < 4 or –8 < – 6
Integral values of are
2, 3, –6, –7
So, we get the co-ordinates as
(–7, –11), (–6, –9) (2, 7) and (3, 9)
5. Answer (1, 2)
L1 : 3x + 4y + 5 = 0
L2 : 3x + 4y + 15
B
L1L2
B1
C
A(4, 3)
A1
Distance between the two lines = 2 2
15 102
3 4
BC = 2 units, let BAB1 =
AB = 2 cosec, BB1 = 2 sec
1 1
1( )
2ar ABB AB BB
= 1
2 cosec 2sec2
= 4
sin2
Ay
x= 2 + 3
B(1, 2) C(2, 3)
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 108 -
Q.No. Solution
Now area of parallelogram AA1 BB1
= 2 ABB1= 8
sin2
Clearly this is least when sin2 = 1
4
Let the slope of the line drawn be ‘m’
3141 7,
3 714
mm
m
Hence the equation of lines
1
( 3) ( 4) or ( 3) 7( 4)7
y x y x
6. Answer (1, 2)
Equation of lines along OA, OB and AB are y = 0, x = 0, and 3
2x y respectively. Now P and B will
lie on the same side of y = 0 if cos > 0. Similarly P and A will lie on the same side of x = 0 if sin > 0
and P and Q will lie on the same side of 3
2x y if
3
sin cos2
Now3
sin cos2
3
sin4 2
04 3
2
3 4
Since sin > 0 and cos > 0
012
or
5
12 2
7. Combined equation of line is
x2 – 2xy – 3y2 + 8y – 4 = 0
(x – y)2 = 4y2 – 8y + 4
x – y = ± 2 (y – 1)
Thus two sides of a triangle are
L1 : 3y – x – 2 = 0 and L2 : y + x – 2 = 0
And these intersects at A (1, 1)
Solutions of Assignment (Set-2) Straight Lines (Solutions)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 109 -
Q.No. Solution
Let the third side be (y + 1) = m(x + 5)
L3 : y = mx + 5 m – 1
Let L3 meets the line L1 and L2 at B and C respectively
15 5 3 5 7 1
, 1 ; ,1 3 1 1
m m mB Cm m m
Now as origin has to be an interior point so
(1 – 5 m) (1 – m + 1 – 5 m ) > 0
1 1
or3 5
m m
Similarly, points ‘O’ and ‘C’ should be on the same side of line AB.
3(7 1) 3 5
2 2 01 1
m mm m
1
13
m
Finally points ‘O’ and B should lie on the same side of AC
15 5
2 ( 1) 2 01 3
mm
2
10 and
3m m R
1
1,5
m
8. An (n, 3n)
(O An)2 = n2 + 9n2 = 10n2
12
1
2
nnOA = 10.
6
12124112.1012
1
2
n
n = 6500
9. OA2 = 2OA1 = 2 – 1 = 2 !
OA3 = 3 . OA2 = 3 . 2! = 3 !, OA4 = 4!
Hence OA8 = 8 !
!81818 22 aa
6a = 8.7 . 6. 5 . 4 . 3 . 2 .1
a = ± 6720
10. 2nOA = n2 + n3
12
1
2
nnOA =
4
1
6
121 22
nnnnn
Put n = 12 = 6734
Straight Lines (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 - 110 -
Q.No. Solution
11. m1 = – 1, m2 =
2
1– , m3 = m
Line L is bisector of angle of other two.
31
31
21
21
1
–
1
–
mmmm
mmmm
3
1 =
1–
1
mm
m = –2
812 (m2) + 3 = 812 (4) + 3 = 3251