straight lines (package solutions)

Solutions of Assignment (Set-2) Straight Lines (Solutions)

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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)


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



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



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



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



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



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)



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)



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



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



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



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)



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



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.



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



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



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



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.



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.



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



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



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



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



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)



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)



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)



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)



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



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



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



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



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



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



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.



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)



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



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.



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



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



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






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)



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)



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)



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



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

straight lines (package solutions)

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