q.no. 1 the bisector of the acute angle between the lines...

Q.No. 1 The bisector of the acute angle between the lines 3x - 4y + 7 = 0 and 12x + 5y - 2 = 0, is:

Option 1 11x + 3y - 9 = 0

Option 2 21x + 77y - 101 = 0

Option 3 11x - 3y + 9 = 0

Option 4 None of these

Correct Answer 3

Explanation L1 : 3x - 4y + 7 = 0

L2 : -12x - 5y + 2 = 0

Check: 3 (-12)+(-4) (-5) 0 <

Eqn. to acute angle bisector

3 - 4 +7 -12 - 5 +2=

5 13

x y x y

11x - 3y + 9 = 0

Q.No. 2 The equation of the bisector of the angle between two lines 3x - 4y + 12 = 0 and

12x - 5y + 7 = 0 which contains the points (-1, 4) is:

Option 1 21x + 27y - 121 = 0

Option 2 21x - 27y + 121 = 0

Option 3 21x + 27y + 191 = 0

Option 4 -3 + 4 -12 12 - 5 +7=

5 13

x y x y

Correct Answer 1

Explanation Eqn. to the bisector containing (0, 0)

3 - 4 +12 12 - 5 +7=

5 13

x y x y

Check: 3 (-1)- 4 1+12) 0 <

And 12 (-1)- 5 4 +7) 0 <

(-1, 4) lies is the opposite region as shown below:

So, eqn. to the bisector containing (-1, 4) is

3 - 4 +12 12 - 5 +7=

5 13

x y x y

21x + 27y - 121 = 0

Q.No. 3 The line x + 3y - 2 = 0 bisects the angle between a pair of straight lines of which one

has equation x - 7y + 5 = 0. The equation of the other line is :

Option 1 3x + 3y - 1 = 0

Option 2 x - 3y + 2 = 0

Option 3 5x + 5y - 3 = 0

Option 4 None

Correct Answer 3

Explanation Let the line is

x + 3y - 2 + (x - 7y + 5) = 0 Therefore,

1 1 + 1 1- + - -

3 3- 7 3 7=1 -1 - 1 1

1 + - 1 + -3 3- 7 3 7

1=

4

Eqn. to line is

5x + 5y - 3 = 0

Q.No. 4 Find the angle between there lines represented by the equation x2 - 2pxy + y

2 = 0

Option 1 sec-1

(p)

Option 2 tan-1

(p)

Option 3 cos-1

(p)


Correct Answer 1

Explanation x2 - 2p xy + y

2 = 0

let the angle be tan

-1tan = 2

2

p

2

tan = - 1 sec =p p 2 2 2 2

sec = = sec ( )p p -1 is one solution.

Q.No. 5 The distance between the lines represented by the equation,

+ 2 2 + 2 + 4 + 4 2 +1 = 0 is :x xy y x y2 2

Option 1 4

3

Option 2 4

Option 3 2

Option 4 2 3

Correct Answer 3

Explanation + 2 2 + 2 + 4 + 4 2 +1 = 0x xy y x y2 2

Let the lines be

+ 2 + A = 0x y 1

+ 2 + A = 0x y 2

A1A2 = 1 and A1 + A2 = 4

Distance between lines

(((( ))))

(A + A ) - 4 A AA = A= =

3L + 2

21 2 1 21 2

22

12= = 2

3

Q.No. 6 If sum and product of the slopes of lines represented by 4x2 + 2hxy - 7y

2 = 0 is equal to

then h is equal to:

Option 1 -6

Option 2 -2

Option 3 -4

Option 4 4

Correct Answer 2

Explanation 7y2 - 2hxy - 4x

2 = 0

2 -4= = -2

7 7

hh

Q.No. 7 If is the angle between the lines represented by - 3 + + 3 - 5 + 2 = 0x xy y x y2 2 , where

is a real number, then cosec2 equals :

Option 1 9

Option 2 3

Option 3 10

Option 4 100

Correct Answer 3

Explanation - 3 + + 3 - 5 + 2 = 0x xy y x y2 2

-3 3 -5= 1, = , = , = , = , = 2

2 2 2a h b g f c

For pair of straight lines : = 0

45 25 9 92 + - - . - 2. = 0

4 4 4 4

= 2

92 - 2

14tan = =1+ 2 3

cot = 3 cosec = 10 2 2 2

Q.No. 8 The difference of the slopes of the lines, (sec - sin ) - 2 tan + sin = 0x xy y 2 2 2 2 2 is

Option 1 -2

Option 2 1

2

Option 3 2

Option 4 1

Correct Answer 3

Explanation sin - 2 tan + (sec - sin ) = 0y xy x 2 2 2 2 2

2tan sec - sin[ - ]= - 4.

sin sinm m

2 2 2

1 2 2 2

2

= tan -(tan - sin )sin

2 2 4

2

sin

= 2 = 2sin

2

2

Q.No. 9 If pairs of straight lines, x2 - 2p xy - y

2 = 0 and x

2 - 2q xy - y

2 = 0 be such that each pair

bisects the angles between the other pair then :

Option 1 1= -

2pq

Option 2 pq = -2

Option 3 pq = -1

Option 4 = -1

p

q

Correct Answer 3

Explanation Eqn. to bisectors of x2 - 2pxy - y

2 = 0

-=

2 -

x y xy

p

2 2

px2 + 2xy - py2 = 0 ____(1) is Equivalent to

x2 - 2qxy - y

2 = 0 ____(2)

2=

1 -2

p

q

= -1pq _____(3)

Q.No. 10 Find the condition that the pair of straight lines joining the origin to the intersections

of the line y = mx + c and the circle x2 + y

2 = a

2 may be at right angles.

Option 1 2a2 = c

2(1 + m

2)

Option 2 2a2 = c

2(1 - m

2)

Option 3 2c2 = a

2(1 + m

2)

Option 4 2c2 = a

2(1 - m

2)

Correct Answer 3

Explanation Homogenising + = with =

-+ = 0+ -x y a y m

y mxx y a

cx c

2 2 22

2

1 2

c2x2 + c2y - a2y - a2m2x2 + 2 amxy = 0 Sum of coeff. of x

2 and y

2 = 0

+ - - = 0c c a a m 2 2 2 2 2

(1 + ) = 2a m c 2 2 2

Q.No. 11 The vertices of OBC are respectively (0, 0), (-3, -1) and (-1, -3). The equation of line

1parallel to BC and at a distance form O which intersects OB and OC is :

2

Option 1 2 + 2 + 2 = 0x y

Option 2 2 - 2 + 2 = 0x y

Option 3 2 + 2 - 2 = 0x y


Correct Answer 1

Explanation -1(-3H)Slope of OD = =1

(-1+3)

Hence, Eq. of line

1+ = -

22 2

x y

2 + 2 + 2 = 0x y

Q.No. 12 If the sum of the distance of a point from two perpendicular lines in a plane is 1 then it

locus is :

Option 1 Square

Option 2 Circle

Option 3 A straight line

Option 4 Two intersecting line

Correct Answer 1

Explanation Let the two lines be

ax + by + c1 = 0 - L1

bx - ay + c2 = 0 - L2

Let the point be h, k

+ + + ++ =1

+ +

ah bk c bh ak c

a b a b

1 2

2 2 2 2

If the axes are along the line L1 and L2 then

+ = 1x y

Clearly, the locus is square.

Q.No. 13 Given the points A(0,4) and B(0, -4), the equation of the locus of the point P(x, y) such

that AP-BP = 6 is :

Option 1 9x2 - 7y

2 + 63 = 0

Option 2 9x2 - 7y

2 - 63 = 0

Option 3 7x2 - 9y

2 + 63 = 0

Option 4 7x2 - 9y

2 - 63 = 0

Correct Answer 1

Explanation +( + 4) - +( + 4) = 6x y x y2 2 2 2

+( - 4) + +( + 4) - 2 +( - 4) +( + 4) = 36x y x y x y x y 2 2 2 2 2 2 2 2

( + -2) = +( - 4) +( + 4)x y x y x y 2 2 2 2 2 2 2

- 4 - 4 + 4 = 32 - 32 + 256x y x y 2 2 2 2

36 - 28 + 252 = 0x y 2 2

9 - 7 + 63 = 0x y 2 2

Q.No. 14 A triangle ABC with vertices A(-1, 0), B(-2, 3/4) and C(-3, -7/6) has its orthocentre H.

Then the orthocentre of triangle BCH will be :

Option 1 (-3, -2)

Option 2 (1, 3)

Option 3 (-1, 2)


Correct Answer 4

Explanation

A will be orthocentre of HBC

Q.No. 15 A variable straight line passes through a fixed point (a, b) intersecting the co-ordinates

axes at A and B. If O is the origin then the locus of the centroid of the triangle OAB is :

Option 1 bx + ay - 3xy = 0

Option 2 bx + ay - 2xy = 0

Option 3 ax + by - 3xy = 0

Option 4 ax + by - 2xy = 0

Correct Answer 1

Explanation Let (y - b) = m(x - a) be the variable line through (a, b).

Let (h, K) be the centroid.

3 = - ;b

h am

3K = b - am;

ab = (a - 3h) (b - 3K)

bx + ay = 3xy - required locus.

Q.No. 16 Two vertices of a triangle are (4, -3) and (-2, 5). If the othocentre of the triangle is at

(1, 2), then the coordinates of the third vertex are

Option 1 (33, 130)

Option 2 (33, 26)

Option 3 (-33, 26)

Option 4 (33, -26)

Correct Answer 2

Explanation BE AC

- 5 2- (-3)

= (-1)+2 1 - 4

y

x

3 - 5 + 3 = 0 ( )x y

Also CF AB gives

x - y - 7 = 0 - (2)

from (1) & (2)

(x, y) (33, 26)

Q.No. 17 Drawn from the origin are two mutually perpendicular straight lines forming an

isosceles triangle together with the straight line, 2x + y = a. Then the area of the

triangle is :

Option 1

2

a2

Option 2

3

a2

Option 3

5

a2

Option 4 none

Correct Answer 3

Explanation

Let the slope of the lines be-1

&mm

.

Then,

12 +

- 2=

2L+ 21 -

m m

m

m

1= or- 3

3m

Let ' ' be the equal angle

1- 2

3= tan = 452

1 +3

-1 0

Length of perpendicular from (0,0) =5

a

1Area of = 2 =

2 55 5

a a a

2

Q.No. 18 The point (a2, a + 1) is a point in the angle between the lines 3x - y + 1 = 0 and

x + 2y - 5 = 0 containing the origin if :

Option 1 1 or -3a a

Option 2 1(-3,0) ,1

3a

Option 3 (0,1)a


Correct Answer 2

Explanation (3 - ( +1)+1) 1 0 and ( + 2( +1) - 5) (-5) 0a a a a 2 2> >

(3- ) 0 & ( +3)( -1) 0a a a a > <

1(-3,0) , 1

3a

Q.No. 19 Area of the quadrilateral formed by the lines + = 2x y is :

Option 1 8

Option 2 6

Option 3 4

Option 4 none

Correct Answer 1

Explanation + = 2x y

Locus of (x, y) is square

Area of square = (((( ))))2 2 = 82

Q.No. 20 The orthocentre of the triangle ABC is B and the circumcentre is S (a, b). If A is the

origin then the co-ordinates of C are :

Option 1 (2a, 2b)

Option 2 ,

2 2

a b

Option 3 (((( ))))+ ,0a b2 2 Option 4 None

Correct Answer 1

Explanation

One of the vertices is orthocentre. So, it is right angle triangle at which is midpoint of

AC.

C (2 ,2 )a b

Q.No. 21 The equation of the line segment AB is y = x. If A and B lie on the same side of the line

mirror 2x - y = 1, the image of AB has the equation

Option 1 x + y = 2

Option 2 8x + y = 9

Option 3 7x - y = 6


Correct Answer 3

Explanation

Clearly, mirror could be bisector of angle between AB & AB. Let m be the slope of

AB.

- 2 1 - 2=

L+ 2 1 + 2

m

m

= 7 or =1m m

Image line passes through point of intersection of AB & mirror.

( - )+ (2 - -1) = 0 - A'B'y x x y

-(2 -1) 6slope = 7 = =

L- 5

Hence image line:

7x - y = 6

Q.No. 22 + 3

4A = 2 is a point on either of two lines at a distance of units from their

3y x

Point of intersection. The co-ordinates of the foot of perpendicular from A on the

bisector of the angle between them are

Option 1 2- ,2

3

Option 2 (0, 0)

Option 3 2,2

3

Option 4 (0, 4)

Correct Answer 2

Explanation

Two lines are + 3 = 2y x

- 3 = 2y x

Intersect at (0, 2)

Equations to bisectors

+ 3 -2 = - 3 - 2y x y x

x = 0 or y = 2

4In PBA, B = 90 & PA =

3 0

4 1 2PB = PA cos60 = =

23 3 0

2B ,2

3

There would be other point of same property on y = 2

-2B' ,2 by symmetry.

3

For bisector x = 0

In POA, O = 90 0

4 3PO = PAcos 30 = = 2

230

Hence, such O points will have coordinates (0, 0) & (0, 4).

Hence all options are correct.

Q.No. 23 On the portion of the straight line, x + 2y = 4 intercepted between the axes, a square is

constructed on the side of the line away from the origin. Then the point of intersection

of its diagonals has co-ordinates :

Option 1 (2, 3)

Option 2 (3, 2)

Option 3 (3, 3)

Option 4 none

Correct Answer 3

Explanation

AB = 2 5

AD = 10

1+

2Let be slope of AD = 1

1 -2

m

mm

1= ,-3

3m

1Slope of AD = = tan

3

3 1To be coordinates of D 0 + 10. , 2 + 10 (3,3)

10 10

Q.No. 24 The equations of three lines, AB, CD and EF are, (b - c)x + (c - a)y + (a - b) = 0, (c - a) x

+ (a - b)y + (b - c) = 0 and (a - b)x + (b - c)y + (c - a) = 0. Which one of the following

inferences is correct.

Option 1 The lines are parallel to each other

Option 2 AB and BC are perpendicular to EF

Option 3 All the lines are coincident

Option 4 The lines are coincident

Correct Answer 4

Explanation AB : (b - c)x + (c - a)y + (a - b) = 0

CD : (c - a)x + (a - b)y + (b - c) = 0

AB : (a - b)x + (b - c)y + (c - a) = 0

Since: AB + CD + EF = 0

Lines are coincident

Q.No. 25 The base BC of a ABC is bisected at the point (p, q) and the equation to the side AB and AC are px + qy = 1 and qx + py = 1. The equation of the median through A is :

Option 1 (p - 2q)x + (q - 2p)y + 1 = 0

Option 2 (p + q) (x + y) - 2 = 0

Option 3 (2pq - 1) (px + qy - 1) = (p2 + q

2 - 1)(qx + py -1)

Option 4 none

Correct Answer 3

Explanation Equation of median through A

px + qy - 1 + (qx + py - 1) = 0 passes through (p, q)

p2 + q

2 - 1 + (pq + pq - 1 = 0)

+ -1=

1 - 2

p q

pq

2 2

median :

+ -1+ -1 = ( + -1) = 0

1 - 2

p qpx qy qx py

pq

2 2

(2pq - 1)(qx + py - 1) = (p2 + q

2 - 1)(qx + py -1)

Q.No. 26 Given the family of lines, a (3x + 4y + 6) + b(x + y + 2) = 0. The line of the family situated

at the greatest distance from the point P(2, 3) has equation :

Option 1 4x + 3y + 8 = 0

Option 2 5x + 3y + 10 = 0

Option 3 15x + 8y + 30 = 0

Option 4 None

Correct Answer 1

Explanation a(3x + 4y + 6) + b(x + y + 2) = 0 always through intersection of (3x + 4y + 6) = 0 and

x + y + 2 = 0 i.e. (-2, 0). Most distant line would be perpendicular to PR.

Eqn. to line

-1=

3- 0+2

2+2

y

x

3y + 4y + 8 = 0

Q.No. 27 Let P = (1, 1) and Q = (3, 2). The point R on the x-axis such that PR + RQ is the minimum

is

Option 1 5,0

3

Option 2 1,0

3

Option 3 (3, 0)


Correct Answer 1

Explanation R can be found out by join p to image of a in x-axis p(1, 1)

PR + RQ PR + PQ

PR + RQ PR + PQ

R minimises the distances eqn. to PQ

+ 2 - 3=

-3 2

y x

5Coordinates of R ,0

3

Q.No. 28 The co-ordinates of a point P on the line 2x - y + 5 = 0 such that PA-PB is maximum

where A is (4, -2) and B is (2, -4) will be :

Option 1 (11, 27)

Option 2 (-11, -17)

Option 3 (-11, 17)

Option 4 (0, 5)

Correct Answer 2

Explanation

AP'-BP' AB

Equality holds P' P Eqn. to AB

x - y = 6 intersects 2x - y + 5 = 0 at (-11, 27)

Q.No. 29 A right angle triangle ABC having C a right angle has AC = a and BC = b units. The

points A and B slide along the cartesian axes (A on x-axis and B on y-axis). Then the

locus of C is :

Option 1 = 0by ax

Option 2 = 0ay bx

Option 3 x2 + y

2 = 0

Option 4 xy = ab

Correct Answer 1

Explanation

We have a

2 + b

2 = a

2 + b

2

In BB'C

- = '- and - = '-b h b k a k a h2 2 2 2

(((( )))) (((( ))))' + ' = - + + + +a b b h k a k h2 2

2 2 2 2 2 2

(((( )))) (((( ))))+ = - + + - +a b b h k a k h2 2

2 2 2 2 2 2

by ax = 0 [after replacing h = x, k = y]

Q.No. 30 A line is drawn through the point A(p, q) in the direction to meet the line,

ax + by + c = 0 in B, then AB =

Option 1 + +

cos + sin

ap bq c

a b

Option 2 + +

cos + sin

aq bp c

a b

Option 3 + +

sin + cos

aq bq c

a b

Option 4 + +

sin + cos

aq bp c

a b

Correct Answer 1

Explanation

Coordinates of B (p + r cos , q + r sin ), B satisfies ax + by + c = 0 (p + r cos )a + b(q + r sin ) + c = 0

+ +=

cos + sin

ap bq cr

a b

Considering the opposite direction

+ +=

cos + sin

ap bq cr

a b

Q.No. 31 The set of values of b for which the origin and the point (1, 1) lie on the same side of

the straight line + + 1 = 0 R, 0 > > > >2 aaby ba x are :

Option 1 (2,4)b

Option 2 (0,2)b

Option 3 [0,2]b

Option 4 [2, ] b

Correct Answer 2

Explanation (a2.0 + ab.0 + 1) (a

2 + ab + 1) > 0

a2 + ab + 1 > 0

For above to hold

b2 - 4 < 0

b (-2, 2), b > 0

Q.No. 32 The acute angle between two straight lines passing through the point M(-6, -8) and the

points in which the line segment 2x + y + 10 = 0 enclosed between the co-ordinate

axes is divided in the ratio 1 : 2 : 2 in the direction from the point of its intersection

with the x-axis to the point of intersection with the y-axis is :

Option 1

3

Option 2

4

Option 3

6

Option 4

12

Correct Answer 2

Explanation AP : PQ : QB = 1 : 2 : 2

m (-6, -8)

-20 -10P ,

5 5

P (-4,- 2)

Q (-2,-6)

-2 + 8slope pf PM = = 3

-4 + 6

-6 + 8 1slope pf QM = =

-2 + 6 2

13-

2Angle between PM and QM = tan3

1 +2

= 45 =4

0

Q.No. 33 The equation of the sides of a square whose each side is of length 4 units and centre is

(1, 1). Given that one pair of sides is parallel to 3x - 4y = 0

Option 1 3x - 4y + 11 = 0, 3x - 4y - 9 = 0, 4x + 3y + 3 = 0, 4x + 3y - 17 = 0

Option 2 3x - 4y - 15 = 0, 3x - 4y + 5 = 0, 4x + 3y + 3 = 0, 4x + 3y - 17 = 0

Option 3 3x - 4y + 11 = 0, 3x - 4y - 9 = 0, 4x + 3y + 2 = 0, 4x + 3y - 18 = 0

Option 4 none

Correct Answer 1

Explanation

Line through (1, 1) parallel

3x - 4y is 3x - 4y + 1 = 0

Let 3x - 4y + d = 0 be eqn. to AD or BC.

-1= 2 -1 = 10 = 11 or - 9

5

dd d

Eqn. to AD or BC are 3x - 4y - 9 = 0 and 3x - 4y + 11 = 0

Line through (1, 1) 3x - 4y = 0 and 4x + 3y - 7 = 0 Eqn. to other sides are 4x + 3y + 3 = 0 or 4x + 3y - 17 = 0

Q.No. 34 If , , , , , 1 2 3 1 2 3 are the values of n for which

-12

=0

x

n

r

r

is divisible by -1

=0

x

n

r

r

, then

The triangle having vertices ( , ), ( , ) and ( , ) 1 1 2 2 3 3 cannot be.

Option 1 An isosceles triangle

Option 2 A right angled isosceles triangle

Option 3 A right angled triangle

Option 4 An equilateral triangle

Correct Answer 4

Explanation is divisible by

-1 -1

2

=0 =0

x x

n n

r r

r r

1 - ( )=

1 -

-1 22

2

=0

xx

x

n n

r

r

1 -=

1 --1

2

=0

xx

x

n n

r

r

1 - 1 - 1 +=

1 +1 - 1 -

2

2

x x x

xx x

n n

n

For perfect division, n = odd positive intege

Since vertices of triangle formed is all rational, cant be equilateral.

Q.No. 35 Consider the straight line ax + by = c, where a, b, c R+ this line meets the coordinates

axes at A and B respectively. If the area of the OAB , O being origin, does not depend upon a, b and c, then

Option 1 a, b, c are in AP

Option 2 a, b, c are in GP

Option 3 a, b, c are in HP


Correct Answer 2

Explanation

1

Area of O AB = OA OB2

1

=2

c c

b a

1

=2

2

c

ab

Area of OAB is independent of , G.P in one way to do it.2

c

ab

Q.No. 36 ABC is an equilateral triangle such that the vertices B and C lie on two parallel lines at a

distance 6. If A lies between the parallel lines at a distance 4 from one of them, then

the length of a side of the equilateral triangle is

Option 1 8

Option 2 88

3

Option 3 4 7

3


Correct Answer 3

Explanation

Let the length of side be d

EB = - 42 2

d

CF = - 22 2

d

EB- CF = BD

BD2 = BC

2 - DC

2

(((( ))))- 4 - -2 = -62

2 2 2 2 2 2d d d

- 4 + - 2 -2 - 4 - - 2 = - 6 2 2 2 2 2 2 2 2 2 2d d d d d

+ 6 - 4 - 2 = 2 - 4 - - 2 2 2 2 2 2 2 2 2d d d

(d2 + 16)

2 = 4(d

4 - 20d

2 + 64)

4 73 -112 = 0 =

3 4 2d d d

Q.No. 37 All the points lying on or inside the triangle formed by the points (1, 3), (5, 6) and

(-1, 2) satisfy

Option 1 3 +2 0x y

Option 2 2 + +1 0x y

Option 3 2 +3 -12 0x y

Option 4 -2 +11 0x

Explanation For the in equation to be satisfied by all points inside or on the triangle, corresponding

line wont intersect with the triangle formed.

Position of all vertices must be on a single side of the lines.

L1 = 3x + 2y

L (1,3), L (5,6),L (-1,2) 0 i.e 3 +2 0 is satisfied by all points.x y 1 1 1

L3(1,3) = 2.1 + 3.3 - 12 < 0 not satisfied.

Other lines are satisfied.

Q.No. 38 Given two straight lines x - y - 7 = 0 and x - y + 3 = 0. Equation of a line which divides

the distance between them in the ratio 3 : 2 can be and parallel to them.

Option 1 x - y - 1 = 0

Option 2 x - y - 3 = 0

Option 3 y = x

Option 4 x - y + 1 = 0

Explanation

Distance between L1 and L2

10= = 3 +2

2k k

= 2, 3 = 3 2k k

Let x - y + d = 0 be L

+ 7= 3 2 = -1 or -13

2

dd

Eqn. to L : x - y - 1 = 0

Q.No. 39 A light beam emanating from the point A(3, 10) reflects from the straight line 2x + y - 6

= 0 and then passes through the point B(4, 3). The equation of the reflected beam is :

Option 1 3x - y + 1 = 0

Option 2 x + 3y - 13 = 0

Option 3 3x + y - 15 = 0

Option 4 x - 3y + 5 = 0

Explanation

Let C (x, 6 - 2x)

1slope of CD =

2

(((( )))) (((( ))))tan ACD = tan DCB

1 6 -2 -10 6 -2 - 3 1- -

2 - 3 - 4 2=1 6 -2 -10 1 6 -2 - 3

1+ . 1+2 - 3 2 - 4

x x

x xx x

x x

+1 2 -=

2 1

x x

x = 1; y = 4 or x = 5; y = -4 Eqn. of AB :

7x + y - 31 = 0

Point(5, -4) lies on AB; it cant be considered. The diagram will make clear.

Required incident point (1, 4) Eqn. to reflected ray is 3x - y + 1 =0

Illustration of (5, -4):

Clearly, the incident ray is passing through (4, 3), which is not to be taken.

Q.No. 40 If + =1 is a line through the intersection of + =1 and + =1 and the lengths of

x y x y x y

c d a b b a

The perpendiculars drawn from the origin to these lines are equal to these lines are

equal in lengths then :

Option 1 1 1 1 1+ = +

2 2 2 2a b c d

Option 2 1 1 1 1- = -

2 2 2 2a b c d

Option 3 1 1 1 1+ = +

a b c d

Option 4 none

Explanation + = 1+ + -1 = 0 is equivalent to

x y x y

a b b a

+ =1x y

c d

1 1+ +

1+= = -(1)

1 1 1

a b b a

c d

Equating the value of ; we get

1 1 1 1- -

=1 1 1 1

- -

a c c b

b d d a

_____(1)

Also distance from origin is equal:

1 1 1 1- - +

1 1 1 1+ = + =

1 1 1 1- +

a c b d

a b c d

b d a c

2 2 2 2 ____(2)

1 1 1 1- - +

Similarly: =1 1 1 1

- +

c b d a

d a c b

____(3)

1 1 1 1 1 1 1 1From (ii) and (iii): - = - and + - + = 0

a b d c a b c d

1 1 1 1I) + = +

a b c d

Q.No. 41 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

Option 1 9x - 7y = 1

Option 2 9x + 7y = 71

Option 3 7x + 9y = 73

Option 4 7x - 9y + 17 = 0

Explanation

The line are inclined to 5y = 12x + 6 - 4 __L1

and 3x = 4y + 7 __L2

so, they would be parallel to angle bisectors of L1 and L2.

Equation to angle bisectors:

12 - 5 6 3 - 4 -7+=

13 5

x y x y

21x + 27y + 121 = 0 and 99x - 77y - 61 = 0

Hence, lines passing through (4, 5) are 7x + 9y = 73 and 9x - 7y = 1

Q.No. 42 Let + + = 0, - + = 0, R 3 3u ax by a b v bx ay b a a b be two straight lines. The

equation of the bisectors of the angle formed by k1u - k2v = 0 and k1u + k2v = 0 for non

zero real k1 and k2 are :

Option 1 u = 0

Option 2 k2u + k1v = 0

Option 3 k2u - k1v = 0

Option 4 v = 0

Explanation ( + + ) = 0u ax by a b 3

( - + ) = 0v bx ay b a 3

Clearly u v

(((( )))) (((( ))))so, X + + ; Y - +ax by a b bx ay b a 3 3 In coordinate system XY, the new lines are K1X - K2Y = 0 and K2X + K2Y = 0

Eqn. to bisectors:

K X-K Y K X-K Y= X = 0 or Y = 0

K +K K +K 1 2 1 2

2 2 2 21 2 1 2

Hence, two lines are u = 0 and v = 0

Q.No. 43 If the equation, 2x2 + k xy - 3y

2 - x - 4y - 1= 0 represents a pair of lines then the value of

k can be:

Option 1 1

Option 2 5

Option 3 -1

Option 4 -5

Explanation Apply condition for a pair of lines.

k = -5, 1

Q.No. 44 Assertion: The lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 are

concurrent if = 0

a b c

a b c

a b c

1 1 1

2 2 2

3 3 3

Reason: The area of the triangle formed by three concurrent lines must be zero.

Option 1 If both assertion and reason are correct and reason is the correct explanation of

assertion.

Option 2 If both assertion and reason are true but reason is not the correct explanation of

assertion.

Option 3 If assertion is true but reason is false.

Option 4 If assertion is false but reason is true.

Correct Answer 1

Explanation Lines are: ai x + bi y + ci = 0, i = 1, 2, 3

- -1

- -

- -1Area of = mod 1 = 0

2 - -

- -1

- -

b c b c a c a c

a b b a a b b a

b c b c a c a c

a b b a a b b a

b c b c a c a c

a b a b a b a b

1 2 2 1 2 1 1 2

1 2 1 2 1 2 1 2

2 3 3 2 3 2 2 3

2 3 2 3 2 3 2 3

3 1 1 3 1 3 3 1

3 1 1 3 3 1 1 3

= 0

a b c

a b c

a b c

2

3 3 3

1 1 1

2 2 2

Q.No. 45 Assertion: Each point on the line y - x + 12 = 0 is equidistant from the lines

4y + 3x - 12 = 0, 3y + 4x - 24 = 0.

Reason: The locus of a point which is equidistant from two given lines is the angular

bisector of the two lines.


assertion.


assertion.



Correct Answer 1

Explanation Angle bisectors:

4 + 3 - 24 = 3 + 4 -12x y x y

x - y - 12 = 0 or 7x + 7y - 36 = 0 Assertion- True

Reason- true and correct explanation

Q.No. 46 Assertion: if (a1x + b1y + c1) + (a2x + b2y + c2) + (a3x + b3y + c3) = 0, then lines

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 can not be parallel.

Reason: If sum of three straight lines equation is identically zero then they are either

concurrent or parallel.


assertion.


assertion.



Correct Answer 4

Explanation (a1x + b1y + c1) + (a2x + b2y + c2) + (a3x + b3y + c3) = 0 i.e. ai x + bi y + c i = 0 are

concurrent and can not be parallel.

Assertion-true

Reason-false

Q.No. 47 Assertion: the four straight lines given by 12x2 + 7xy - 12y

2 = 0 and

12x2 + 7xy - 12y

2 - x + 7y - 1 = 0 are the sides of a square.

Reason: The lines represented by general equation of second degree

ax2 + 2hxy + by

2 + 2gx + 2fy + c = 0 are perpendicular if a + b = 0


assertion.


assertion.



Correct Answer 2

Explanation 12x2 + 7xy - 12y

2 = 0 and 12x

2 + 7xy - 12y

2 - x + 7y - 1 = 0 gives four straight lines

3x + 4y = 0; 4x - 3y = 0; 3x + 4y - 1 = 0 and 4x - 3y + 1 = 0. They are pair of parallel lines

having same distance between them.

Assertion-True

Reason-True, but not correct explanation.

Q.No. 48 Assertion: The diagonals of parallelogram formed by the lines

ax + by + c = 0, ax + by + c = 0,

ax + by + c = 0, ax + by + c = 0 will be perpendicular if aa + bb = 0

Reason: the diagonals of rhombus are always perpendicular.


assertion.


assertion.



Correct Answer 4

Explanation ax + by + c = 0, ax + by + c = 0,

ax + by + c = 0, ax + by + c = 0

if aa + bb = 0, they are sides of rectangle.

Assertion-False

Reason-True

Passage Text Let A(0, ), B(-2, 0) and C (1, 1) be the vertices of a triangle then

Q.No. 49 Angle A of the triangle ABC will be obtuse if lies in

Option 1 (-1, 2)

Option 2 52,

2

Option 3 2 2-1, ,2

3 3


Correct Answer 3

Explanation

Eqn. to BC: 3y - x - 2 = 0

A dont lie on BC

2B

3

A 90 > > > > 0

+ -0

2 .

b c a

b c

2 2 2

<

+ -0 1+(1- ) + 4 + -10 0

2

b c a

bc

2 2 22 2

< <

2(-1,2),

3

2 2

-1,3 3

Q.No. 50 If I1, is the interval of values of for which A is obtuse and I2 be the interval of value of

for which A is largest angle of ABC , then

Option 1 I1 = I2

Option 2 I1 is a subset of I2

Option 3 I2 is a subset of I1


Correct Answer 2

Explanation I1 I2 because, even in acute or right triangle, A may be greatest angle.

Q.No. 51 All the values of for which angle A of the triangle ABC is largest lie in interval.

Option 1 (-2, 1)

Option 2 2 2-2, ,1

3 3

Option 3 2 2-2, , 6

3 3


Correct Answer 3

Explanation 2 2Case I : A is obture : -1, ,2

3 3

____(i)

Case II: A = 900 = -1,2 ____(ii)

Case III: A is greatest in acute ABC anda b a c> >

10 1 + (1 - ) and 10 2 + > > > > 2 2 2>

- 2 - 8 0 and 6 < < < < 2 2 < ___(iii)

2and

3

Combining all cares:

2 2-2, , 6

3 3

Passage Text Let OX and OY be two fixed lines inclined at a constant angle , A variable line cuts OX

at P and OY at Q. From P and Q perpendicular PM and PN are drawn to OY and OX

respectively.

Q.No. 52 Let the axes be chosen as OX and OY. If OP = a, OQ = b, then equation of PQ must be

Option 1 + =1

cos sin

x y

a b

Option 2 + =1

sin cos

x y

a b

Option 3 + = 1

x y

a b


Correct Answer 3

Explanation

In XY coordinate, equation to PQ

+ = 1x y

a b

Q.No. 53 The equation of MN must be

Option 1 + =1

cos sin

x y

a b

Option 2 + =1

cos cos

x y

b a

Option 3 + =1

sin sin

x y

a b


Correct Answer 2

Explanation

M cos a

N cos b

Eqn. to MN:

+ =Lcos cos

x y

b a

Q.No. 54 If AB passes through a fixed point (h, k) then MN passes through a fixed point

Option 1 ( cos , sin ) k h

Option 2 ( cos , cos ) k h

Option 3 ( sin , cos ) k h


Correct Answer 2

Explanation

Let (y - k) = m(x - b) be line AB and (h, k) be the fixed point.

-M ( - ) cos ; N cos

mh kk mh

m

MN= + =1( - )cos ( - )cos

x my

k mh mh k

( - cos )+ ( cos - ) = 0 x k m h y passes through ( cos , cos ) k h

Q.No. 55 Match the values of k for which origin and (k, 3) lie

No. Column A Column B Column C Id of Additional

Answer

1 (A) In the same angle formed

by lines x - 2y + 3 = 0 and

2x + y + 5 = 0

(P) (-4, 3) A-R

2 (B) Opposite angles formed


2x + y + 5 = 0

(Q) (- ,-4) B-Q

3 (C) Adjacent angles formed


2x + y + 5 = 0

(R) (3, ) C-P

Explanation (A) (0, 0) and (k, 3) lie in the same angle 3( -6+ 3) 0 and 5(2 +3+5) 0 3 > > >k k k

(B) (0, 0) and (k, 3) lie in opposite angle 3( -6+3) 0 and 5(2 +3+5) 0 -4 < < k k

Or ( -6+ 3)3 0 and 5(2 +3+5) 0>

Q.No. 56 A lines cuts X-axis at A and Y-axis at B such that AB = l. Match the following loci:

No. Column A Column B Column C Id of Additional

Answer

1 (A) Circumcentre of

triangle ABC (P) + = 9

22 2 l

x y A-Q

2 (B) Orthocentre of

triangle ABC (Q) + = 4

22 2 l

x y B-R

3 (C) Incentre of

triangle ABC

(R) x2 + y

2 = 0 C-S

4 (D) Centroid of

triangle ABC

(S) y = x D-P

Explanation (A) Circumcentre (u, k) (0, b)

a

2 + b

2 + l

2

4h2 + 4k2 = l2

(B) origin is always orthocentre.

x2 + y2 = 0

(C) .0 + . + 0. .0 + 0. + .

Incentre ,+ + + +

l a b a l b b a

l a b l b a

x = y (D) centroid (x, y) 3x = 1 and 3y = b

+ =9

2

2 2 lx y

Q.No. 57 For 0 2 < , if the point (2cos , 2sin ) lies in the angle between the lines = - 2y x in

which origin lies, then lies in the interval of length k , then k must be

Explanation 0001

Lines are: y = x - 2 and y + x - 2 = 0

(2cos ,2sin ) lies in the region of (0, 0)

(2sin -2cos + 2) 0 and(2sin +2cos -2) 0 > <

-1 1sin - and sin +

4 42 2

> > > >

<

3, and ,2

4 2 2

3, = =1

2 2

k k

Q.No. 58 The co-ordinates of A, B, C are (6, 3) , (-3, 5), (4, -2) respectively. For any point P(x, y) if

the ratio of the areas of the + - 2

PBC and ABC istriangles .

x y

Then the numerical quantity must be equal to

Explanation 0007

A (6,3); B (-3,5); C (4,-2); P ( , ) x y

6 3 1

2A ( ABC) = mod -3 5 1 = 49

4 -2 1

r

1

2A ( PBC) = mod -3 5 1

4 -2 1

x y

r

= 7 +7 -14x y

+ - 2 + -2A ( PBC)= =

A ( ABC) 7

x y x yr

r

= 2

Q.No. 59 Two sides of a rhombus lying in the first quadrant are given by 3x - 4y = 0 and

12x - 5y = 0. The length of the longer diagonal is 12. If the equations of other two sides

180 468are 3 - 4 = - and 12 - 5 = then the numerical quantity should bex y x y k

k k

Explanation 0130

OB = 12

OA : 3x - 4y = 0

OC = 12x - 5y = 0

12 3-

335 4tan = =36 56

1 +20

In = OCB :

2 -12cos(180 - ) =

2

2 2

2

a

a

-56 2 -12=

65 2

2 2

2

a

a

6 130=

11 a

3A ( cos , sin ) tan =

4 a a

24 130 18 130B ,

55 55

30 130 72 130C ,

143 143

42 54D ,

130 130

Eqn. to CB:

-198 130 -1803 - 4 = =

143 130

x y

Eqn. to AB:

46812 - 5 = = 130

130x y k

Q.No. 60 The base of a triangle passes through a fixed point (1, 1) and its sides are bisected at

right angle by the lines y2 - 8xy - 9x

2 = 0. If the locus of its vertex is a circle of radius

. Then the numerical quantity l must be equal.32

Explanation 0041

Let A (h, k) BC y - 1 = m(x - 1) of y + x = 0 of (y - 9x) = 0 Eqn. to AB:

9y + x = 9k + h

Eqn. to AC:

y - x = k - h

9 + 81 + 9F , is midpoint of AB

82 82

k h k h

9 - 40 40 + 9B ,

41 41

k h k h

- -E , is midpoint of AC

2 2

h k k h

C (- ,- ) k h

Form eqn. of BC, eliminating m

9 - 40 40 + 9-1 -1

41 41=- -1 - -1

k h k h

k h

(9 - 40 - 41) 40 + 9 - 41=

+1 +1

k h k h

k h

4 + 4 +9 - = 0 2 2h k h k

41which is a circle of radius =

32

q.no. 1 the bisector of the acute angle between the lines...

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