Jee 2014 Booklet2 Hwt Solutions Straight Line

Vidyamandir Classes

Solutions to Home Practice Test/Mathematics Straight Line

3.(B)

HWT - 1

 b  a cos    a  b cos   ba sin   a  b sin    M  x, y    ,  ba ba  

               2ab  sin 2ab  sin   sin 2    cos 2  2 2   ,y    x ba b  a  

      x cos    y sin  2   0 2    

4.(D)

 a  12 2

 a  12 2

 

2  0   1 x  3 2  0   1 x  3



x 3



y 3



x  a  1  y  a  1 2

2

6.(A)



Mid point of diagonals will lie on L1

 3b  a 5  2   2 2   c     and (a, b) also lies on L1.  b = 2a + c From two equations we get c = –7 

7.(C)

If every point on the line is equidistant from the two given points.  The line is the perpendicular bisector of the line formed by joining these two points.  centre of line joining two points lie on the given line 

10.(C)

 a1  a2   

a1  a2  b b   b1  b2   1 2   c 2    2 



a12  b12  a22  b22  2c

From Pythagoras theorem we have

 k  12   h  2 2   2  12  1  12   h  12   k  12 

h 2  4  4h  1  h 2  1  2h



h=1

VMC/Straight Line



2 = 2h

1

HWT-Solutions/Mathematics

Vidyamandir Classes Now area = 1  

1 1 (base) × height = 1  k  1  1 2 2 

k 1  2

k  3,  1

Straight Line 3.(C)

HWT - 2

Slope of diagonal passing through origin = tan   45   

slope of other diagonal 

tan   1 1  tan 

1 tan   1 sin   cos     tan   1  tan   1 sin   cos   1  tan    

The other diagonal passes through A  a cos  , a sin   

 y  a sin  Equation of line (slope form) is   x  a cos 

 sin   cos    sin   cos  

x  cos   sin    y  sin   cos    a

4.(D)

6.(A)

Let G.P. x1 x2 x3 be x1 , x1r, x1r 2

y1 , y2 , y3 be y1 , y1r, y1r 2

and





points are  x1 , y1   x1r, y1r  x1r 2 , y1 x 2



points lie on the line



xy1  x1 y

(where x1 and y1 are the first terms of the G.P.’s)

Line is ax  by  c  0 But 3a  2b  c  0 

line is ax  by  3a  2b  0



a  x  3  b  y  2   0



b x  3     y  2  0 a



 x  3  k  y  2   0



This is a family of line concurrent at (3, 2)

b  k  a   

10.(CD) Since mid point of AC = Mid point of BD Quadrilateral is definitely a parallelogram. Now since diagonals AC and BD are perpendicular also  this is a rhombus. Since sides are not perpendicular it is not a square/rectangle.

Straight Line 2.(C)

Let the point as y-axis be (0, ) and ratio be k : 1 

5.(C)

7.(A)

HWT - 3

 0,    

k  3 2k  4  , k  1 

 k 1



k=3

In this triangle AB is  to BC  the triangle is right angled at B. 

circumcentre is mid point of AC (as AC is the diameter of the circumcircle)



circumcentre  0 ,  2

Let the line passing from the point (–2, 3) be y 3  m [in slope form] x2

VMC/Straight Line



y  mx  2m  3

2

HWT-Solutions/Mathematics

Vidyamandir Classes Now using distance formulae we have 4m  2m  3  5 m 1





 12

2

36m 2  64  96m  144m 2  144

108m 2  96m  80  0

Now discriminant of this equation is –ve  No real solution.

Straight Line

HWT - 4

3.(D)

Diagonals are perpendicular  PQRS is a rhombus.

6.(B)

Graph of | x | + | y | = 1 is

 8.(B)

Area = (Side)2 =



SA  a 2 1  t 2



2

 2

2

 2 sq. units.



 4a 2t 2  a 1  t 2



a 1  t2 1  Similarly SB  a 1  2   t  t2 



2

 a (1  t 2 )



  

 

1  t2 1 1 1 1 1      2 2 2 SA SB a a (1  t ) a 1 t 1 t





t2 

dependent on a but independent of 't'.

Straight Line 1.(B)

HWT - 5

f  m   2m3  3m 2  3m  2 f  1  0 

1 is one of its roots now let other roots be  , 

 3      1      and   1  1    1  2 1 ,   , 2  2 1 Now X intercepts =  mi 2

Y intercept =  

2.(B)

1 mi

 1  2  21 2 2 A      2    1    2 4     7 A   B  2

Since

a1 b c  1  1 a2 b2 c2



 and v are parallel lines

VMC/Straight Line

1 3  2 1  2 2



B



2 A   7B  2 A  7B  0



v + kv is family of parallel lines.

3

HWT-Solutions/Mathematics

Vidyamandir Classes 5.(C)

Since lines passes through origin let a line be y = mx y = mx y/x = m  Now

x3  6 x 2 y  11xy 2  6 y 3  0 2

3

 y  y  y 1  6    11   6    0 x x x f 1  0



1  6m  11m 2  6m3  0  f  m 

 1 is its roots. Let other roots be  ,  .

11      5/ 6 6 1 1  1      6 6 1 1 Roots 1, , are in harmonic Progression. 2 3 1   



1 1 ,  2 3

Straight Line 5.(D)

Since a1 b1 c are in H.P. 2ac We have b  ac Now

7.(D)

HWT - 6

x y 1   0 a b c

x a  c 1  y  0 a  2ac  c



2cx  y  a  c   2a  0



c 2 x  y   a  y  2  0



a 2 x  y     y  2  0 c



A family of line concurrent at 1,  2  .

These points are collinear.

8.

Area of rhombus =

Straight Line 1.(D)

This is an equilateral triangle centroid is incentre. 

3.(D)

This is right angled triangle.

HWT - 7  x  y  c    x  y  c   0 is possible only if x  y  c  0 and x  y  c  0 x=0 y=c   a point (0, c)  2

5.

orthocentre  0 , 0 a b circumcentre   ,  2 2

10.

Straight Line 3.(D)

2

Same as Question No.6 in HWT-2.

HWT - 8

Let ratio be k = 1  2k  1  k  2  ,    lies on 3x + 4y = 7 k 1   k 1 

3  2k  1

 k  1



42  k 

 k  1

7 

2k  11  7 k  7

4 9 If you observe this is an equilateral triangle therefore orthocenter and incentre are the same point. 

9.(A)

1 (product of 2 diagonals) 2

4 = 9k

VMC/Straight Line



k

4

HWT-Solutions/Mathematics

Vidyamandir Classes

Straight Line

HWT - 9



 2  3 3 1  2   2  2   1 2   1           1  

3.(BD) 

2  3 2  1 

8.(C)



2 2  3  2  0 



3

9  16 4

2



2

x 2  y 2  2 xr  2 yr  r 2  0

Now distance of

x y  1 from circle is r. a b

r r  1 a b r 1 1  2 a2 b

r  a  b   ab





a 2  b2

r







r 2 a 2  b 2  2ab  a 2 b 2  2ab  r  a  b   r 2 a 2  b 2



2r  2r  a  b   ab  0



r  a  b 

Reject

r  a b 



r  a  b a  b =

2



2  a  b   4  a  b   8ab 2

r



4

a 2  b2

a 2  b2 2

as r < a and r < b

 a  b 2  a 2  b 2 

2

ab a b 2

2



2ab a  b  a 2  b2

.



   P  1   P  1

P P2  1 1

Let the line be

2

2

2



P  1

x y  1 a b



a b Mid point of intercepts is  ,  2 2



a = 6, b = 8



equation is

x y  1 6 8

This represents two lines passing through origin. y = mx 

y m x



1  2cm  7 m 2  0

2

 1  2c 4   7  7

10.(A)

HWT - 10

These lines are perpendicular to a common line  these lines are parallel to each other. 

9.(C)

2r 2 ab  2abr  a  b   a 2b 2  0



Straight Line

6.(B)

3 5 1  2, . 4 2

Let equation of incircle be  x  t    y  r   r 2 

3.(A)





 y  y 1  2c    7    0 x x



c=2

Let that point be P (x, y)  

 x  a1 2   y  b1 2



x a    y  b  2 2

2

2 2

2 xa1  2a2 x  2b1 y  2b2 y  a22  b22  a12  b22

VMC/Straight Line

2



5

c

1 2 a2  b22  a12  b12   2

HWT-Solutions/Mathematics