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lavjeet (0)

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1. A variable straight line, drawn through the point of intersection of the straight lines  x/a + y/b = 1  and      


x/b + y/a = 1  meets the coordinate axes in A and B.


Show that the locus of mid point of AB is the curve


2(a + b)xy = ab(x + y).


 


 


2. If sum of the distances of a point from two perpendicular lines  in a plane is 1, then its locus is


a) square                   b)circle


c)straight line             d)two intersecting lines


 


 


3. The line ( p + 2q)x + (p-3q)y = p-q for different values  of p and q passes through rhe point


a) (3/2 , 5/2)                 b) (2/3  , 2/5)


c) (3/5 , 3/5)                 d) (2/5 , 3/5 )


 


4. If x/c + x/d = 1 be any line through the intersection of lines x/a + x/b = 1 and x/b + x/a = 1, then


a) 1/a + 1/d = 1/b + 1/c                     b) 1/b + 1/d = 1/c + 1/a


c) 1/c + 1/d = 1/a + 1/b                     d)n.o.t


 


5.  The eqn. to the pair of straight lines through the origin and perpendicular


to 2 - 4xy - 7 = 0 is ...................................?


 


6. If a + 2hxy + b + 2gx + 2fy + c = 0 represents a pair of lines, equidistant from the origin, then


a) = c( bf - ag )                          b) - = c( b - a )


c)  + = c( b + a )             d) n.o.t


 


 


7. The eqn. of bisector of that angle b/w the lines x + y + 1 = 0   and


2x - 3y - 5 =0 which contains the point ( 10 , -20 ) is ..............................?


 


8. A point moves so that sum of squares of its distances from the four sides of a  square  is constant. This point lies on a


a) straight line                                       b) circle


c) ellipse                                               d) parabola


 


 


9. Given the system of straight lines a( 2x + y - 3 ) + b( 3x + 2y - 5 ) = 0, the


line of the system situated farthest from the point ( 4, -3 ) has the eqn.  ................................?


 


10. If point (4 , 5 ) is rotated by anticlockwise though the origin the new point is ..............................?


 


 


11. All points lying inside the triangle formed by the points ( 1,3 ) , ( 5 , 0 ) and ( -1 , 2 ) satisfy


a) 3x + 2y 0                                 b) 2x + y - 13 0


c) 2x - 3y - 12 0                           d) -2x + y 0


 


12. The line L  has intercepts  a and b on the co-ordinate axes. When keeping the origin fixed, the co-ordinate axes are rotated through a fixed angle, then the same line has intercepts p and q on the rotated axes, then


a) + = +                           b) + = +


c) + = +                          d) + +  


 


13. The area of triangle is 5sq. units and two of its vertices are(2 ,1 ) and 


( 3 , -2 ). The third vertex which lies on yhe line y = x + 3 is


a) (-3/2,3/2) correct                                                                b) (5/2,11/2)


c) (7/2.13/2) correct                                      d) (0,0)


 


 


14. If one of the diag0nals of a square is along the line x = 2y and one of its vertices is (3,0), then the eqns. of the sides through this vertex are given


by ...........................?


 


15. The eqn. h + ( 1- )xy  - h = 0 represents a pair of straight lines, then


a) h =2 necessarily                                   b) h = -2  necessarily


c) h may be any real number                    d) n.o.t


 


 

    
deedee (1944)

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for  d 5th q




 


d ansewr is 7x^2+4xy-2y^2=0




 


c if ther's ne combined equation of straigth lines ax^2-2hxy+by^2=0.....which passes thru d origin then d equation of line perpendicular 2 it is given by bx^2+2hxy+ay^2=0




 


yeh raata mar lo.......wat u hav 2 do is just change d sign of d middle term and change d coefficients of d other two...




 


note .this is only and only if they pass thru origin


don't walk as if u rule d world
walk as if u dont care who rules d world

-this is knw as attitude


B who u r and say wat u feel ......
coz those who mind don't matter ........
and those who matter dont mind ......... :)


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iitjee_2009 (0)

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for the 15th question..............................


                                                       if they should reperesnt a pair of lines then they shud satisfy....


                                            h square greater than ab......caluculate tat and u find is satisfies for all real values of h


 


for 13th the answer is c draw the traingle and find the length of its base....then height to caluculate the area will be perpendicular distance from the third vertex to the line u have got.....so multiply 1/2 base*height and u will get it


 


 

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iitjee_2009 (0)

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and for 12th one though rotate it,,...perpendicular distance remains the same.....so its answer is c

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anchitsaini (4352)

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1) \mbox{Point of intersection = } [\frac{ab}{a+b},\frac{ab}{a+b}] \\ \\<br/>\mbox{Let the equation of line be }->y=mx + c \\ \\<br/>\mbox{Now since it passes through} \frac{ab}{a+b},\frac{ab}{a+b} ,\mbox{ we get relation as }-> \\ \\<br/>c = \frac{[1-m]ab}{a+b} \\ \\<br/>Also , A -> \frac{-c}{m}, 0 \\ \\<br/>B -> 0, c \\ \\<br/>Mid-Point -> x=\frac{-c}{2m} , y=\frac{c}{2} \\ \\<br/>\mbox{which implies }m = \frac{-y}{x} \\ \\<br/>Hence -> \\ \\<br/>y=\frac{c}{2}=\frac{[1-m]ab}{2(a+b)}=\frac{[x+y]ab}{2x(a+b)}\\ \\<br/>or ->\\ \\<br/>\mbox{2xy(a+b)=ab(x+y)}


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anchitsaini (4352)

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3)\mbox{The equation can be rewritten as }-> \\ \\<br/>(x+y-1) + \frac{q}{p}(2x - 3y + 1)=0 .........or \\ \\<br/>L_1 + \lambda L_2 = 0 .....\\ \\<br/>\mbox{Hence it passes through point of intersection of }-> \\ \\<br/>x+y-1=0 , \ and \ ,2x - 3y + 1=0 \\ \\<br/>which \ is \ \\ \\<br/>\frac{2}{5},\frac{3}{5}


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anchitsaini (4352)

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4)\mbox{The point of intersection as calculated in qn 1 is}-> \\ \\<br/>\frac{ab}{a+b} , \frac{ab}{a+b}  \\ \\<br/>\mbox{Putting it in the eqn , we get }-> \\ \\<br/>\frac{1}{c} + \frac{1}{d}=\frac{a+b}{ab}=\frac{1}{a}+\frac{1}{b}


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krish1092 (528)

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For 13th question,take the third vertex as (x , x+3) and substitute in the area of triangle formula


[y = x+3 is given]..on solving |4x-4| = 10 will be obtained


4x-4 = 10                                                                        4x-4 = -10


x = 7/2                                                                              x = -3/2 


substitute x=7/2 in y=x+3                                          substitute x= -3/2 in y= x+3


y = 13/2                                                                        y = 3/2


3rd vertex is ( 7/2,13/2)                                                3rd vertex is (-3/2,3/2)  

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allamraju (3422)

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The questions 1,3,4,5,13,15 are already answered.Let me answer some of the remaining questions.

2)Let us assume the lines to be X,Y axes for convenience.Then,P(x0,y0) belongs to the locus iff Ix0I+Iy0I=1.Thus the locus is a square formed by the lines IxI+IyI=1 with centre as origin.

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8)Let us assume the square to be formed by the lines x=0,y=0,x=a,y=a.Then the sum of squares of distances of a point (x,y) from the four sides is

x2+y2+(x-a)2+(y-a)2=k,a constant.

2x2+2y2-2ax-2ay+2a2-k=0. which is a circle for ka2.

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10)When rotated thro' an angle k,the point (x,y) becomes (x',y')

where x'=xcosk+ysink and y'=-xsink+ycosk.

Here,k=pi/4 and (x,y)=(4,5) and so,(x',y')=(9/rt2,1/rt2).

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krishna.gopal (2397)

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SOrry friend but in this forum we entertain only one question per post. For first question anchit's answer is right.

Krishna Gopal Singh
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IIT Delhi 2002
Currently doing PhD from IIT Delhi
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