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aankurverma (1310)

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Olaaa!! Perrrfect answer. 222  [322 rates]

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                                          PAIR OF STRAIGHT LINES
 
1. ax2 + 2hxy + by2 = 0 is called a homogeneous equation of 2nd degree 
    passing through the origin.
 
a) two lines r real/distinct if h2 > ab
 
b) two lines r imaginary if  h2 < ab
 
c) two lines r coincident if h2 = ab 
 
 
2. If ax2 + 2hxy + by2 = 0 = b(y - m1x)(y - m2x)
   
    then m1 + m2 = -2h/b   & m1m2 = a/b
 
 
3. Angle b/w two above lines is theta = tan-12h2 - ab/ a + b
   
    a)  theta = 90o a+ b = 0           b) theta = 0o if h2 = ab
 
 
4. Equation of pair of straight lines perpendicular to ax2 + 2hxy + by2 = 0 is
   
     bx2 - 2hxy + ay2
 
 
5. Equation of bisectors of ax2 + 2hxy + by2 = 0  is (x2 - y2)/a - b = xy/h
   
    if a = b then bisectors  : y =  x
    
    if h = 0,then bisectors  : x = 0, y = 0
 
 
6. General equation of 2nd degree ax2 + 2hxy + by2 +2gx + 2fy + c = 0        
 
    represents a pair of straight lines if
 
    abc + 2fgh - a f2 - bg2 - ch2 = 0
 
 
7. The angle b/w above two lines is theta =   tan-12h2 - ab/ a + b
 
 
    a) lines parallel : h2 = ab
 
    b) lines perpendicular : a + b = 0
    
    c) lines coincident if h2 = ab , g2 = ac , f2 = ac
 
 
8. Pt. of intersection of lines S = ax2 + 2hxy + by2 +2gx + 2fy + c = 0 can b 
 
    obtained by solving
 
    delta s/delta x = 0  & delta s/delta y = 0
 
 
9. If (x1,y1) is the point of intersection of such lines then the eqn of bisectors of
 
   S = 0 is
 
   [(x - x1)2 - (y - y1)2] a - b  = (x - x1)(y - y1)/h   
  
 
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