Lecture 3 [17 Feb, 2020] MATH 127 (Prof Farid)

 

MATH -127 (Coordinate Geometry)
Chapter-2
(Pair of Straight Lines)





General equation of second degree,
  ax2+2hxy+by2+2hx+2fy+c=0  
If we multiply two linear functions we get one non-linear function.
To remove the first-degree term from (1), we must shift the origin to  (α,β)   where,
  α=bgfhh2ab=fhbgabh2β=afghh2ab=ghafabh2  
and to remove the term containing xy, we must rotate the axes through,  θ=12tan12hab  

Homogenous quadratic equation (where the total power of variables are same in every term):  ax2+2hxy+by2=0  
we know, if m1 and m2 are two roots of an equation, 
  x2(m1+m2)x+m1m2=0  
Now, take b0 and divide throughout of (2) by bx2
  ab+2hy6x+y2x2=0 is a quadratic equation in yx

we know, for every quadratic equation,  AX2+BX+C=0  
Here,  A=1,   B=2hb,   C=ab,   X=yx  
If  m1, m2 are the roots of (3), then we get, 
  m1+m2=2hb   and   m1m2=ab  
From (2) we get, 
  ax2+2hxy+by2=(yxm1)(yxm2)=0ym1x=0   and   ym2x=0  
Equation (2) represents two straight lines that pass through the origin.

        ♦ Generalization: 
an equation of the form,  aoyn+a1yn1x1+a2yn2x2++anxn=0   is a homogenous equation in xy  of degree n
  Pn(x)=ao+a1x++anxn  
if θ be the angle between  ym1x=0   and   ym2x=0   then,
  tanθ=m1m21+m1m2=(m1m2)21+m1m2=(m1+m2)24m1m21+m1m2=2h2aba+b θ=tan12h2aba+b  




        ♦ Corollary:
1) if the lines  ym1x=0   and   ym2x=0 are parallel:
  θ=2h2aba+b=0 h2=ab  

2)if the line represented by equation (2) are perpendicular:
  tan90=2h2aba+b=10  a+b=0  


•Under which condition equation (1) represents a pair of straight lines:
  |ahghbfgfc|=0abc+2fghaf2bg2ch2=0  
•angle between the lines represented by (1)  lx+my+n=0and   l1x+m1y+n1=0  
They will not pass through the origin. In case of homogenous, they will pass through the origin.




    Question 1: Prove that the equation x2+6xy+9y2+12y5=0  represents a pair of straight lines.
Solution: Given that  x2+6xy+9y2+12y5=0  
comparing equation (1.1) with the standard equation we get, 
  ax2+2hxy+by2+2gx+2fy+c=0 a=1,   b=9,   c=5,   h=3,   g=2,   f=6  
the equation will represent a pair of  straight lines if,abc+2fghaf2bg2ch2=0So now,  1×9×(5)+2×6.2×31×369×4+5×9=45+723636+45=0                                                                                                    
The condition is satisfied.


    Question 2: Find the values of k, so that the equation 3x2+10xy+9y2kx26y+21=0 represents a pair of straight lines and also find the angle between them.
Solution: The given equation is,3x2+10xy+9y2kx26y+21=0                                                                                                            
comparing equation (2.1) with the standard equation,
  ax2+2hxy+by2+2gx+2fy+c=0we get,a=3,   b=8,   c=21,   h=5,   g=k2,   f=13.  now, 3×8×21+2×(13)×(k2)×53×(13)28(k2)221×52=02k265k+528=0  




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