Show that the line y=mx bisects the angle between lines #ax^2-2hxy+by^2=0# if #h(1-m^2)+m(a-b)=0#?

1 Answer
P dilip_k
Aug 6, 2018

Let the equations #y-m_1x=0andy-m_2x=0# are two straight lines represented by the given equation of pair of straight lines.Here #m_1=tanalphaandm_2=tanbetaand beta>alpha#

Hence

#(y-m_1)(y-m_2x)=y^2-(2h)/bxy+a/bx^2#

So

#m_1+m_2==(2h)/bandm_1m_2=a/b#

If #theta# be the angle subtended by angle bisector (#y=mx#) of the pair of straight line with the positive direction of X-axis ,then #m=tantheta#

Now it is obvious that

#theta-alpha=beta-theta#

So

#alpha+beta=2theta#

#=>tan(alpha+beta)=tan(2theta)#

#=>(tanalpha+tanbeta)/(1-tanalphatanbeta)=(2tantheta)/(1-tan^2theta)#

#=>(m_1+m_2)/(1-m_1m_2)=(2m)/(1-m^2)#

#=>((2h)/b)/(1-a/b)=(2m)/(1-m^2)#

#=>h/(b-a)=m/(1-m^2)#

#=>h(1-m^2)=(b-a)m#

#=>h(1-m^2)+(a-b)m=0#

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