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Lines L1-=a x+b y+c=0 and L2-=l x+m y+n=...

Lines `L_1-=a x+b y+c=0` and `L_2-=l x+m y+n=0` intersect at the point `P` and make an angle `theta` with each other. Find the equation of a line different from `L_2` which passes through `P` and makes the same angle `theta` with `L_1dot`

Text Solution

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From the figure. Image of point P(h,k) on required line in the line `L_(1) " lies on line " L_(2)`.
So, equation of required line is locus of point P such that its image in line `L_(1) " lies on line " L_(2)`.
Let image of point P in line `L_(1) " be Q"(x_(1),y_(1))`.
`rArr (x_(1)-h)/(a) = (y_(1)-k)/(b) = (2(ah+bk+c))/(a^(2)+b^(2))`
`rArr x_(1) =-(2a(ah+bk+c))/(a^(2)+b^(2))+h`
`" and " y_(1) =-(2b(ah+bk+c))/(a^(2)+b^(2))+k`
`" Now ",Q(x_(1),y_(1)) "lies on the line" L_(2).`
`therefore l[-(2a(ah+bk+c))/(a^(2)+b^(2))+h]+m[-(2b(ah+bk+c))/(a^(2)+b^(2))+k]+n=0`
Therefore, equation of required line is
`2(al+mb)(ax+by+c)-(a^(2)+b^(2))(lx+my+n)=0`
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