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The line x+3y-2=0 bisects the angle between a pair of straight lines of which one has eqn x-7y+5=0.The eqn of the other line is:-
ANS-5x+5y-3=0
Please give complete explaination....
let new line has slope - m
the angle between bisector & new line will be same as btn bisector and line x-7y+5=0
tan x = ((1/7) +(1/3))/(1- (1/21)) = |m + 1/3|/(1 -m/3)
1/2 = |m + 1/3|/(1 -m/3)
|3m + 1|=(3 -m)
m = -1 or 1/7
so the new line has slope of -1
the pt. of intersection of line and bisecotr is : (-1/10 , 7/10 )
hence the line through (-1/10 , 7/10 ) with slope of -1 is : 5x+5y-3=0
--
regards
Ramesh
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