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

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regards

Ramesh

Answer after solving your equation is coming 10x+10y +3=0 I am not getting the answer sir as mentioned by you please recheck it ........... .

No. It`s correct , while solving the equation we will gety-(7/10) = -1(x+1/10)(Since the first line has the slope 1/7 .thus new line would have the slope -1)10y-7 =-10x-1 10x+10y-6=05x+5y-3=0