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Sahil Kochar Grade: 12
        

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....

7 years ago

Answers : (1)

Ramesh V
70 Points
										

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





7 years ago
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