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Find the equation of the line passing through the inersection of lines x-y+1=0 and 2x-3y+5=0 and whose distance from the point (3,2) is 7/5.

Find the equation of the line passing through the inersection of lines  x-y+1=0   and  2x-3y+5=0  and whose distance from the point (3,2) is 7/5.

Grade:

1 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
6 years ago
Hello Student, Please find the solution

Here you can use the concept of family of lines
The equation of the line passing through the inersection of lines x-y+1=0 and 2x-3y+5=0 can be assumed as (x-y+1) + \lambda ( 2x-3y+5) = 0

Now the distance of this line from (3,2) is 7/5
so, \frac{7}{5} =| \frac{(3-2+1)+\lambda (2*3-3*2+5)}{\sqrt{(1+2\lambda )^{2}+(1+3\lambda )^2}}|

Solve the above equation and find out \lambda
Once you get \lambda, put it back in equation
(x-y+1) + \lambda ( 2x-3y+5) = 0
And you will get the final answer

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