Q. 114.0( 4 Votes )

Find the equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 whose distance from the point (3, 2) is .

Answer :

Given:


Lines x – y + 1 = 0 and 2x – 3y + 5 = 0


To find:


The equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 whose distance from the point (3, 2) is 7/5.


Explanation:


The equation of the straight line passing through the point of intersection of x – y + 1 = 0 and 2x – 3y + 5 = 0 is given below:


x y + 1 + λ(2x – 3y + 5) = 0


(1 + 2λ)x + ( 3λ – 1)y + 5λ + 1 = 0 … (1)


The distance of this line from the point is given by




25(5λ + 2)2 = 49(13λ2 + 10λ + 2)


6λ2 – 5λ – 1 = 0


λ


Substituting the value of λ in (1), we get the equation of the required line.


3x – 4y + 6 = 0 and 4x – 3y + 1 = 0


Hence, equation of required line is 3x – 4y + 6 = 0 and 4x – 3y + 1 = 0.


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