# Find the equation

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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