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(13λ^{2} + 10λ + 2)

⇒ 6λ^{2} – 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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