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 .
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.