Given:
2x + 3y + 1 = 0 and 3x – 5y – 5 = 0
To find:
The equation of the straight line passing through the point of intersection of 2x + 3y + 1 = 0 and 3x – 5y – 5 = 0 and equally inclined to the axes.
Explanation:
The equation of the straight line passing through the points of intersection of 2x + 3y + 1 = 0 and 3x− 5y − 5 = 0 is
2x + 3y + 1 + λ(3x − 5y − 5) = 0
⇒ (2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0
⇒ y = - \(\Big(\frac{2+3λ}{3-5λ}\Big)-\Big(\frac{1-5λ}{3-5λ}\Big)\)
The required line is equally inclined to the axes. So, the slope of the required line is either 1 or − 1.
∴ - \(\Big(\frac{2+3λ}{3-5λ}\Big)\) =1 and - \(\Big(\frac{2+3λ}{3-5λ}\Big)\) = -1
⇒ -2 – 3λ = 3 – 5λ and 2 + 3λ = 3 – 5λ
⇒ λ = \(\frac{5}{2}\) and \(\frac{1}{8}\)
Substituting the values of λ in (2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0, we get the equations of the required lines.
⇒ 19x – 19y – 23 = 0 and 19x + 19y + 3 = 0
Hence, required equation is 19x – 19y – 23 = 0 and 19x + 19y + 3 = 0
