Question

For what value of λ are the three lines 2x – 5y + 3 = 0, 5x – 9y + λ = 0 and x – 2y + 1 = 0 concurrent?

Answer 1

Given: 2x − 5y + 3 = 0 … (1) 

5x − 9y + λ = 0 … (2) 

x − 2y + 1 = 0 … (3) 

To find:

Value of λ. 

Concept Used: 

Determinant of equation is zero. 

Explanation:

It is given that the three lines are concurrent.

⇒ 2(-9 + 2λ) + 5(5 – λ) + 3(-10 + 9) = 0 

⇒ -18 + 4λ + 25 – 5λ – 3 = 0 

⇒ λ = 4 

Hence, λ = 4.