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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 -

Given:

2x − 5y + 3= 0 … (1)

5x − 9y+ λ = 0 … (2)

x − 2y + 1 =0 … (3)

It is given that thethree lines are concurrent.

Now, consider thefollowing determinant:

2(-9 + 2λ) + 5(5– λ) + 3(-10 + 9) = 0

-18 + 4λ + 25 –5λ – 3 = 0

λ = 4

The value ofλ is 4.

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