Engineering
Mathematics
Introduction to Determinants
Question
lf the lines 3x + 2y – 5 = 0, 2x – 5y + 3 = 0, 5x + by + c = 0 are concurrent then b + c =
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Solution
We know that concurrent lines have same intersection points.Then
∴ 3x + 2y – 5 = 0 --(1)
2x – 5y + 3 = 0 ---(2)
5x + by + c = 0 ---(3)
From 1 & 2,
6x + 4y – 10 = 0
6x – 15y + 9 = 0
⇒ 19y – 19 = 0
y = 2 & x = 1
So, substituting x = 1 & y = 1, in equation (3),
5x + by + c = 0
5 + b + c = 0
∴ b + c = – 5
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