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Question
Prove that the lines 4x + 3y = 10, 3x - 4y = - 5 and 5x + y = 7 are concurrent.
Solution
Given lines are
4x + 3y = 10
⇒ 4x + 3y - 10 = 0 ⇒ 3x - 4y = - 5
⇒ 3x- 4y + 5 = 0 ⇒ 5x + y = 7
⇒ 5x + y - 7 = 0
The condition for the given lines to be concurrent is
`|(4,3,-10),(3,-4,5),(5,1,-7)|` = 0
Consider `|(4,3,-10),(3,-4,5),(5,1,-7)|`
`=> 4|(-4,5),(1,-7)| - 3|(3,5),(5,-7)| - 10|(3,-4),(5,1)|` ...[Expanding along R1]
⇒ 4(28 - 5) - 3(- 21 - 25) - 10(3 + 20)
⇒ 4(23) - 3(- 46) - 10(23)
⇒ 92 + 138 - 230
⇒ 230 - 230 = 0
Hence, the given lines are concurrent.
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