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Condition of Concurrency of 3 Straight Lines

If the lines ...

Question

If the lines $4 x + 3 y - 1 = 0; x - y + 5 = 0$ and $k x + 5 y - 3 = 0$ are concurrent then $k =$

A

4

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B

5

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C

6

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D

7

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Solution

The correct option is B $6$
Let $L_{1} \equiv 4 x + 3 y - 1 = 0; L_{2} \equiv x - y + 5 = 0 L_{3} \equiv k x + 5 y - 3 = 0$
Solving equation $L_{1}$ and $L_{2}$ we their point of intersection which will be satisfied by $L_{3}$ also.
Since the $3$ lines are concurrent.
$4 x + 3 y = 1$ -(i)
$x - y = - 5$ -(ii)
Adding (i) & (ii)
$7 x = - 14 x = - 2$
$- 2 - y = - 5 \Rightarrow y = - 2 + 5 = 3 x = - 2, y = 3$
Putting this in $L_{3}$
$k x + 5 y - 3 = 0 \Rightarrow - 2 k + 15 - 3 = 0 \Rightarrow - 2 k = - 12 k = 6$

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