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Question

lf the lines 3x+2y5=0, 2x5y+3=0, 5x+by+c=0 are concurrent then b+c=

A
7
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B
-5
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C
6
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D
9
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Solution

The correct option is B -5

We know that concurrent lines have same intersection points.Then

3x+2y5=0 --(1)

2x5y+3=0 ---(2)

5x+by+c=0 ---(3)

From 1 & 2,

6x+4y10=0

6x15y+9=0

19y19=0

y=1 & 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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