lf the lines $3 x + 2 y - 5 = 0, 2 x - 5 y + 3 = 0, 5 x + b y + c = 0$ are concurrent then $b + c =$

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-5

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Solution

The correct option is B -5
We know that concurrent lines have same intersection points.Then
$∴ 3 x + 2 y - 5 = 0$ --(1)
$2 x - 5 y + 3 = 0$ ---(2)
$5 x + b y + c = 0$ ---(3)
From 1 & 2,
$6 x + 4 y - 10 = 0$
$6 x - 15 y + 9 = 0$
$\Rightarrow 19 y - 19 = 0$
$y = 1$ & $x = 1$
So, $s u b s t i t u t i n g x = 1$ & $y = 1$ , in equation (3),
$5 x + b y + c = 0$
$5 + b + c = 0$
$∴ b + c = - 5$

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