If the straight line $(x + y + 1) + K (2 x - y - 1) = 0$ is perpendicular to $2 x + 3 y - 8 = 0$ then K =

5

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- 5

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4

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Solution

The correct option is B $- 5$
Line $1 : x + y + 1 + k (2 x - y - 1) = 0$
$\Rightarrow (2 k + 1) x + (1 - k) y + (1 - k) = 0$
Slope $= \frac{- (2 k + 1)}{1 - k} = \frac{2 k + 1}{k - 1} = m_{1}$
Slope of line $a x + b y + c = o$ is $(\frac{- a}{b})$
Line $2 : 2 x + 3 y - 8 = 0$
Slope $= \frac{- 2}{3} = m_{2}$
Lines are $⊥$ circular $\Rightarrow m_{1} m_{2} = - 1$
$\Rightarrow \frac{2 k + 1}{k - 1} \times + \frac{2}{3} = + 1$
$\Rightarrow 4 k + 2 = 3 k - 3$
$\Rightarrow k = - 5$

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