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The line L is given by $\frac{x}{5} + \frac{y}{b} = 1$ passes through the point $(13, 32)$ . The line K is parallel to $L$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$ Then the distance between $L$ and $K$ is

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Correct option is B

As L:x5+yb=1 passes through (13,32), we get 135+32b=1⇒b=−20As K:xc+y3=1 is parallel to L−b5=−3c⇒c=−34Now equation of L and K are respectively 4x−y−20=0 and 4x−y+3=0and the distance between them is|−20−3|16+1=2317

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The line L is given by x5+yb=1 passes through the point (13, 32). The line K is parallel to L and has the equation xc+y3=1 Then the distance between L and K is

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The line L is given by $\frac{x}{5} + \frac{y}{b} = 1$ passes through the point $(13, 32)$ . The line K is parallel to $L$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$ Then the distance between $L$ and $K$ is