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  • A straight line meets the parallel lines <img alt="\mathrm{3 x-4 y=6\, and \, 6 x-8 y+c=0}"

A straight line \mathrm{x= y} meets the parallel lines \mathrm{3 x-4 y=6\, and \, 6 x-8 y+c=0} at points \mathrm{P} and \mathrm{Q} respectively such that \mathrm{\frac{\mathrm{OP}}{\mathrm{OQ}}=\frac{4}{3}} (where O is the origin). Then value of c (c > 0) is

Option: 1

10


Option: 2

14


Option: 3

16


Option: 4

9


Answers (1)

best_answer

Given parallel lines are \mathrm{3 x-4 y=6 .........(i)}
and \mathrm{3 x-4 y+\frac{c}{2}=0}\quad \ldots(ii)

Now the equation of line OP is \mathrm{\frac{\frac{x}{1}}{\sqrt{2}}=\frac{\frac{y}{1}}{\sqrt{2}}=\alpha}
any point on this line is given by \mathrm{\left(\frac{\alpha}{\sqrt{2}}, \frac{\alpha}{\sqrt{2}}\right)}. If a=OP, then the point \mathrm{\left(\frac{\alpha}{\sqrt{2}}, \frac{\alpha}{\sqrt{2}}\right)} will also satisfy the equation (i). Hence \mathrm{\mathrm{OP}=-6 \sqrt{2}}. Similarly \mathrm{\mathrm{OQ}=\frac{c}{\sqrt{2}}}

\begin{aligned} &\left|\frac{+6 \sqrt{2}}{\frac{c}{\sqrt{2}}}\right|=\frac{4}{3} \\ & \Rightarrow \quad C=9 \end{aligned}

Hence (D) is the correct answer.

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