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What is the perpendicular distance between the parallel lines $3 \mathrm{x}+4 \mathrm{y}=9$ and $9 \mathrm{x}+12 \mathrm{y}+28=0 ?$
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Verified Answer
The correct answer is:
$\frac{11}{3}$ units
The given lines are:$3 x+4 y=9$
$\Rightarrow \quad y=\frac{9}{4}-\frac{3}{4} x$ ...(i)
and $9 x+12 y+28=0 \Rightarrow y=\frac{-7}{3}-\frac{3 x}{4}$ ...(ii)
We have,
$$
\mathrm{m}=\frac{-3}{4} ; \mathrm{C}_{1}=\frac{9}{4} ; \mathrm{C}_{2}=\frac{-7}{3}
$$
$$
\text { Now, Distance }=\frac{\left|C_{1}-C_{2}\right|}{\sqrt{1+m^{2}}}=\frac{\left|\frac{9}{4}-\left(-\frac{7}{3}\right)\right|}{\sqrt{1+\frac{9}{16}}}=\frac{11}{3} \text { units }
$$
$\Rightarrow \quad y=\frac{9}{4}-\frac{3}{4} x$ ...(i)
and $9 x+12 y+28=0 \Rightarrow y=\frac{-7}{3}-\frac{3 x}{4}$ ...(ii)
We have,
$$
\mathrm{m}=\frac{-3}{4} ; \mathrm{C}_{1}=\frac{9}{4} ; \mathrm{C}_{2}=\frac{-7}{3}
$$
$$
\text { Now, Distance }=\frac{\left|C_{1}-C_{2}\right|}{\sqrt{1+m^{2}}}=\frac{\left|\frac{9}{4}-\left(-\frac{7}{3}\right)\right|}{\sqrt{1+\frac{9}{16}}}=\frac{11}{3} \text { units }
$$
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