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The Perpendicular distance between the straight lines $6 x+8 y+15=0$ and $3 x+4 y+9=0$ is
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Verified Answer
The correct answer is:
$\frac{3}{10}$ unit
$6 \mathrm{x}+8 \mathrm{y}+15=0$...(i)
and $3 x+4 y+9=0$ ...(ii)
Multiply equation (ii) by 2 , we get $6 x+8 y+18=0$
Distance between the straight lines
$\frac{\left|c_{2}-c_{1}\right|}{\sqrt{a^{2}+b^{2}}}=\frac{18-15}{\sqrt{(6)^{2}+(8)^{2}}}=\frac{3}{10}$ unit
Option (b) is correct.
and $3 x+4 y+9=0$ ...(ii)
Multiply equation (ii) by 2 , we get $6 x+8 y+18=0$
Distance between the straight lines
$\frac{\left|c_{2}-c_{1}\right|}{\sqrt{a^{2}+b^{2}}}=\frac{18-15}{\sqrt{(6)^{2}+(8)^{2}}}=\frac{3}{10}$ unit
Option (b) is correct.
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