MHT CET · Maths · Straight Lines
The distance between the lines given by \(3 x+4 \mathrm{y}=9\) and \(6 x+8 \mathrm{y}=15\) is
- A \(5\) units
- B $ 3 units
- C \(0\cdot 5\) units
- D \(0\cdot 3\) units
Answer & Solution
Correct Answer
(D) \(0\cdot 3\) units
Step-by-step Solution
Detailed explanation
Given parallel lines are
\(3 x+4 y=9 \Rightarrow 6 x+8 y=18\) and \(6 x+8 y=15\)
Distance between them is
\(d=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|=\frac{|18-15|}{\sqrt{36+64}}=\frac{3}{10}=0.3\)
\(3 x+4 y=9 \Rightarrow 6 x+8 y=18\) and \(6 x+8 y=15\)
Distance between them is
\(d=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|=\frac{|18-15|}{\sqrt{36+64}}=\frac{3}{10}=0.3\)
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