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InterviewSolution
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Find the distance between 2x+y+4=0 and 2x+y+8=0.(a) \(\frac{4}{\sqrt{5}}\)(b) \(\frac{3}{\sqrt{5}}\)(c) \(\frac{9}{\sqrt{5}}\)(d) \(\frac{3}{\sqrt{2}}\)I had been asked this question in final exam.My question is from Distance of a Point from a Line in chapter Straight Lines of Mathematics – Class 11 |
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Answer» RIGHT OPTION is (a) \(\frac{4}{\sqrt{5}}\) For EXPLANATION: Distance between PARALLEL lines ax +by +c1=0 and ax+by+c2=0 is \(|\frac{c1-c2}{\sqrt{a^2+b^2}}|\). So, distance 2x + y+4=0 and 2x+y+8=0 is \(|\frac{8-4}{\sqrt{2^2+1^2}}|=\frac{4}{\sqrt{5}}\). |
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