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Find the perpendicular distance between the lines 9x + 40y – 20 = 0 and 9x + 40y + 21 = 0. |
Answer» Putting \(x\) = 0 in equation of one of the lines say 9\(x\) + 40y –20 = 0, we get y = \(\frac{1}{2}\) ∴ A point on 9\(x\) + 40y – 20 = 0 is \(\big(0,\frac{1}{2}\big)\) ∴ Distance of \(\big(0,\frac{1}{2}\big)\) from 9\(x\) + 40y + 21 = 0 is \(\frac{\big|9\times0+40\times\frac{1}{2}+21\big|}{\sqrt{9^2+40^2}}\) = \(\frac{|41|}{\sqrt{1681}}\) = \(\frac{41}{41}\) = 1. |
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