1.

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.



Discussion

No Comment Found

Related InterviewSolutions