The point whose abscissa is equal to its ordinate and which is equidistant from A(–1, 0) and B(0, 5) is -Maths 9th

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Answer :

Putting \(x\) = 0 in equation of one of the lines say 9\(x\) + 40y –20 = 0, we get y = \(rac{1}{2}\)∴ A point on 9\(x\) + 40y – 20 = 0 is \(\big(0,rac{1}{2}\big)\)∴ Distance of \(\big(0,rac{1}{2}\big)\) from 9\(x\) + 40y + 21 = 0 is \(rac{\big|9 imes0+40 imesrac{1}{2}+21\big|}{\sqrt{9^2+40^2}}\) = \(rac{|41|}{\sqrt{1681}}\) = \(rac{41}{41}\) = 1.

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