Question

If the distance between the points (na, nb) and (a, b) is 4 times the distance
between the points (5a, 5b) and (a, b), then 'n' is equal to

Answer 1

$\textbf{Given:}$

$\text{Distance between (na,nb) and (a,b) = 4 times distance between (5a,5b) and (a,b)}$

$\textbf{To find:}$

$\text{The value of n}$

$\textbf{Solution:}$

$\text{Formula used:}$

$\text{The distance between two points (x_1,y_1) and (x_2,y_2) is}$

$\boxed{\bf\:d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

$\text{Consider,}$

$\textbf{Distance between (na,nb) and (a,b) = 4 times distance between (5a,5b) and (a,b)}$

$\sqrt{(na-a)^2+(nb-b)^2}=4{\times}\sqrt{(5a-a)^2+(5b-b)^2}$

$\sqrt{a^2(n-1)^2+b^2(n-1)^2}=4{\times}\sqrt{(4a)^2+(4b)^2}$

$\sqrt{a^2(n-1)^2+b^2(n-1)^2}=4{\times}\sqrt{16a^2+16b^2}$

$(n-1)\sqrt{a^2+b^2}=4{\times}4\sqrt{a^2+b^2}$

$(n-1)\sqrt{a^2+b^2}=16\sqrt{a^2+b^2}$

$n-1=16$

$n=16+1$

$\implies\bf\,n=17$

$\therefore\textbf{The value of n is 17}$

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