Question

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

Answer :

We know that the distance between two points (a₁, a₂) and (b₁, b₂) is

= $\sqrt{(a_1 - b_1)^{2} + (a_2 - b_2)^{2}}$

By the given condition,

$\sqrt{ {(na - a)}^{2} + {(nb - b)}^{2} } = 4 \sqrt{ {(5a - a)}^{2} + {(5b - b)}^{2} } \\ \\ \implies {(n - 1)}^{2} {a}^{2} + {(n - 1)}^{2} {b}^{2} = 16(16 {a}^{2} + 16 {b}^{2} ) \\ \\ \implies {(n - 1)}^{2} ( {a}^{2} + {b}^{2} ) = {16}^{2} ( {a}^{2} + {b}^{2} ) \\ \\ \implies {(n - 1)}^{2} = {16}^{2} \\ \\ \implies n - 1 = 16 \\ \\ \implies n = 16 + 1 \\ \\ \implies n = 17$

Therefore, the value of n is 17

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