The distance of the point P (6, 8) from the origin is

8

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2 \sqrt{7}

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10

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6

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Solution

The correct option is C $10$
$T h e d i s t a n c e p o f a p o i n t P (x, y) f r o m t h e o r i g i n e i s p = \sqrt{x^{2} + y^{2}} . H e r e P (x, y) = (6, 8) . ∴ p = \sqrt{x^{2} + y^{2}} = \sqrt{6^{2} + 8^{2}} u n i t s = \sqrt{100} u n i t s = 10 u n i t s . A n s - 10 u n i t s .$

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2 \sqrt{7}

The distance of the point P(-6, 8) from the origin is

(a) $8$ (b) $2 \sqrt{7}$ (c) $6$ (d) $10$

2 \sqrt{7}

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