If P is a point on the line $y = 2 x$ in the first quadrant, and the distance from the origin to point P is $5$ , find the approximate coordinates of point P.

(2.24, 4.47)

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(3.00, 6.00)

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(4.00, 8.00)

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(4.47, 2.24)

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(4.72, 2.36)

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Solution

The correct option is B $(2.24, 4.47)$
Any point on the line $y = 2 x$ can be represented as $(x, 2 x)$ . Now the distance of the point from the origin is 5 units.
Hence $\sqrt{(x^{2} - 0) + ((2 x)^{2} - 0)} = 5$
$\Rightarrow$ $\sqrt{5} x = 5$ or $x = \sqrt{5}$ .
Hence the required point is $(x, 2 x) = (\sqrt{5}, 2 \sqrt{5})$ $\approx (2.24, 4.47)$ .

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If P is a point on the line y=2x in the first quadrant, and the distance from the origin to point P is 5, find the approximate coordinates of point P.

The correct option is B (2.24,4.47)Any point on the line y=2x can be represented as (x,2x). Now the distance of the point from the origin is 5 units. Hence √(x2−0)+((2x)2−0)=5 ⇒ √5x=5 or x=√5. Hence the required point is (x,2x)=(√5,2√5) ≈(2.24,4.47).

If P is a point on the line $y = 2 x$ in the first quadrant, and the distance from the origin to point P is $5$ , find the approximate coordinates of point P.