The distance between $(0, 4)$ and $(x, 0)$ is $5$ units.
Find $x$ .

\pm 2

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\pm 3

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\pm 4

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\pm 5

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Solution

The correct option is C $\pm 3$
Let $(0, 4) = (x_{1}, y_{1})$ and $(x, 0) = (x_{2}, y_{2})$
By distance formula,
$5 = \sqrt{(x - 0)^{2} + (0 - 4)^{2}}$
Squaring on both sides, we get
$25 = x^{2} + 16$
$x^{2} = 9$
$x = \pm 3$

So, option B is correct.

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The distance between (0,4) and (x,0) is 5 units.Find x.

The correct option is C ±3 Let (0,4)=(x1,y1) and (x,0)=(x2,y2)By distance formula, 5=√(x−0)2+(0−4)2Squaring on both sides, we get25=x2+16x2=9x=±3So, option B is correct.

The distance between $(0, 4)$ and $(x, 0)$ is $5$ units.
Find $x$ .

The correct option is C $\pm 3$
Let $(0, 4) = (x_{1}, y_{1})$ and $(x, 0) = (x_{2}, y_{2})$
By distance formula,
$5 = \sqrt{(x - 0)^{2} + (0 - 4)^{2}}$
Squaring on both sides, we get
$25 = x^{2} + 16$
$x^{2} = 9$
$x = \pm 3$

So, option B is correct.