Question

The perpendicular distance of a point P(5,
8) from the y-axis is:
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Answer 1

5 units

Given point is (5,8) and its x-coordinate(abscissa) is 5. Therefore, perpendicular distance of point P(5,8) from y-axis is 5 units

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Answer 2

Hey there!

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Answer:

★ The perpendicular distance of a point P (5, 8) from the y-axis is [We can find the distance between these points by using formula].

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Using distance formula:-

★ $\sf{Distance = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}}$

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Calculations:-

$\sf{\longrightarrow Distance = \sqrt{(x_{5} - x_{0})^{2} + (y_{8} - y_{8})^{2}}}$

$\sf{\longrightarrow Distance = \sqrt{(5 - 0)^{2} + (8 - 8)^{2}}}$

$\sf{\longrightarrow Distance = \sqrt{5^{2} + 0^{2}}}$

$\sf{\longrightarrow Distance = \sqrt{5^{2} + 0}}$

$\sf{\longrightarrow Distance = \sqrt{5^{2}}}$

${\longrightarrow{\boxed{\bold{Distance = 5}}}}$

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Therefore, 5 units is the required distance between them.

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Thanks!