the perpendicular distance of a point p 5;8 from y axis ​

Question

the perpendicular distance of a point p 5;8 from y axis

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Camila 3 weeks 2021-11-07T00:06:30+00:00 2 Answers 0 views 0

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    0
    2021-11-07T00:07:45+00:00

    Answer:

    5 units

    Step-by-step explanation:

    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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    0
    2021-11-07T00:08:13+00:00

    Hey there!

    ㅤㅤ

    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].

    Using distance formula:-

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

    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}}}}

    Therefore, 5 units is the required distance between them.

    Thanks!

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