Find the distances of the point P (-4, 3, 5) from the coordinate axes.

1.	Find the distances of the point P (-4, 3, 5) from the coordinate axes.
Answer» Given: Point P(-4, 3, 5) To find: distances of the point P from coordinate axes The distance of the point from x-axis will be given by, = \(\sqrt{y^2+z^2}\) = \(\sqrt{3^2+5^2}\) = \(\sqrt{9+25}\) = \(\sqrt{34}\) The distance of the point from y-axis will be given by, = \(\sqrt{x^2+z^2}\) = \(\sqrt{(-4)^2+5^2}\) = \(\sqrt{16+25}\) = \(\sqrt{41}\) The distance of the point from z-axis will be given by, = \(\sqrt{x^2+y^2}\) = \(\sqrt{(-4)^2+3^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5

Answer»

Given: Point P(-4, 3, 5)

To find: distances of the point P from coordinate axes

The distance of the point from x-axis will be given by,

= \(\sqrt{y^2+z^2}\)

= \(\sqrt{3^2+5^2}\)

= \(\sqrt{9+25}\)

= \(\sqrt{34}\)

The distance of the point from y-axis will be given by,

= \(\sqrt{x^2+z^2}\)

= \(\sqrt{(-4)^2+5^2}\)

= \(\sqrt{16+25}\)

= \(\sqrt{41}\)

The distance of the point from z-axis will be given by,

= \(\sqrt{x^2+y^2}\)

= \(\sqrt{(-4)^2+3^2}\)

= \(\sqrt{16+9}\)

= \(\sqrt{25}\)

= 5

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