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Question
Find the distances of the point P(–4, 3, 5) from the coordinate axes.
Solution
Let PQ be the perpendicular to the xy-plane and QA be perpendicular from Q to the y-axis.
PA will be perpendicular to the x-axis.
Also, QA = \[\left| 3 \right|\]and PQ =\[\left| 5 \right|\]
Now, distance of P from the x-axis:
PB =\[\sqrt{B Q^2 + Q P^2}\]
\[= \sqrt{3^2 + 5^2}\]
\[ = \sqrt{9 + 25} = \sqrt{34}\]
Similarly,
From the right-angled \[∆ PAQ\]distance of P from the y-axis:
PA =\[\sqrt{A Q^2 + Q P^2}\]
\[ = \sqrt{16 + 25} = \sqrt{41}\]
\[ = \sqrt{25} = 5\]
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