Question

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

Answer 1

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 =\[\sqrt{A Q^2 + Q P^2}\]

\[= \sqrt{\left( - 4 \right)^2 + \left( 5 \right)^2}\]
\[ = \sqrt{16 + 25} = \sqrt{41}\]

Similarly, the length of the perpendicular from P to the z-axis =\[\sqrt{\left( - 4 \right)^2 + \left( 3 \right)^2}\]

\[= \sqrt{16 + 9}\]
\[ = \sqrt{25} = 5\]