Question

Find the distance 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. PA will be perpendicular to the x-axis.

Also, QA = |3| and PQ = |5| Now, distance of P from the x-axis :

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 $△$: distance of P from the y-axis :

$PA=\sqrt{A{Q}^2+Q{P}^2}=\sqrt{(-4)3^2+(5{)}^2}=\sqrt{16+25}=\sqrt{41}$

Similarly, the length of the perpendicular from P to the z-axis = $\sqrt{(-4{)}^2+(3{)}^2}=\sqrt{16+9}=\sqrt{25}=5$