Distance of a Point from the Origin

We will discuss here how to find the distance of a point from the origin.

The distance of a point A (x, y) from the origin O (0, 0) is given by OA = \(\sqrt{(x - 0)^{2} + (y - 0)^{2}}\)

i.e., OP = \(\sqrt{x^{2} + y^{2}}\)


Consider some of the following examples:

1. Find the distance of the point (6, -6) from the origin.

Solution:

Let M (6, -6) be the given point and O (0, 0) be the origin.

The distance from M to O = OM

                                     = \(\sqrt{(6 - 0)^{2} + (-6 - 0)^{2}}\)

                                     = \(\sqrt{(6)^{2} + (-6)^{2}}\)

                                     = \(\sqrt{36 + 36}\)

                                     = \(\sqrt{72}\)

                                     = \(\sqrt{2 × 2 × 2 × 3 × 3}\)

                                     = 6\(\sqrt{2}\) units.


2. Find the distance between the point (-12, 5) and the origin.

Solution:

Let M (-12, 5) be the given point and O (0, 0) be the origin.

The distance from M to O = OM = \(\sqrt{(-12 - 0)^{2} + (5 - 0)^{2}}\) = \(\sqrt{(-12)^{2} + (5)^{2}}\)

                                           = \(\sqrt{144 + 25}\)

                                           = \(\sqrt{169}\)

                                           = \(\sqrt{13 × 13}\)

                                           = 13 units.

 

3. Find the distance between the point (15, -8) and the origin.

Solution:

Let M (15, 8) be the given point and O (0, 0) be the origin.

The distance from M to O = OM = \(\sqrt{(15 - 0)^{2} + (-8 - 0)^{2}}\) = \(\sqrt{(15)^{2} + (-8)^{2}}\)

                                           = \(\sqrt{225 + 64}\)

                                           = \(\sqrt{289}\)

                                           = \(\sqrt{17 × 17}\)

                                           = 17 units.

 Distance and Section Formulae




10th Grade Math


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