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**

**Distance Formula****Distance Properties in some Geometrical Figures****Conditions of Collinearity of Three Points****Problems on Distance Formula****Distance of a Point from the Origin****Distance Formula in Geometry****Section Formula****Midpoint Formula****Centroid of a Triangle****Worksheet on Distance Formula****Worksheet on Collinearity of Three Points****Worksheet on Finding the Centroid of a Triangle****Worksheet on Section Formula**

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