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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