Geometry | Coordinate Geometry

Coordinate Geometry – Straight Lines

In this article, we look at Coordinate Geometry, particularly, about Straight Lines.

Douglas Tawiah Dwumor

January 03, 2024

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MyStudentClass-Coordinate Geometry - Straight Lines

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Coordinate Axes
Cartesian Coordinates
Magnitude of a point
Distance between two points
Division Of a Line

Important Points of Triangle

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Coordinate Geometry is the study of Geometry using algebra. Because of this, it’s also called Algebraic or Analytic Geometry. In coordinate geometry, we are interested in points, lines and curves.

Rectangular or Coordinate Axes

The coordinate axes refer to two intersecting lines which cut at right angle. If the two lines do not cut at 90⁰, the axes is called oblique.

The Cartesian Coordinates

The Cartesian Coordinates is a pair of values, normally denoted as $\left(x,y\right)$ , which represent horizontal and vertical distances from point of intersection of two perpendicular lines, the $X$ and $Y$ axes. This forms the Cartesian Coordinates System.

Magnitude of a point

The magnitude of a single point in a plane is the distance of the point from the origin. Consider the point $P \left(x,y \right)$ in the Cartesian plane.

By Pythagoras theorem, $|PO|^2=|OB|^2+|BP|^2$ .

$=> |PO|^2=x^2 + y^2$

Therefore, the magnitude of a point with reference to the origin is $x^2 + y^2$ .

Example 1: Find the magnitude of the point A(5,12).

Solution

$|OA|= \sqrt{x^2 + y^2}$

$= \sqrt{5^2 + 12^2}$

$= \sqrt{25 +144}$

$=\sqrt{169}$

$=13$ units

Example 2: Calculate the distance of B(3,7) from the origin.

Solution

$|OB|= \sqrt{x^2 + y^2}$

$= \sqrt{3^2 + 7^2}$

$= \sqrt{9 +49}$

$=\sqrt{58}$

$= \sqrt{58}$ units

Distance between two points

The distance between two points A and B is the magnitude of the line that connects A to B.

$|AB|=\sqrt{ \left(x_2-x_1\right)^2 + \left(y_2 - y_1 \right)^2$

Example 1: Calculate the distance between $P(-1,4)$ and $Q(5,9)$ .

Solution

$|PQ|=\sqrt{ \left(x_2-x_1\right)^2 + \left(y_2 - y_1 \right)^2$

$= \sqrt{ \left(-1-5\right)^2 + \left(4 - 9\right)^2$

$= \sqrt{ \left(-6\right)^2 + \left(-5\right)^2$

$= \sqrt{36 + 25}$

$= \sqrt{61}$ units

Example 2: Find |MN| of the points $M(-6,0)$ and $N(-3, -0.5).$

Solution

$|PQ|=\sqrt{ \left(x_2-x_1\right)^2 + \left(y_2 - y_1 \right)^2$

$= \sqrt{ \left(-6--3\right)^2 + \left(0--0.5\right)^2$

$= \sqrt{ \left(-3\right)^2 + \left(0.5\right)^2$

$= \sqrt{9 + 0.25}$

$= \sqrt{9.25}$ units

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