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

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• 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.

\begin{tikzpicture}
    % Draw grid lines
    \draw[help lines, color=gray!30] (-5,-4) grid (5,4);
    
    % Draw the x-axis
    \draw[->, thick, blue] (-5,0) -- (5,0) node[below] {$x$};
    
    % Draw the y-axis
    \draw[->, thick, red] (0,-4) -- (0,4) node[left] {$y$};
    
    % Add labels and ticks
    \foreach \x in {-4,-3,-2,-1,1,2,3,4}
        \draw (\x,0) -- (\x,-0.1) node[below] {$\x$};
    \foreach \y in {-3,-2,-1,1,2,3}
        \draw (0,\y) -- (-0.1,\y) node[left] {$\y$};
        
    % Add origin label
    \node[below left] at (0,0) {O};
    
    % Add labels for quadrants
    \node[above right, blue] at (2,2) {Quadrant I};
    \node[above left, red] at (-2,2) {Quadrant II};
    \node[below left, blue] at (-2,-2) {Quadrant III};
    \node[below right, red] at (2,-2) {Quadrant IV};
    
    % Plot a point in each quadrant with labels
    \fill[green] (1,1) circle (2pt) node[above right] {$(1,1)$};
    \fill[orange] (-3,1) circle (2pt) node[above left] {$(-3,1)$};
    \fill[purple] (-4,-1) circle (2pt) node[below left] {$(-4,-0.5)$};
    \fill[brown] (3,-1) circle (2pt) node[below right] {$(3,-1)$};
\end{tikzpicture}

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.

\begin{tikzpicture}
    % Draw the x-axis
    \draw[->] (-5,0) -- (5,0) node[below] {$x$};
    
    % Draw the y-axis
    \draw[->] (0,-1) -- (0,5) node[left] {$y$};
    
    % Add labels and ticks
    \foreach \x in {-4,-3,-2,-1,1,2,3,4}
        \draw (\x,0) -- (\x,-0.1) node[below] {$\x$};
    \foreach \y in {1,2,3,4}
        \draw (0,\y) -- (-0.1,\y) node[left] {$\y$};
        
    % Add origin label
    \node[below left] at (0,0) {O};
    
    % Plot points A and B
    \fill[red] (3,4) circle (2pt) node[above right] {A(3,4)};
    \fill[blue] (-4,2) circle (2pt) node[above left] {B(-4,2)};
\end{tikzpicture}

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.

\begin{tikzpicture}
    % Draw the x-axis
    \draw[->] (0,0) -- (5,0) node[below] {$x$};
    
    % Draw the y-axis
    \draw[->] (0,0) -- (0,5) node[left] {$y$};
    
    % Plot point P
    \fill[red] (3,2) circle (2pt) node[above right] {P$(x,y)$};
    
    % Draw line from origin to point P
    \draw[blue, thick] (0,0) -- (3,2);
    
    % Draw vertical line from point P to point M on OX
    \draw[green, thick, dashed] (3,0) -- (3,2) node[midway, right] {$h$};
    
    % Add labels for points
    \node[below left] at (0,0) {O};
    \node[below] at (3,0) {B};
\end{tikzpicture}

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.

\begin{tikzpicture}
    % Draw the x-axis
    \draw[->] (0,0) -- (8,0) node[below] {$x$};
    
    % Draw the y-axis
    \draw[->] (0,0) -- (0,5) node[left] {$y$};
    
    % Plot point A
    \fill[red] (2,3) circle (2pt) node[above right] {A$(x_1, y_1)$};
    
    % Plot point B
    \fill[blue] (6,1) circle (2pt) node[above right] {B$(x_2, y_2)$};
    
    % Draw the distance line
    \draw (2,3) -- (6,1);
    
    % Draw the horizontal dotted line from A
    \draw[dashed] (2,3) -- (0,3) node[left] {$y_1$};
    
    % Draw the vertical dotted line from A
    \draw[dashed] (2,3) -- (2,0) node[below] {$x_1$};
    
    % Draw the horizontal dotted line from B
    \draw[dashed] (6,1) -- (0,1) node[left] {$y_2$};
    
    % Draw the vertical dotted line from B
    \draw[dashed] (6,1) -- (6,0) node[below] {$x_2$};
    
    % Add labels for points
    \node[below right] at (2,3) {A};
    \node[below right] at (6,1) {B};
    
    % Add label for the origin
    \node[below left] at (0,0) {O};
\end{tikzpicture}

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

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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