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Coordinate Geometry

A visual-first chapter page covering the coordinate plane, quadrants, distance formula, midpoint, section formula, area of a triangle, and coordinate-based shape reasoning.

Chapter Roadmap

01The Coordinate Plane 02Quadrants and Sign Logic 03Distance Between Two Points 04Midpoint and Section Formula 05Area of a Triangle 06Shapes Through Coordinates 07Applications and Visual Thinking

The Coordinate Plane

Every point is an address made from two numbers in a fixed order

Address Rule

Read $(4,3)$ as a route: move 4 units parallel to the x-axis, then move 3 units parallel to the y-axis. Because order matters, $(4,3)$ and $(3,4)$ are different points.

The coordinate plane is formed by two perpendicular number lines: the x-axis and the y-axis.

Their point of intersection is the origin

O(0,0)

. A point written as

(x,y)

tells us how far to move horizontally first and vertically after that.

P(x,y)

The first number is the x-coordinate and the second number is the y-coordinate.

\text{Abscissa}=x,\quad \text{Ordinate}=y

Axes and Origin

The x-axis and y-axis meet at the origin, which is the reference point for all plotting.

Plotting P(4,3)

Move along x first, then move parallel to y. That fixed order makes coordinates meaningful.

Worked Example

Read a coordinate as a location

What does the point

P(4,3)

mean on the coordinate plane?

Show solution

Start at the origin.

Move 4 units to the right along the x-direction.

From there, move 3 units upward parallel to the y-axis.

The point reached is $P(4,3)$ .

Practice

Identify the abscissa and ordinate of the points

(5,-2)

(-4,7)

, and

(0,6)

Quadrants and Sign Logic

The signs of x and y tell where a point lies

Axis Trap

If a point lies on an axis, then it does not belong to any quadrant. For example, $(0,-2)$ lies on the y-axis and $(5,0)$ lies on the x-axis.

The two axes divide the plane into four regions called quadrants.

The sign of the x-coordinate tells left or right; the sign of the y-coordinate tells above or below the x-axis.

\text{Quadrant I }=(+,+)

\text{Quadrant II }=(-,+)

\text{Quadrant III }=(-,-)

\text{Quadrant IV }=(+,-)

Quadrants

The sign pattern becomes easy once you read it as right/left and above/below.

Points in Different Regions

A point on an axis is not in any quadrant, which is a common exam trap.

Worked Example

Locate a point by signs

In which quadrant does the point

(-5,4)

lie?

Show solution

The x-coordinate is negative, so the point lies on the left side of the y-axis.

The y-coordinate is positive, so the point lies above the x-axis.

Left and above together means Quadrant II.

Practice

State the quadrant or axis for

(3,-7)

(-2,-6)

(0,5)

, and

(-9,0)

Distance Between Two Points

The distance formula is Pythagoras on graph paper

Take two points

P(x_1,y_1)

and

Q(x_2,y_2)

. The horizontal change is

x_2-x_1

and the vertical change is

y_2-y_1

These form the two perpendicular sides of a right triangle, so the direct distance comes from the Pythagoras theorem.

PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

OP=\sqrt{x^2+y^2}

Special case: distance of

(x,y)

from the origin.

Worked Example

Use the distance formula

Find the distance between

(2,3)

and

(8,11)

Show solution

Here $x_1=2$ , $y_1=3$ , $x_2=8$ , and $y_2=11$ .

So $\Delta x=8-2=6$ and $\Delta y=11-3=8$ .

PQ=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10

Hence the distance is 10 units.

Practice

Find the distance between

(-6,7)

and

(-1,-5)

. Also find the distance of the point

(5,-12)

from the origin.

Midpoint and Section Formula

Midpoint means equal sharing; section formula means weighted sharing

The midpoint of a segment is the point that divides it into two equal parts, so both coordinates are averaged.

If a point divides the line internally in the ratio

m:n

, then its coordinates are weighted averages of the endpoints.

M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

P\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)

Section formula for internal division in the ratio

m:n

Worked Example

Find a midpoint

Find the midpoint of

(2,-1)

and

(8,5)

Show solution

Use the midpoint formula:

M=\left(\frac{2+8}{2},\frac{-1+5}{2}\right)=\left(\frac{10}{2},\frac{4}{2}\right)=(5,2)

So the midpoint is $(5,2)$ .

Practice

A point divides the segment joining

(1,2)

and

(7,8)

in the ratio

1:2

. Find it. Also find the midpoint of

(-4,6)

and

(10,-2)

Area of a Triangle

Coordinates can give the area directly even when the triangle is tilted

When a triangle is drawn on the coordinate plane, the area can be found directly from the coordinates of its vertices.

This formula is also extremely useful for testing whether three points are collinear, because collinear points give zero area.

\text{Area}=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

\text{Area}=0 \Rightarrow \text{points are collinear}

Worked Example

Find the area of a triangle

Find the area of the triangle with vertices

(1,1)

(5,1)

and

(3,4)

Show solution

Use the coordinate area formula:

\text{Area}=\frac12\left|1(1-4)+5(4-1)+3(1-1)\right|

=\frac12\left|-3+15+0\right|=\frac12(12)=6

Hence the area is 6 square units.

Practice

Find whether the points

(1,-1)

(5,2)

and

(9,5)

are collinear using the area idea. Also find the area of the triangle formed by

(2,1)

(6,1)

and

(4,5)

Shapes Through Coordinates

Distances, midpoints, and slopes in disguise help us prove properties of figures

Useful Checks

Rectangle: opposite sides equal and diagonals equal
Square: all sides equal and diagonals equal
Rhombus: all sides equal
Parallelogram: diagonals bisect each other
Right triangle: Pythagoras relation between side lengths

Coordinate geometry lets us prove whether points form a rectangle, a square, a rhombus, a parallelogram, or a right triangle using distances and midpoint relations.

For collinearity, either area becomes zero or one distance becomes the sum of the other two.

Worked Example

Use distances to prove collinearity

Show that

(1,-1)

(5,2)

and

(9,5)

are collinear using distances.

Show solution

AB=\sqrt{(5-1)^2+(2+1)^2}=\sqrt{16+9}=5

BC=\sqrt{(9-5)^2+(5-2)^2}=\sqrt{16+9}=5

AC=\sqrt{(9-1)^2+(5+1)^2}=\sqrt{64+36}=10

Since $AB+BC=AC$ , the three points are collinear.

Practice

The points

(3,2)

(-2,-3)

and

(2,3)

form which type of triangle? Use distances to decide.

Applications and Visual Thinking

Coordinates are really about position, movement, sharing, and measurement

Real-Life Models

Maps: locations are plotted as points
Games and UI: every icon and character position is coordinate-based
Cricket and sports analysis: player positions can be tracked with coordinates
Navigation: midpoint helps place a stop or marker between two locations

Coordinate geometry appears naturally in maps, screens, design layouts, seat plans, cricket fields, and delivery routes.

If you see a point as an address, the midpoint as a balanced location, and the distance formula as a straight-path tool, the whole chapter becomes intuitive.

Worked Example

Map-style midpoint problem

A cafe is at

(2,4)

and a bus stop is at

(10,10)

. Find the midpoint where a signboard should be placed.

Show solution

Use the midpoint formula:

M=\left(\frac{2+10}{2},\frac{4+10}{2}\right)=\left(\frac{12}{2},\frac{14}{2}\right)=(6,7)

So the signboard should be placed at $(6,7)$ .

Practice

1. A camera is at the origin and a drone is at

(8,15)

. Find the direct distance from the camera.
2. Two fielders are at

(6,4)

and

(-6,4)

. On which axis does their midpoint lie?
3. Find the area of the triangle with vertices

(1,1)

(5,1)

and

(3,4)

Quick Summary

Concept	Key Idea
Ordered pair	A point is written as $(x,y)$ and order matters.
Quadrants	Signs of $x$ and $y$ decide the quadrant.
Distance	$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Midpoint	$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
Section formula	Weighted average of endpoint coordinates.
Area of triangle	Coordinate formula works for any orientation.
Collinearity	Area equal to zero means the points are collinear.

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