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Orienting Yourself: The Use of Coordinates

Class 9 . Mathematics . Ganita Manjari . Orienting Yourself the Use of Coordinates

Chapter 1: Orienting Yourself: The Use of Coordinates

Ch 1

Mathematics

Class 9

Lecture 1

Introduction

Reference video · Lecture 1Introduction

Coordinate Systems

Representing a point

In a two-dimensional plane, the coordinates \((x,y)\) specify a point's positions relative to two perpendicular axes. The origin is \((0,0)\); negative coordinate values lie in the negative direction of the corresponding axis.

Coordinates of a point
$$(x,y)$$

Write the horizontal coordinate first and the vertical coordinate second.

Reading an ordered pair

  1. Start at the origin.
    $$(0,0)$$
  2. Use the first coordinate to move horizontally.
    $$x$$
  3. Use the second coordinate to move vertically; the endpoint is the point.
    $$y$$
Worked example

How is the point \((3,2)\) located?

  1. Locate the point represented by \((3,2)\), starting from the origin.
    $$(0,0)$$
  2. Move \(3\) units horizontally, then \(2\) units vertically.
    $$(0,0)\to(3,0)\to(3,2)$$
  3. The endpoint is the required point.
    $$(3,2)$$
Answer: Move \(3\) units horizontally and then \(2\) units vertically from the origin.
⚠ Common mistake

Do not reverse the coordinates: \((x,y)\) means horizontal position first, vertical position second.

Coordinates, Algebra, and Geometry

Why coordinates are useful

  • Precise location: Coordinates allow points and objects to be identified exactly.
  • Connection: Coordinates help represent algebraic equations as geometric shapes, and describe geometric shapes using algebra and equations.
Practice

Starting at the origin, what point is reached by moving \(4\) units horizontally and then \(1\) unit vertically?

Hint: Write the horizontal movement first.
Key step: Use the order \((x,y)\).
Solution: The point is \((4,1)\).
Practice

For the point \((-2,3)\), which coordinate gives the vertical position?

Hint: In an ordered pair, the vertical coordinate comes second.
Key step: Identify the second entry.
Solution: The vertical coordinate is \(3\).

Quick recap

  • A coordinate system uses numbers and reference axes to locate points precisely.
  • The Cartesian plane has two perpendicular axes that meet at the origin \((0,0)\).
  • An ordered pair \((x,y)\) gives the horizontal position first and vertical position second.
  • Negative values indicate positions in the negative direction of the corresponding axis.
  • Coordinates connect algebraic equations with geometric shapes.
Lecture 2

Settling In

Reference video · Lecture 2Settling In

Representing a Room on a Grid

Room floor plan showing the bathroom, bedroom, doors and dimensions.
Room floor plan showing the bathroom, bedroom, doors and dimensions.
Scale conversion
$$\frac{d_{\mathrm{map}}}{d_{\mathrm{room}}}=\frac{1\,\mathrm{cm}}{1\,\mathrm{ft}}$$

Use the map scale to convert a measured map distance into the corresponding room distance.

Using the map scale

  1. Measure the distance between the points on the map.
  2. Use the scale: each map centimetre represents one room foot.
    $$1\,\mathrm{cm}=1\,\mathrm{ft}$$
  3. State the corresponding distance in the room.
Worked example

What room distance does \(3\,\mathrm{cm}\) on the map represent?

  1. A distance measures \(3\,\mathrm{cm}\) on the map. Apply the scale.
    $$3\,\mathrm{cm}\times\frac{1\,\mathrm{ft}}{1\,\mathrm{cm}}$$
  2. Cancel the map-length units.
    $$3\,\mathrm{ft}$$
Answer: \(3\,\mathrm{ft}\)
⚠ Common mistake

Do not assume a floor map shows every part of a room: it represents the floor layout, not vertical features such as windows.

Practice

On the room map, a distance is \(6\,\mathrm{cm}\). What is the corresponding room distance?

Hint: Use \(1\,\mathrm{cm}=1\,\mathrm{ft}\).
Key step: \(6\,\mathrm{cm}\times\frac{1\,\mathrm{ft}}{1\,\mathrm{cm}}=6\,\mathrm{ft}\)
Solution: \(6\,\mathrm{ft}\)
Practice

Why can the room’s windows not be marked on this floor map?

Hint: Consider which part of the room the map represents.
Key step: The map represents the floor plane; windows are on vertical walls.
Solution: The floor map shows the floor layout, not positions on vertical walls, where the windows are.

Quick recap

  • A grid can represent the layout of a room’s floor.
  • Pins mark important points, and wool can connect them to make the layout tactile.
  • The map scale is \(1\,\mathrm{cm}\) to \(1\,\mathrm{ft}\).
  • A floor map shows positions on the floor, not vertical features such as windows.
Lecture 3

The 2-D Cartesian Coordinate System

Reference video · Lecture 3The 2-D Cartesian Coordinate System
Definition

2-D Cartesian coordinate system — A system of two perpendicular number lines used to locate points in a plane.

Axes and the origin

  • \(x\)-axis: The horizontal number line; values to the right are positive and to the left are negative.
  • \(y\)-axis: The vertical number line; values above are positive and below are negative.
  • Origin: The axes intersect at \(O\), whose coordinates are \((0,0)\).
  • Scale: Mark equal units on both axes to measure distances from the origin.
Perpendicular x- and y-axes intersect at the origin O.
Perpendicular \(x\)- and \(y\)-axes intersect at the origin \(O\).

Coordinates and positions

Definition

Coordinates — The ordered pair \((x,y)\) gives a point’s horizontal position first and its vertical position second.

Coordinate notation
$$(x,y)$$

Write the horizontal coordinate before the vertical coordinate.

Points on the axes
PositionCoordinate formExample
On the \(x\)-axis\((x,0)\)\(B=(4.5,0)\) is 4.5 units right of \(O\).
On the \(y\)-axis\((0,y)\)\(G=(0,-4.5)\) is 4.5 units below \(O\); \(H=(0,4)\) is 4 units above \(O\).
  • On the \(x\)-axis: The second coordinate is \(0\); a positive first coordinate is right of \(O\), and a negative one is left.
  • On the \(y\)-axis: The first coordinate is \(0\); a positive second coordinate is above \(O\), and a negative one is below.
  • Writing points: \(P=(x,y)\) is often shortened to \(P(x,y)\) when labelling a graph.

Locate a point from its coordinates

  1. Start at the origin \(O\).
    $$O=(0,0)$$
  2. Use the first coordinate to move horizontally: right if positive, left if negative.
    $$x>0\Rightarrow\text{right},\quad x<0\Rightarrow\text{left}$$
  3. Then use the second coordinate to move vertically: up if positive, down if negative.
    $$y>0\Rightarrow\text{up},\quad y<0\Rightarrow\text{down}$$
Worked example

Where is the point \(P=(-3,2)\)?

  1. Locate \(P=(-3,2)\) from the origin.
    $$P=(-3,2)$$
  2. The first coordinate is negative, so move 3 units left.
    $$x=-3$$
  3. The second coordinate is positive, so move 2 units up.
    $$y=2$$
Answer: \(P\) is 3 units left and 2 units above \(O\).
⚠ Common mistake

Do not reverse the coordinate order: \((x,y)\) means horizontal first, vertical second. On an axis, the coordinate perpendicular to it is \(0\).

Practice

Write the coordinates of a point 6 units left of \(O\) on the \(x\)-axis.

Hint: On the \(x\)-axis, the second coordinate is \(0\).
Key step: Left of \(O\) gives a negative first coordinate.
Solution: \((-6,0)\)
Practice

Write the coordinates of a point 2.5 units below \(O\) on the \(y\)-axis.

Hint: On the \(y\)-axis, the first coordinate is \(0\).
Key step: Below \(O\) gives a negative second coordinate.
Solution: \((0,-2.5)\)
Practice

A point is 4 units right and 3 units below \(O\). Write its coordinates.

Hint: The horizontal coordinate comes first.
Key step: Right is positive and below is negative.
Solution: \((4,-3)\)

Quick recap

  • The \(x\)-axis is horizontal and the \(y\)-axis is vertical; they meet at \(O=(0,0)\).
  • Coordinates are ordered \((x,y)\): horizontal position first, vertical position second.
  • Right and up are positive; left and down are negative.
  • Points on the \(x\)-axis have the form \((x,0)\); points on the \(y\)-axis have the form \((0,y)\).
  • A point label may be written as \(P(x,y)\) instead of \(P=(x,y)\).
Lecture 4

Distance Between Two Points in the 2-D Plane

Reference video · Lecture 4Distance Between Two Points in the 2-D Plane

Distance Between Two Points

A coordinate grid showing points A(3,4), D(7,1), and M(9,6).
A coordinate grid showing points \(A(3,4)\), \(D(7,1)\), and \(M(9,6)\).
Distance formula

For points \(P(x_1,y_1)\) and \(Q(x_2,y_2)\), the Pythagorean theorem gives the distance from their horizontal and vertical coordinate changes.

$$\displaystyle PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
Distance between two points
$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Use for any two points in the coordinate plane.

Finding the distance

  1. Identify the coordinates of the two points.
  2. Find the horizontal and vertical coordinate differences.
    $$\Delta x=x_2-x_1,\qquad \Delta y=y_2-y_1$$
  3. Square both differences, add them, and take the square root.
    $$d=\sqrt{(\Delta x)^2+(\Delta y)^2}$$
Worked example

Find the lengths of \(AD\), \(DM\), and \(MA\) for \(A(3,4)\), \(D(7,1)\), and \(M(9,6)\).

  1. Find the lengths of all three sides of triangle \(ADM\), where \(A(3,4)\), \(D(7,1)\), and \(M(9,6)\).
  2. For \(AD\), the coordinate changes are \(4\) and \(-3\).
    $$AD=\sqrt{(7-3)^2+(1-4)^2}=\sqrt{4^2+(-3)^2}=5$$
  3. For \(DM\), the coordinate changes are \(2\) and \(5\).
    $$DM=\sqrt{(9-7)^2+(6-1)^2}=\sqrt{2^2+5^2}=\sqrt{29}$$
  4. For \(MA\), the coordinate changes are \(-6\) and \(-2\).
    $$MA=\sqrt{(3-9)^2+(4-6)^2}=\sqrt{(-6)^2+(-2)^2}=\sqrt{40}=2\sqrt{10}$$
Answer: \(AD=5\) units, \(DM=\sqrt{29}\) units, and \(MA=\sqrt{40}=2\sqrt{10}\) units.

Negative Coordinates and Reflection

A triangle and its reflected image across the y-axis.
A triangle and its reflected image across the \(y\)-axis.
⚠ Common mistake

Do not treat coordinate differences as distances or ignore their signs before calculating. Square both differences, add them, and take the square root.

Memory hook Prepzy tip

Square, add, root

Square the coordinate changes, add the squares, then take the square root.

Practice

Find the distance between \(P(-3,4)\) and \(Q(1,1)\).

Hint: Use the changes in both coordinates.
Key step: \(\Delta x=4\) and \(\Delta y=-3\).
Solution: \(PQ=\sqrt{4^2+(-3)^2}=5\) units.
Practice

Find the distance between \(A(2,-1)\) and \(B(2,5)\).

Hint: The points have the same \(x\)-coordinate.
Key step: The horizontal change is \(0\) and the vertical change is \(6\).
Solution: \(AB=\sqrt{0^2+6^2}=6\) units.
Practice

Find the distance between \(M(9,6)\) and \(D(7,1)\).

Hint: Find both coordinate differences.
Key step: \(\Delta x=2\) and \(\Delta y=5\).
Solution: \(MD=\sqrt{2^2+5^2}=\sqrt{29}\) units.

Quick recap

  • The distance between two points follows from the Pythagorean theorem.
  • For \(P(x_1,y_1)\) and \(Q(x_2,y_2)\), use \(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
  • Coordinate differences may be negative; squaring removes their effect on the distance.
  • The formula works for points with negative coordinates and in every quadrant.
  • Reflection across an axis changes some coordinate signs but preserves distances.
  • For \(A(3,4)\), \(D(7,1)\), and \(M(9,6)\), the side lengths are \(AD=5\), \(DM=\sqrt{29}\), and \(MA=\sqrt{40}\).
Lecture 5

Summary

Reference video · Lecture 5Summary

The Cartesian Plane and Coordinates

Definition

Coordinates — A point’s \(x\)-coordinate gives its horizontal distance from the \(y\)-axis, and its \(y\)-coordinate gives its vertical distance from the \(x\)-axis; the signs indicate the side of each axis.

Coordinates of points on the axes
$$x\text{-axis}: (x,0),\qquad y\text{-axis}: (0,y),\qquad \text{origin}: (0,0)$$

Use these forms to recognise points lying on either axis.

Identify a point's quadrant

  1. Check the signs of the point's coordinates.
  2. Match the sign pair to the quadrant.
Worked example

Locate \(P=(-3,2)\) and state its distances from the axes.

  1. For the point, the first coordinate is negative and the second is positive.
    $$P=(-3,2)$$
  2. The sign pair is \((-,+)\), so the point lies in the second quadrant.
    $$(-,+)\longrightarrow \text{Quadrant II}$$
  3. Its horizontal and vertical distances from the axes are \(3\) and \(2\) units.
    $$|-3|=3,\qquad |2|=2$$
Answer: \(P\) lies in Quadrant II; it is \(3\) units from the \(y\)-axis and \(2\) units from the \(x\)-axis.
⚠ Common mistake

Do not swap the coordinates: \((x,y)\) gives the horizontal position first and the vertical position second.

Distance Between Two Points

Distance formulas

For points on the same horizontal or vertical line, use the absolute difference. For any two points, combine the horizontal and vertical differences using the Pythagorean theorem.

$$d_{\text{horizontal}}=|x_2-x_1|,\qquad d_{\text{vertical}}=|y_2-y_1|,\qquad d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Find the distance between two points

  1. Identify the coordinates \((x_1,y_1)\) and \((x_2,y_2)\).
  2. Substitute into the distance formula.
    $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
  3. Simplify and give the non-negative distance.
Worked example

Find the distance between \(A=(1,2)\) and \(B=(4,6)\).

  1. Substitute the coordinates of the two points.
    $$A=(1,2),\qquad B=(4,6)$$
  2. Apply the distance formula.
    $$AB=\sqrt{(4-1)^2+(6-2)^2}$$
  3. Evaluate the squares and simplify.
    $$AB=\sqrt{3^2+4^2}=\sqrt{25}=5$$
Answer: \(AB=5\) units.
Practice

Find the distance between \((-2,5)\) and \((4,5)\).

Hint: The points have the same \(y\)-coordinate.
Key step: Use the horizontal distance formula: \(|x_2-x_1|\).
Solution: \(|4-(-2)|=6\) units.
Practice

Find the distance between \((3,-1)\) and \((3,7)\).

Hint: The points have the same \(x\)-coordinate.
Key step: Use the vertical distance formula: \(|y_2-y_1|\).
Solution: \(|7-(-1)|=8\) units.

Quick recap

  • The Cartesian plane, coordinate plane, or \(xy\)-plane has perpendicular \(x\)- and \(y\)-axes meeting at \((0,0)\).
  • A point’s \(x\)-coordinate measures its distance from the \(y\)-axis; its \(y\)-coordinate measures its distance from the \(x\)-axis.
  • Coordinates are ordered as \((x,y)\); the \(x\)-coordinate comes first.
  • The signs of the coordinates identify a point's quadrant.
  • Points on the \(x\)-axis have form \((x,0)\); points on the \(y\)-axis have form \((0,y)\).
  • Horizontal and vertical distances are absolute coordinate differences; the general distance formula follows from the Pythagorean theorem.

Formula sheet

Coordinates of a point
$$(x,y)$$
Scale conversion
$$\frac{d_{\mathrm{map}}}{d_{\mathrm{room}}}=\frac{1\,\mathrm{cm}}{1\,\mathrm{ft}}$$
Coordinate notation
$$(x,y)$$
Distance between two points
$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
Coordinates of points on the axes
$$x\text{-axis}: (x,0),\qquad y\text{-axis}: (0,y),\qquad \text{origin}: (0,0)$$

Mathematics — ALL CHAPTERS

1

Orienting Yourself: The Use of Coordinates

2

Introduction to Linear Polynomials

3

The World of Numbers

4

Exploring Algebraic Identities

5

I’m Up and Down, and Round and Round

6

Measuring Space: Perimeter and Area

7

The Mathematics of Maybe: Introduction to Probability

8

Predicting What Comes Next: Exploring Sequences and Progression