Learning Objectives
After studying this chapter you should be able to:
- explain why we need a coordinate system to describe a location;
- describe the 2-D Cartesian coordinate system and name the x-axis, y-axis and origin;
- state which directions on each axis are positive and which are negative;
- write the coordinates of a point as an ordered pair and read the x-coordinate and y-coordinate correctly;
- plot a point on the coordinate plane using a fixed step method;
- name the quadrant in which a point lies using the signs of its coordinates;
- recognise points that lie on the x-axis, on the y-axis, and at the origin;
- read the coordinates of a plotted point from a graph;
- find the horizontal distance, the vertical distance and the distance between two points;
- find the midpoint of a line segment;
- apply coordinates to simple real-life situations such as floor plans and maps.
1. Why Do We Need Coordinates?
Suppose your friend says, "The library is near the temple." That is a description, not a location — there may be several libraries. But if your friend says, "The library is at 4 units east and 3 units north of the temple", the place is fixed exactly. Giving two numbers, one horizontal and one vertical, fixes a position precisely. This is the whole idea behind coordinates.
A coordinate system is a fixed framework that uses numbers to describe exact locations of points or objects. The numbers work as distances measured from fixed reference lines, so anyone reading them can find the same spot.
On the number line you needed a single number, because everything lay on one straight line. That is one-dimensional (1-D). But a flat surface such as a floor, a map or a graph sheet has width as well as length. To describe a position on it you need two numbers — one for each direction. This is two-dimensional (2-D) space.
A short historical note
The habit of using perpendicular reference lines is old. In the Indian tradition, towns of the Sindhu-Sarasvati civilisation were laid out with streets running north-south and east-west at regular intervals, which is grid thinking of a practical kind. Baudhayana (about 800 CE) used such perpendicular directions for constructions and arrived at a result equivalent to the Pythagoras theorem. Later, Aryabhata (about 499 CE) used sine to compute positions of stars, and Brahmagupta (about 628 CE) gave zero and negative numbers a firm mathematical status — both are needed before a four-quadrant plane can exist at all. In 1637 CE, Rene Descartes formalised the system we now use: every point in a plane can be pinned down by just two numbers.
2. The 2-D Cartesian Coordinate System
Take two number lines and place them at right angles (perpendicular) to each other. For convenience one is drawn horizontal and the other vertical.
- The horizontal line is the x-axis.
- The vertical line is the y-axis.
- Together they are called the coordinate axes.
- The point where they meet is the origin, written \(O\), and its coordinates are \((0, 0)\).
- The plane in which the axes lie is the Cartesian plane, the coordinate plane, or the xy-plane.
Equal units are marked off from \(O\) on both axes, and the directions are fixed as follows.
| Direction of travel | Sign |
|---|---|
| To the right of \(O\) on the x-axis | positive \(x\) |
| To the left of \(O\) on the x-axis | negative \(x\) |
| Above \(O\) on the y-axis | positive \(y\) |
| Below \(O\) on the y-axis | negative \(y\) |
Figure 1: The two axes meet at the origin \(O(0, 0)\), and equal units are marked on both axes. The point \(P(3, 2)\) lies 3 units to the right of \(O\) and 2 units above it.
3. Coordinates of a Point
The position of a point \(P\) is written as an ordered pair
\(P(x, y)\)
where
- \(x\) is the x-coordinate: the perpendicular distance of \(P\) from the y-axis, measured along the x-axis;
- \(y\) is the y-coordinate: the perpendicular distance of \(P\) from the x-axis, measured along the y-axis.
The pair is ordered, so \((x, y)\) and \((y, x)\) are different points unless \(x = y\). For example, \((4, 1)\) lies 4 units right and 1 unit up; \((1, 4)\) lies 1 unit right and 4 units up. They are not the same point.
Points on the axes and the origin
- A point on the x-axis has the form \((x, 0)\) — its distance from the x-axis is zero.
- A point on the y-axis has the form \((0, y)\) — its distance from the y-axis is zero.
- The origin is \((0, 0)\).
4. Plotting a Point: A Fixed Method
Never plot "roughly". Use the same five steps every time — this removes almost all plotting errors.
Plot \(T(-4, 3)\).
Step 1: start at \(O\). Step 2: \(x = -4\) is negative, so move 4 units left along the x-axis. Step 3: \(y = 3\) is positive, so move 3 units up. Step 4: mark the dot. Step 5: label it \(T(-4, 3)\).
Final answer: \(T\) lies in Quadrant II, because \(x\) is negative and \(y\) is positive.
Figure 2: The axes cut the plane into four quadrants. The point \(M(-2, 4)\) lies in Quadrant II and \(N(5, -3)\) lies in Quadrant IV, but \(W(0, 3)\) is on the y-axis and therefore in no quadrant.
5. Quadrants and the Signs of Coordinates
The two axes cut the plane into four regions called quadrants. They are numbered anti-clockwise, starting from the region above the x-axis and to the right of the y-axis. Each quadrant has a fixed sign pattern, so the signs alone tell you the quadrant.
| Quadrant | Region | Sign of \(x\) | Sign of \(y\) | Pattern |
|---|---|---|---|---|
| I | top right | positive | positive | \((+, +)\) |
| II | top left | negative | positive | \((-, +)\) |
| III | bottom left | negative | negative | \((-, -)\) |
| IV | bottom right | positive | negative | \((+, -)\) |
A point that lies on the x-axis or on the y-axis is not in any quadrant, because one of its coordinates is zero — and zero is neither positive nor negative. The origin \((0,0)\) belongs to no quadrant either.
6. Reading Coordinates from a Graph
To find the coordinates of a plotted point, work back along the same lines you used to plot it.
From Figure 2, point \(M\) is 2 units to the left of the y-axis and 4 units above the x-axis, so its x-coordinate is \(-2\) and its y-coordinate is \(4\).
Final answer: \(M(-5, 4)\), which lies in Quadrant II.
7. Distance Between Two Points
Two easy cases come first. If the two points lie on a line parallel to the x-axis (same \(y\)), the distance is the horizontal gap. If they lie on a line parallel to the y-axis (same \(x\)), the distance is the vertical gap.
- Distance between \((x_1, y)\) and \((x_2, y)\) is \(\lvert x_2 - x_1 \rvert\).
- Distance between \((x, y_1)\) and \((x, y_2)\) is \(\lvert y_2 - y_1 \rvert\).
The absolute value bars matter: a distance is never negative.
When the segment is tilted
If the joining segment is not parallel to either axis, imagine completing it into a right-angled triangle. Drop a perpendicular from one point and draw a line parallel to an axis from the other point. The two legs of the triangle are exactly the horizontal and vertical gaps you already know how to find, and the segment itself is the hypotenuse. Now apply the Baudhayana-Pythagoras Theorem you studied in Class 8.
For points \(A(x_1, y_1)\) and \(D(x_2, y_2)\),
\(AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
It does not matter whether the differences are positive or negative, because each one is squared. The formula therefore works in every quadrant, including where the coordinates are negative.
Figure 3: The horizontal gap is 4 units and the vertical gap is 3 units, so the distance is \(\sqrt{4^2 + 3^2} = \sqrt{25} = 5\) units.
8. Midpoint of a Line Segment
The midpoint of a segment is the point that divides it into two equal halves. It lies exactly halfway in each direction, so its coordinates are the averages of the matching coordinates of the end points.
If \(M\) is the midpoint of \(A(x_1, y_1)\) and \(B(x_2, y_2)\), then
\(M\left(\dfrac{x_1 + x_2}{2},\; \dfrac{y_1 + y_2}{2}\right)\)
The smallest coordinate formula below is useful for checking, or for finding a missing end point:
\(x_M = \dfrac{x_1 + x_2}{2} \Rightarrow x_2 = 2x_M - x_1\), and similarly \(y_2 = 2y_M - y_1\).
A quick justification: because the midpoint is halfway, moving from \(A\) to \(M\) horizontally covers half the total horizontal span. Half of \((x_2 - x_1)\) added to \(x_1\) gives \(x_1 + \dfrac{x_2 - x_1}{2} = \dfrac{x_1 + x_2}{2}\), exactly as stated.
Reflections in the axes
Reflecting a point in an axis flips only one sign, and this is a very common examination step.
| Reflection | Image of \((x, y)\) |
|---|---|
| in the x-axis | \((x, -y)\) |
| in the y-axis | \((-x, y)\) |
| in the origin | \((-x, -y)\) |
Reflection is a rigid motion: it slides every point to a new place but does not stretch or shrink anything. So the image triangle has exactly the same side lengths as the original — the distance between two reflected points equals the distance between the original points.
9. Solved Examples
A. Reading coordinates and quadrants
Example 1
Question: For the points \(A(4, 7)\), \(B(-2, 6)\), \(C(-5, -3)\) and \(D(6, -1)\), state the quadrant (or axis) each point lies on.
Concept: The sign pattern of \((x, y)\) decides the region.
Solution:
- \(A\): \(x > 0\), \(y > 0\) → signs \((+, +)\) → Quadrant I
- \(B\): \(x < 0\), \(y > 0\) → signs \((-, +)\) → Quadrant II
- \(C\): \(x < 0\), \(y < 0\) → signs \((-, -)\) → Quadrant III
- \(D\): \(x > 0\), \(y < 0\) → signs \((+, -)\) → Quadrant IV
Final answer: A in I, B in II, C in III, D in IV.
Example 2
Question: The point \(P\) has coordinates \((-9, 4)\). State its x-coordinate, its y-coordinate, its distance from the y-axis and its distance from the x-axis.
Concept: x is measured from the y-axis; y is measured from the x-axis.
Solution: x-coordinate \(= -9\); y-coordinate \(= 4\). The distance from the y-axis is the size of the x-coordinate, \(9\) units. The distance from the x-axis is the size of the y-coordinate, \(4\) units.
Final answer: x-coordinate \(-9\), y-coordinate \(4\); distance from y-axis \(9\) units, from x-axis \(4\) units.
B. Points on the axes
Example 3
Question: Which axis does each point lie on, and in which quadrant (if any)? (a) \((0, 7)\) (b) \((-6, 0)\) (c) \((0, 0)\)
Concept: A zero coordinate means the point lies on an axis.
Solution:
- (a) \(x = 0\), so it lies on the y-axis, \(7\) units above \(O\). Not in a quadrant.
- (b) \(y = 0\), so it lies on the x-axis, \(6\) units to the left of \(O\). Not in a quadrant.
- (c) both are zero, so it is the origin. Not in a quadrant.
Final answer: (a) y-axis, (b) x-axis, (c) origin; none belongs to a quadrant.
C. Plotting points
Example 4
Question: Plot \((3, -5)\) and \((-2, -1)\) on the same coordinate plane.
Concept: horizontal first, then vertical.
Solution:
- \((3, -5)\): move 3 units right along the x-axis, then 5 units down. The point is in Quadrant IV.
- \((-2, -1)\): move 2 units left along the x-axis, then 1 unit down. The point is in Quadrant III.
Final answer: \((3, -5)\) in Quadrant IV and \((-2, -1)\) in Quadrant III.
D. Ordered pairs compared
Example 5
Question: Explain the difference between \((3, 5)\) and \((5, 3)\), and find the distance between the two points.
Concept: Order matters; the points lie on a horizontal line here.
Solution: \((3, 5)\) is 3 units right and 5 units up. \((5, 3)\) is 5 units right and 3 units up. They are different points because the y-coordinates differ. Both have \(y = 3\), so the segment joining them is parallel to the x-axis, and the distance is \(\lvert 5 - 3 \rvert = 2\) units.
Final answer: different points; distance \(= 2\) units.
Example 6
Question: Is \((-4, -4)\) the same point as \((-4, 4)\)? What is the distance between them?
Concept: a reflection in the x-axis; use the distance formula.
Solution: No. The points lie on a line parallel to the y-axis (same \(x = -4\)), so the distance is \(\lvert 4 - (-4) \rvert = 8\) units. Checking with the full formula: \(\sqrt{(0)^2 + (8)^2} = \sqrt{64} = 8\) units.
Final answer: not the same point; distance \(= 8\) units.
E. Distances
Example 7
Question: Find the distance between \(A(-1, 2)\) and \(B(3, 6)\).
Concept: build a right triangle and use the Baudhayana-Pythagoras Theorem.
Solution: Horizontal leg: \(\lvert 3 - (-1) \rvert = 4\). Vertical leg: \(\lvert 6 - 2 \rvert = 4\). Therefore \(AB = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.66\) units.
Final answer: \(AB = 4\sqrt{2} \approx 5.66\) units.
Example 8
Question: A room is 8 m long and 6 m wide. Taking one corner as the origin, with the length along the x-axis and the width along the y-axis, find the distance between two diagonally opposite corners.
Concept: real-life application of the distance formula.
Solution: The corners are \((0, 0)\), \((8, 0)\), \((8, 6)\) and \((0, 6)\). The diagonal joins \((0, 0)\) and \((8, 6)\), so the distance is \(\sqrt{(8-0)^2 + (6-0)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\) m.
Final answer: diagonal \(= 10\) m.
Example 9
Question: \(A(3, 4)\) is reflected in the x-axis to \(A'\), in the y-axis to \(A''\) and in the origin to \(A'''\). Find the distances \(AA'\), \(AA''\) and \(A'A''\).
Concept: reflections change one sign; verify with the distance formula.
Solution: \(A' = (3, -4)\), \(A'' = (-3, 4)\), \(A''' = (-3, -4)\).
- \(AA'\): same \(x\), so distance \(= \lvert -4 - 4 \rvert = 8\) units
- \(AA''\): same \(y\), so distance \(= \lvert -3 - 3 \rvert = 6\) units
- \(A'A''\): horizontal gap \(6\), vertical gap \(8\), so distance \(= \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\) units
Final answer: 8 units, 6 units and 10 units. Since \(6^2 + 8^2 = 10^2\), the angle at \(A\) is a right angle.
F. Midpoint
Example 10
Question: Find the midpoint of the segment joining \(P(3, 7)\) and \(Q(11, -1)\), and verify it.
Concept: average the matching coordinates, then check with distances.
Solution: \(x_M = \dfrac{3 + 11}{2} = \dfrac{14}{2} = 7\), \(y_M = \dfrac{7 + (-1)}{2} = \dfrac{6}{2} = 3\). So \(M(7, 3)\).
Verification: \(PM = \sqrt{(7-3)^2 + (3-7)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\), and \(MQ = \sqrt{(11-7)^2 + (-1-3)^2} = \sqrt{16 + 16} = 4\sqrt{2}\). The two halves are equal.
Final answer: midpoint \(M(7, 3)\).
Example 11
Question: \(M(-2, 5)\) is the midpoint of \(A(3, y)\) and \(B(x, 7)\). Find \(x\) and \(y\).
Concept: the midpoint is the average of the end points, so solve for the missing values.
Solution: The midpoint averages the matching coordinates. Using the x-coordinates: \(-2 = \dfrac{3 + x}{2} \Rightarrow -4 = 3 + x \Rightarrow x = -7\). Using the y-coordinates: \(5 = \dfrac{y + 7}{2} \Rightarrow 10 = y + 7 \Rightarrow y = 3\). So \(B(-7, 7)\).
Check: average of \(x\)-values \(= \dfrac{3 + (-7)}{2} = \dfrac{-4}{2} = -2\) and of \(y\)-values \(= \dfrac{3 + 7}{2} = \dfrac{10}{2} = 5\). Correct.
Final answer: \(x = -7\), \(y = 3\), so \(B(-7, 7)\).
G. Reasoning and application
Example 12
Question: A triangle has vertices \(A(1, 1)\), \(B(5, 1)\) and \(C(1, 4)\). Name the type of triangle, and find its area.
Concept: compare coordinate differences to detect right angles and equal sides.
Solution: \(AB\) is horizontal with length \(\lvert 5 - 1 \rvert = 4\); \(AC\) is vertical with length \(\lvert 4 - 1 \rvert = 3\). So \(AB \perp AC\) and the triangle is right-angled at \(A\). Also \(BC = \sqrt{4^2 + 3^2} = 5\), so it is a 3-4-5 right triangle. Area of a right triangle \(= \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 4 \times 3 = 6\) square units.
Final answer: right-angled at \(A\), area \(6\) square units.
Example 13
Question: On a floor plan with axes drawn along the walls, the class door is at \((-3, 2)\) and the blackboard centre is at \((4, 2)\), each in metres. How far apart are they, and what kind of path joins them?
Concept: same y-coordinate means the path is parallel to the x-axis.
Solution: Both have \(y = 2\), so the joining segment is parallel to the x-axis and its length is \(\lvert 4 - (-3) \rvert = 7\) m. One must walk 3 m left of the origin, then 7 m along the wall direction, so the path is straight and level.
Final answer: 7 m apart, joined by a line parallel to the x-axis.
10. Common Mistakes
What went wrong: writing \((5, 3)\) when the point is \((3, 5)\).
Correction: x first, y second, always. Say the pair aloud as "x comma y" before writing it.
What went wrong: treating the x-coordinate as the distance from the x-axis.
Correction: the x-coordinate is measured along the x-axis, so it is the distance from the y-axis. Remember: x is measured from the y-axis.
What went wrong: going up 3 first and then right 4, which lands on a wrong point.
Correction: horizontal first (x), then vertical (y). Trace the path in your head before marking.
What went wrong: calling \((-5, 4)\) Quadrant III.
Correction: read the signs in order: \(x\) negative, \(y\) positive is \((-, +)\), which is Quadrant II. The quadrants are numbered anti-clockwise.
What went wrong: saying \((0, -4)\) lies in Quadrant IV.
Correction: a point on an axis belongs to no quadrant, because one coordinate is zero.
What went wrong: starting the count of units from the first tick mark instead of from \(O\).
Correction: zero sits at \(O\). Count units from the origin, not from the "1" mark.
What went wrong: writing \(x_2 - x_1 = -9\) as a distance.
Correction: a distance is never negative, so use \(\lvert x_2 - x_1 \rvert\), or simply note that squaring removes the sign in the full formula.
11. Important Results at a Glance
Axes and origin
The x-axis is horizontal, the y-axis is vertical, and they meet at the origin \(O(0, 0)\) at right angles.
Signs
Right and up are positive; left and down are negative.
Coordinates
\(x\) is the distance from the y-axis; \(y\) is the distance from the x-axis.
Order matters
\((x, y) = (y, x)\) only when \(x = y\).
Axes points
x-axis: \((x, 0)\). y-axis: \((0, y)\). Origin: \((0, 0)\).
Quadrants
I \((+,+)\), II \((-,+)\), III \((-,-)\), IV \((+, -)\). Axis points are in no quadrant.
Straight-line cases
Same \(y\): distance \(\lvert x_2 - x_1 \rvert\). Same \(x\): distance \(\lvert y_2 - y_1 \rvert\).
Reflections
In the x-axis \((x, -y)\); in the y-axis \((-x, y)\); in the origin \((-x, -y)\). Distances are unchanged.
12. Formulae and Quick Reference
| What you need | Formula or rule |
|---|---|
| Origin | \(O = (0, 0)\) |
| Point on the x-axis | \((x, 0)\) |
| Point on the y-axis | \((0, y)\) |
| Horizontal distance | \(\lvert x_2 - x_1 \rvert\) |
| Vertical distance | \(\lvert y_2 - y_1 \rvert\) |
| Distance between two points | \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) |
| Midpoint of \(AB\) | \(\left(\dfrac{x_1 + x_2}{2},\; \dfrac{y_1 + y_2}{2}\right)\) |
13. Think and Understand
Q1.
What happens to a point if you interchange its coordinates? When are the two points the same?
Q2.
Why can a point on an axis not be placed in any quadrant? Use the idea of positive and negative numbers.
Q3.
How can you decide the quadrant of a point using only its signs, without drawing it?
Q4.
Why is the origin a special point? What would change if we chose a different origin?
Q5.
The distance between two points is unchanged if both points are reflected in the x-axis. Explain why.
Q6.
If \(M\) is the midpoint of \(AB\), how would you find the coordinates of \(B\) when \(A\) and \(M\) are known?
14. Exam Tips
- Always write the pair in the order \((x, y)\); say "x comma y" to yourself.
- Check signs before anything else — the sign decides the quadrant.
- Label both axes and mark a consistent scale; a wrong scale distorts every later step.
- Plot horizontally first, then vertically, and dot the point clearly.
- Join the plotted points with a ruler and extend the line with an arrowhead.
- For distances, show the horizontal leg, the vertical leg and then the Pythagoras step. Marks are given for working.
- Write the unit with every distance, for example 5 units or 10 m.
- Use the absolute value, or square the differences, so the answer is never negative.
- Name the quadrant in words as well as by number, if the question asks for it.
15. Quick Revision
- Coordinate system: two number lines at right angles; used to fix positions in 2-D space.
- x-axis horizontal, y-axis vertical, meeting at the origin \(O(0,0)\).
- Positive right and up; negative left and down.
- Coordinates \((x, y)\): \(x\) from the y-axis, \(y\) from the x-axis. Order matters.
- Axes: x-axis \((x, 0)\); y-axis \((0, y)\); origin \((0, 0)\).
- Quadrants: I \((+,+)\), II \((-,+)\), III \((-,-)\), IV \((+, -)\); axis points are in none.
- Plotting: origin → horizontal to \(x\) → vertical to \(y\) → mark → label.
- Reading: vertical down to the x-axis for \(x\); horizontal across to the y-axis for \(y\).
- Distance: \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\), from the Baudhayana-Pythagoras Theorem.
- Midpoint: averages of the matching coordinates.
- Reflection preserves all lengths.
16. Chapter Checklist
- I can explain why coordinates are needed to describe a position.
- I understand the difference between the number line and the 2-D Cartesian plane.
- I can identify the x-axis and the y-axis.
- I can identify the origin and state its coordinates as (0, 0).
- I know which directions on each axis are positive and which are negative.
- I can read and write ordered pairs in the form (x, y).
- I can give the x-coordinate and the y-coordinate of a point.
- I can plot a point using the five-step method.
- I can identify the quadrant of a point from the signs.
- I know the sign pattern in all four quadrants.
- I can identify points on the x-axis and on the y-axis.
- I understand that points on the axes are not in any quadrant.
- I can read coordinates of a plotted point from a graph.
- I can find the distance between two points using the distance formula.
- I can find the midpoint of a line segment.
- I can apply coordinates to simple real-life situations.