1. Introduction to Coordinate Geometry
Coordinate geometry answers a simple question: where is this point, and how far apart are two points? Instead of describing positions with words ("roughly in the middle, a bit to the left"), it pins every location to a pair of numbers measured from fixed reference lines. Geometry done this way becomes arithmetic, and arithmetic can be done quickly, exactly and automatically.
The same tool runs the world around you: the grid on a city map that gives every street a number and letter, the floor plan of a house drawn on a grid, the search-by-coordinates of a navigation app, the square-by-square layout of a spreadsheet, and the position of a ship or an aircraft reported as a latitude and longitude. In every case, two numbers locate one point, and nothing about the shape is lost in the process.
Cartesian plane
A flat grid made by two perpendicular number lines that cross at the origin.
Ordered pair \((x,y)\)
Two numbers that name one point; the order matters.
Distance
The straight-line gap between two points, found with the distance formula.
Midpoint
The point exactly halfway between two points, found by averaging coordinates.
What you will learn in this chapter:
- Describe the Cartesian plane, the axes, the origin and ordered pairs.
- Name the four quadrants and recall the sign pattern of each.
- Plot a point from its coordinates and read coordinates back from a graph.
- Apply the distance formula, including points with negative coordinates and horizontal or vertical segments.
- Find the midpoint of a segment and recover an unknown endpoint when the midpoint is known.
- Recognise vertical and horizontal lines from the coordinates of their points.
- Use distance and midpoint calculations to check simple geometric shapes like right triangles and rectangles.
2. Cartesian Coordinate System
The Cartesian coordinate system places every point of the plane on a grid made of two number lines that cross at right angles at their zero marks. The horizontal number line is the \(x\)-axis; the vertical number line is the \(y\)-axis; and their crossing point is the origin \(O(0,0)\). Positive numbers lie to the right on the \(x\)-axis and upwards on the \(y\)-axis; negative numbers lie to the left and downwards.
A point is named by an ordered pair written as \(P(x,y)\). The first number \(x\) is the x-coordinate (also called the abscissa) and tells how far the point is from the \(y\)-axis, moving right for positive values and left for negative values. The second number \(y\) is the y-coordinate (also called the ordinate) and tells how far the point is from the \(x\)-axis, moving up for positive values and down for negative values.
\(P(x,y)\) locates a point whose x-coordinate (abscissa) is \(x\) and whose y-coordinate (ordinate) is \(y\). The pair is ordered: \((x,y)\) and \((y,x)\) are generally different points. For example \(P(2,3)\) is two units right of the \(y\)-axis and three units above the \(x\)-axis, while \(P(3,2)\) is three units right and two units above.
Always read "across, then up": the first coordinate is the horizontal step \((x)\), the second is the vertical step \((y)\). This is why \((2,3)\) and \((3,2)\) are two different places on the plane.
3. Quadrants
The two axes divide the plane into four regions called quadrants, numbered anticlockwise starting from the top-right region. The sign pattern of a point tells you instantly which quadrant it lives in.
| Quadrant | x-coordinate | y-coordinate | Example |
|---|---|---|---|
| First (I) | Positive | Positive | \((3,4)\) |
| Second (II) | Negative | Positive | \((-3,4)\) |
| Third (III) | Negative | Negative | \((-3,-4)\) |
| Fourth (IV) | Positive | Negative | \((3,-4)\) |
Points on the axes
A point never belongs to a quadrant if it sits on an axis:
- A point of the form \((a,0)\) lies on the x-axis; its ordinate is 0.
- A point of the form \((0,b)\) lies on the y-axis; its abscissa is 0.
- The point \((0,0)\) is the origin \(O\), where both axes meet.
Scan the signs of \((x,y)\). "Positive, negative" means fourth quadrant, "negative, positive" means second. Zero in either slot means the point is on an axis, not in any quadrant.
4. Plotting Points
Plotting means converting a pair of numbers into a marked dot on the plane. The rule is to move exactly like a set of driving directions: turn right or left for the x-step, then up or down for the y-step.
Plot \(P(-3,4)\). Start at \(O\). The x-coordinate is \(-3\), so move 3 units to the left along the x-axis. The y-coordinate is \(+4\), so move 4 units up parallel to the y-axis. Mark and label the dot \(P(-3,4)\). This point lies in the second quadrant, which is exactly what the signs \(( -, + )\) predict.
5. Reading Coordinates from a Graph
Reading coordinates is the reverse of plotting: start at the point, and travel straight to the axes to find the two numbers. Drop a vertical line from the point to the \(x\)-axis: the number where it lands is the x-coordinate. Drop a horizontal line to the \(y\)-axis: the number where it lands is the y-coordinate.
Read the x-coordinate first, then the y-coordinate second, and write the answer as the ordered pair \((x,y)\). For example, the points above read as \(\mathbf{P}(3,2)\), \(\mathbf{Q}(-2,3)\), \(\mathbf{R}(2,-3)\) and \(\mathbf{S}(-4,1)\).
The horizontal step is always recorded first. Swapping gives a different point: reading \(P\) as \((2,3)\) instead of \((3,2)\) moves the point to a completely different location.
6. Distance Between Two Points
The distance between two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is the length of the straight segment joining them. To find it without measuring, we borrow a right triangle from the grid and use the Pythagoras theorem.
\[ \boxed{AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \]
What each part means:
- \((x_2 - x_1)\) is the horizontal step between the points; squaring it makes it non-negative even when the step is leftwards.
- \((y_2 - y_1)\) is the vertical step between the points; squaring removes the sign.
- The two squares are the two legs of a right triangle, so their sum is the square of the hypotenuse.
- The square root at the end converts the squared length back into an actual distance.
The point \(C\) shares \(A\)'s y-coordinate and \(B\)'s x-coordinate, so \(AC\) is horizontal and \(CB\) is vertical, making \(ABC\) a right triangle with \(AB\) as its hypotenuse. The horizontal leg has length \(|x_2 - x_1|\), the vertical leg has length \(|y_2 - y_1|\), and the Pythagoras theorem turns those two legs into the hypotenuse \(AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
7. Derivation of the Distance Formula
We derive the formula from scratch so you can see exactly why it works. Take two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) and introduce the helper point \(C(x_2,y_1)\), the corner where the horizontal step from \(A\) meets the vertical step from \(B\).
Squares wipe out minus signs, so \(AB\) computed as "from A to B" equals \(BA\) computed as "from B to A". You may subtract either way round within each bracket.
8. Distance Formula — Worked Examples
Each example follows the same skeleton: Given → Formula → Substitution → Simplification → Answer.
Given: \(A(2,3)\) and \(B(5,7)\). Find \(AB\).
Formula: \(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Substitution: \(AB = \sqrt{(5 - 2)^2 + (7 - 3)^2}\)
Simplification: \(= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25}\)
Answer: \(AB = 5\) units. ✓
Given: \(P(-2,-1)\) and \(Q(2,2)\). Find \(PQ\).
Formula: \(PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Substitution: \(PQ = \sqrt{(2 - (-2))^2 + (2 - (-1))^2}\)
Simplification: \(= \sqrt{(2 + 2)^2 + (2 + 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25}\)
Answer: \(PQ = 5\) units. ✓
Given: \(A(-1,2)\) in the second quadrant and \(B(3,-1)\) in the fourth quadrant. Find \(AB\).
Formula: \(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Substitution: \(AB = \sqrt{(3 - (-1))^2 + (-1 - 2)^2}\)
Simplification: \(= \sqrt{(3 + 1)^2 + (-3)^2} = \sqrt{4^2 + 9} = \sqrt{16 + 9} = \sqrt{25}\)
Answer: \(AB = 5\) units. ✓
Given: \(P(-2,1)\) and \(Q(4,1)\). Find \(PQ\).
Formula: \(PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Substitution: \(PQ = \sqrt{(4 - (-2))^2 + (1 - 1)^2}\)
Simplification: \(= \sqrt{6^2 + 0^2} = \sqrt{36}\)
Answer: \(PQ = 6\) units. The two points share the same y-coordinate, so the segment is horizontal and the distance is simply \(|4 - (-2)| = 6\). ✓
Given: \(M(3,-1)\) and \(N(3,5)\). Find \(MN\).
Formula: \(MN = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Substitution: \(MN = \sqrt{(3 - 3)^2 + (5 - (-1))^2}\)
Simplification: \(= \sqrt{0^2 + 6^2} = \sqrt{36}\)
Answer: \(MN = 6\) units. The two points share the same x-coordinate, so the segment is vertical and the distance is simply \(|5 - (-1)| = 6\). ✓
Given: The distance between \(A(1,2)\) and \(P(x,-1)\) is 5 units. Find \(x\).
Formula: \(AP = \sqrt{(x - 1)^2 + (-1 - 2)^2} = 5\)
Substitution: \(\sqrt{(x - 1)^2 + (-3)^2} = \sqrt{(x - 1)^2 + 9} = 5\)
Simplification: square both sides: \((x - 1)^2 + 9 = 25\), so \((x - 1)^2 = 16\), giving \(x - 1 = \pm 4\).
Answer: \(x = 5\) or \(x = -3\). Both \((5,-1)\) and \((-3,-1)\) are exactly 5 units from \(A(1,2)\). ✓
9. Midpoint of a Line Segment
The midpoint of a segment \(AB\) is the single point \(M\) that lies exactly halfway between \(A\) and \(B\). Split the horizontal step in half and the vertical step in half, and the two halves meet at the midpoint.
\[ \boxed{M\left( \frac{x_1 + x_2}{2}\,,\ \frac{y_1 + y_2}{2} \right)} \]
Why averaging? To reach the middle of the segment you must cover half of the total horizontal travel and half of the total vertical travel. Halfway between \(x_1\) and \(x_2\) on the number line is their average \(\frac{x_1+x_2}{2}\), and likewise for the y-coordinates.
10. Midpoint Formula — Derivation
Let \(A(x_1,y_1)\) and \(B(x_2,y_2)\) be the endpoints, and let \(M(x,y)\) be their midpoint. Project \(A\), \(M\) and \(B\) straight down onto the x-axis. They land on the x-axis at \((x_1,0)\), \((x,0)\) and \((x_2,0)\) in that order, and projection preserves "halfway between": the projected point of \(M\) is exactly the midpoint of the two projected endpoints on the axis.
Verify with the diagram above: \(A(1,2)\) and \(B(7,6)\) differ by 6 in x and 4 in y. The midpoint moves 3 right and 2 up from \(A\), landing at \((4,4)\), and 3 left and 2 down from \(B\) lands at the same \((4,4)\).
11. Midpoint Formula — Worked Examples
Given: \(A(2,4)\) and \(B(6,8)\). Find the midpoint \(M\).
Formula: \(M\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Substitution: \(M\left( \frac{2 + 6}{2}, \frac{4 + 8}{2} \right)\)
Calculation: \(= \left( \frac{8}{2}, \frac{12}{2} \right) = (4,6)\)
Answer: \(M(4,6)\). ✓
Given: \(P(-4,-2)\) and \(Q(2,6)\). Find the midpoint \(M\).
Formula: \(M\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Substitution: \(M\left( \frac{-4 + 2}{2}, \frac{-2 + 6}{2} \right)\)
Calculation: \(= \left( \frac{-2}{2}, \frac{4}{2} \right) = (-1,2)\)
Answer: \(M(-1,2)\). ✓
Given: \(A(5,-3)\) in the fourth quadrant and \(B(-1,7)\) in the second quadrant. Find the midpoint \(M\).
Formula: \(M\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Substitution: \(M\left( \frac{5 + (-1)}{2}, \frac{-3 + 7}{2} \right)\)
Calculation: \(= \left( \frac{4}{2}, \frac{4}{2} \right) = (2,2)\)
Answer: \(M(2,2)\). ✓
Given: \(M(3,4)\) is the midpoint of \(A(1,2)\) and \(B\). Find \(B\).
Formula: \(M\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\) with \(A\) as \((x_1,y_1)\) and \(B\) as \((x_2,y_2)\).
Substitution: \(\frac{1 + x_2}{2} = 3\) and \(\frac{2 + y_2}{2} = 4\).
Calculation: \(1 + x_2 = 6 \implies x_2 = 5\); \(2 + y_2 = 8 \implies y_2 = 6\).
Answer: \(B(5,6)\). ✓ (Check: \(M\) of \((1,2)\) and \((5,6)\) is \(\left(3,4\right)\), correct.)
Given: Two towns stand at \(A(-3,1)\) and \(B(7,5)\) on a district map. A water tank is to be built at the point exactly halfway between them. Find the tank's position.
Formula: \(M\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Substitution: \(M\left( \frac{-3 + 7}{2}, \frac{1 + 5}{2} \right)\)
Calculation: \(= \left( \frac{4}{2}, \frac{6}{2} \right) = (2,3)\)
Answer: The tank should be built at \((2,3)\). ✓
12. Applications of Coordinate Geometry
The same two ideas — distance and midpoint — run a surprising amount of the world. Whenever a position can be pinned to a grid, coordinate geometry can measure it without a tape.
Maps
Every street and landmark has a grid cell; distance and midpoint help locate the real gap between two places.
Floor plans
A house plan drawn on a grid lets builders read exact placements and find the centre of a room or the middle of a wall.
Position tracking
A delivery rider or a ship is a live point on a map; distance-between-points answers "how far away is it now?".
Grid layouts
Games, chessboards, spreadsheets and printed layouts all use number-pairs to place objects precisely.
Measuring distances
Distance on a scaled map, straight-line gaps in a park, and rejection distances in sports can all be computed from coordinates.
Midpoint of a path
Finding the halfway point of a route, the centre of a plot, or the balance point of two ends is a midpoint calculation.
Simple modelling
Ideas like "is this corner square?" or "is this plot a rectangle?" are geometry questions the distance formula can answer with coordinates.
Navigation apps compute straight-line ("as the crow flies") distances with the distance formula, then combine the midpoint idea with road data to give the shortest practical route between two locations.
13. Important Conceptual Relationships
Two coordinates can tell you a geometric fact without any calculation. When the two coordinates of a pair match, the points line up neatly.
| Condition | Meaning | Distance simplifies to |
|---|---|---|
| \(x_1 = x_2\) | The segment is vertical (points share one x-coordinate) | \(AB = |y_2 - y_1|\) |
| \(y_1 = y_2\) | The segment is horizontal (points share one y-coordinate) | \(AB = |x_2 - x_1|\) |
| \(x_1 = x_2\) and \(y_1 = y_2\) | The "two" points are the same point | \(AB = 0\) |
A vertical line is not "x is the same as y". It is "x is the same as each other for every point on it", i.e. all the points share one x-value. Penned the other way: a horizontal line shares one y-value.
14. Coordinate Geometry and Geometric Shapes
Coordinates turn shape-properties into number-checks. Because the distance formula gives every side and diagonal exactly, we can verify whether a figure is a right triangle, a rectangle, a square or a set of collinear points by pure calculation.
Given: \(A(1,2)\), \(B(4,2)\), \(C(4,6)\). Show that \(\triangle ABC\) is right-angled.
Working: find all three side lengths with the distance formula.
\(AB = \sqrt{(4-1)^2 + (2-2)^2} = \sqrt{9 + 0} = 3\)
\(BC = \sqrt{(4-4)^2 + (6-2)^2} = \sqrt{0 + 16} = 4\)
\(AC = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = 5\)
Conclusion: \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\), so the Pythagoras relation holds and \(\triangle ABC\) is right-angled at \(B\) (where the horizontal side \(AB\) meets the vertical side \(BC\)). ✓
Given: \(A(1,1)\), \(B(5,1)\), \(C(5,3)\), \(D(1,3)\). Show that \(ABCD\) is a rectangle.
Working: check sides and diagonals.
Sides: \(AB\) has \(y_1 = y_2 = 1\), so it is horizontal with \(AB = 4\). \(BC\) has \(x_1 = x_2 = 5\), so it is vertical with \(BC = 2\). \(CD\) is horizontal with \(CD = 4\) and \(DA\) is vertical with \(DA = 2\). Every corner joins a horizontal side to a vertical side, giving four right angles.
Diagonals: \(AC = \sqrt{(5-1)^2 + (3-1)^2} = \sqrt{16 + 4} = \sqrt{20}\) and \(BD = \sqrt{(1-5)^2 + (3-1)^2} = \sqrt{16 + 4} = \sqrt{20}\). They are equal, confirming the rectangle. ✓
Three points \(A, B, C\) are collinear (on one straight line) precisely when the larger distance equals the sum of the two smaller ones, i.e. \(AB + BC = AC\) for the correct ordering. For example, \(P(-1,2)\), \(Q(1,4)\), \(R(3,6)\) satisfy \(PQ = 2\sqrt{2}\), \(QR = 2\sqrt{2}\) and \(PR = 4\sqrt{2}\), so they lie on one line.
15. Common Mistakes
| Mistake | Why it happens | Fix |
|---|---|---|
| Writing y before x, e.g. \((4,3)\) instead of \((3,4)\) | Reading up before across | Always write \((x,y)\): across first, up second |
| Wrong quadrant for a point | Ignoring one of the signs | Check both signs; \(( -, + )\) means left and up, quadrant II |
| Sign errors inside the brackets | Subtracting a negative incorrectly | Bracket the negative number: \((x_2 - (-3))^2\) |
| Forgetting brackets around negative differences | Squaring a negative difference sloppily | Square the whole difference: \((-4)^2\), never \(-4^2 = -16\) |
| Wrong subtraction order | Mixing \(x_1 - x_2\) with \(y_2 - y_1\) | Squaring fixes the sign, but stay consistent: both as \(x_2-x_1\) and \(y_2-y_1\) |
| Forgetting to square | Adding differences without squaring | The formula needs squares: write each bracket squared |
| Ignoring the square root | Stopping at the squared number | Finish with \(\sqrt{\,}\) and report the distance |
| Using the wrong endpoint's coordinates in midpoint | Averaging mismatched pairs | Average the x's together and the y's together, never x with y |
| Dividing instead of averaging | Misremembering the midpoint rule | Add the two coordinates first, then divide each sum by 2 |
| Arithmetic slips in \(\sqrt{0 + 36}\) | Rushing the final steps | Write each simplification line and check squares before adding |
\((3,-4)\) is not the same as \((-3,4)\). The first is right and down (quadrant IV); the second is left and up (quadrant II). One careless sign moves the point across two quadrants.
16. Exam Tips
- Always write coordinates in the \((x,y)\) order, and label plots with their ordered pair.
- Check the signs of both coordinates before substituting.
- Write the formula first, then substitute values underneath it.
- Use brackets for negative numbers: \((x_2 - (-3))^2\) removes sign confusion.
- Simplify the squares carefully and add them before taking the root.
- Confirm the final distance is non-negative; a distance can never be negative.
- For midpoints, average the x-coordinates separately and the y-coordinates separately.
- Label points clearly on diagrams so the examiner can follow your working.
- Show enough steps: each substitution line earns marks, even if the final answer is wrong.
When the coordinates share an x or a y, skip the full formula mentally: vertical distance is \(|y_2 - y_1|\), horizontal distance is \(|x_2 - x_1|\). Spotting this can save a full minute per question.
17. Quick Revision
| Concept | Key result |
|---|---|
| Ordered pair for a point | \((x,y)\) — x first, y second |
| x-coordinate | Abscissa |
| y-coordinate | Ordinate |
| Origin | \((0,0)\) |
| Distance between \(A(x_1,y_1)\), \(B(x_2,y_2)\) | \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) |
| Midpoint of \(A(x_1,y_1)\), \(B(x_2,y_2)\) | \(\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)\) |
| Same x-coordinate | Vertical line; distance \(= |y_2-y_1|\) |
| Same y-coordinate | Horizontal line; distance \(= |x_2-x_1|\) |
| Quadrant | Sign of x | Sign of y |
|---|---|---|
| I | \(+\) | \(+\) |
| II | \(-\) | \(+\) |
| III | \(-\) | \(-\) |
| IV | \(+\) | \(-\) |
18. Coordinate Geometry Formula Box
\[ \boxed{AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}} \]
for \(A(x_1,y_1)\) and \(B(x_2,y_2)\).
\[ \boxed{M\left( \frac{x_1+x_2}{2}\,,\ \frac{y_1+y_2}{2} \right)} \]
for the midpoint \(M\) of the segment joining \(A(x_1,y_1)\) and \(B(x_2,y_2)\).
\[ \boxed{P(x,y)} \]
Every point is named by the ordered pair of its coordinates: \(x\) the abscissa, \(y\) the ordinate.
These two formulas cover the coordinate geometry taught in Class 10: distance and midpoint. Recognise that a "midpoint" is just the special halfway-ratio case of dividing a segment, which you meet fully in later classes.
19. Worked Examples — Consolidated Section
The ten examples below travel through the whole chapter: quadrants, plotting, reading, distances (positive, negative, horizontal, vertical), midpoints, unknown endpoints and a geometric figure.