Class 9 Mathematics Chapter 1: Orienting Yourself: The Use of Coordinates

Finding a shop without a map

A friend tells you their shop is "in the market". You will not find it.

They tell you instead: from the clock tower, walk 4 blocks east, then 3 blocks north. Now you will find it, and so will anybody else, first time, with no guesswork.

That pair of numbers is the whole idea of this chapter. Fix a starting point, fix two directions at right angles, and any position in a flat space can be pinned down by exactly two numbers. Nothing more is needed, and nothing less will do.

The two axes

To turn that idea into mathematics, draw two number lines crossing at right angles.

• The horizontal one is the x-axis.
• The vertical one is the y-axis.
• They cross at the origin, written $O$, whose coordinates are $(0, 0)$.

Together they are the coordinate axes, and the flat surface they sit in is the Cartesian plane — also called the coordinate plane, or simply the $xy$-plane.

Both axes are marked off in equal units from $O$. Right of $O$ and above $O$ count as positive; left of $O$ and below $O$ count as negative. That single convention is what lets two numbers describe every point.

What each coordinate actually measures

A point $P$ is written $P(x, y)$, and the two numbers are not interchangeable labels — each one measures something specific:

• $x$, the x-coordinate, measures how far $P$ lies from the y-axis, taken at right angles to it.
• $y$, the y-coordinate, measures how far $P$ lies from the x-axis, taken at right angles to it.

Read that twice, because it is the reverse of what most people guess. The x-coordinate is measured from the y-axis. It has to be: to say how far right you are, you must measure from the vertical line.

Reading a point off the grid

Given a marked point, drop a perpendicular to each axis and read the two values.

Example. A point sits 4 units right of the y-axis and 3 units above the x-axis. Its coordinates are $(4, 3)$.

If instead it sits 4 units left and 3 units above, the horizontal reading is negative: $(-4, 3)$.

Plotting a point

Going the other way, plotting $(x, y)$ is two moves from the origin:

1. Move $x$ units along the x-axis — right if $x$ is positive, left if $x$ is negative.
2. From there move $y$ units parallel to the y-axis — up if $y$ is positive, down if $y$ is negative.

Mark the point and label it. The order of the two moves does not change where you land, but doing the x-move first is the habit that stops you swapping the numbers.

The order of the two numbers matters

$(x, y)$ is an ordered pair: swapping the entries usually gives a different point.

$(2, 7)$ is 2 right and 7 up. $(7, 2)$ is 7 right and 2 up. Two different places on the plane.

The one exception is when the two numbers are equal. $(5, 5)$ swapped is still $(5, 5)$. So:

$$(x, y) = (y, x) \quad\text{exactly when}\quad x = y$$

and whenever $x \neq y$, the two pairs are different points.

Points that sit on an axis

A point on an axis has a zero in it, and which entry is zero tells you which axis.

• On the x-axis, the point is not above or below it, so its distance from the x-axis is $0$. Every such point has the form $(x, 0)$.
• On the y-axis, the distance from the y-axis is $0$, so every such point has the form $(0, y)$.
• The origin is on both at once, so it is $(0, 0)$.

A quick test: $(-6, 0)$ lies on the x-axis, 6 units left of $O$. $(0, -6)$ lies on the y-axis, 6 units below $O$. Same digits, completely different positions.

The four quadrants

The two axes cut the plane into four regions, called quadrants. They are numbered anticlockwise, starting from the one at the top right:

Learn the direction of the numbering, not just the table: start top right and go anticlockwise. Then you can rebuild the signs whenever you need them.

Naming a quadrant from the signs

The signs alone decide it, so no picture is needed.

Example. Where is $(-9, 14)$? Negative x, positive y — left and up — so Quadrant II.

Example. Where is $(0.5, -12)$? Positive x, negative y — right and down — so Quadrant IV.

A point on an axis is in no quadrant at all. The quadrants are the four open regions between the axes; the axes themselves are the boundaries. Calling $(0, 5)$ "Quadrant I" is a common and costly slip.

Distances along a grid line

Two points that share a coordinate are easy to separate, with no formula worth the name.

If two points have the same y-coordinate, they lie on a horizontal line, and the gap between them is the difference of the x-coordinates:

$$\text{distance between } (x_1, y) \text{ and } (x_2, y) = |x_2 - x_1|$$

If they share the same x-coordinate, they lie on a vertical line and the gap is $|y_2 - y_1|$.

Example. From $(-2, 5)$ to $(6, 5)$ the distance is $|6 - (-2)| = 8$ units.

The bars matter. Distance is a length, so it is never negative — take the difference and drop the sign.

A room drawn on the grid

Here is the kind of question this chapter is really for.

A rectangular room is drawn with one corner at the origin, the floor running east along the x-axis and north along the y-axis, and $1$ unit representing $1$ foot. Three corners are at $O(0, 0)$, $A(12, 0)$ and $B(12, 9)$.

Where is the fourth corner? The side from $O$ to $A$ runs 12 units across; the opposite side from the fourth corner to $B$ must do the same. So the fourth corner is $C(0, 9)$.

How wide is the room, and how deep? Width $= |12 - 0| = 12$ feet, depth $= |9 - 0| = 9$ feet.

A door runs from $(4, 0)$ to $(7, 0)$. How wide is it? $|7 - 4| = 3$ feet — wide enough for a wheelchair, which needs about $2$ feet $9$ inches of clear opening.

Notice that every one of those answers came from subtracting two numbers. That is the payoff of putting a plan on a grid.

Why negative numbers are not optional

Suppose we refused to use them. Then $x$ and $y$ could never be less than zero, and the only region we could describe would be Quadrant I — a quarter of the plane. Three quarters of every map would be unreachable.

This is not a small historical footnote. Brahmagupta (c. 628 CE) was the mathematician who set out rules for zero and for negative quantities as numbers in their own right, and the four-quadrant plane rests on that work.

The grid is an old idea

Coordinates feel modern, but grid-based thinking is ancient. The streets of the Sindhu-Sarasvatī cities were laid out on a striking north-south and east-west plan thousands of years ago, so a trader really could locate a warehouse by counting blocks in two directions. Later, Āryabhaṭa (c. 499 CE) mapped positions in the sky, and René Descartes (1637 CE) set out the full algebraic version we use today — which is why the plane carries his name.

Slips that cost marks

• Writing the coordinates the wrong way round. Say "along, then up" to yourself every time.
• Measuring the x-coordinate from the x-axis. It is measured from the y-axis.
• Putting a point on an axis into a quadrant.
• Dropping a minus sign while plotting in Quadrant II or III.
• Giving a distance as a negative number.

What to write in the exam

Name the quadrant by its number — Quadrant III, not "bottom left". When you plot, draw and label the axes, mark the unit scale, and write each point's name beside its dot. For a distance along a grid line, show the subtraction and take the positive value. Small habits, and they are worth whole marks.

The gap that is not along a grid line

Two friends are marked on a map of a park: Meera at $(2, 1)$ and Nikhil at $(6, 4)$, with one unit standing for one hundred metres.

How far apart are they?

If they shared a row or a column we could subtract and be done. They do not. The line between them runs at a slant, and no subtraction on its own will measure it.

What we can already do

Two facts from the last topic still hold, and everything here is built on them.

• Points sharing a y-coordinate lie on a horizontal line, and the gap is $|x_2 - x_1|$.
• Points sharing an x-coordinate lie on a vertical line, and the gap is $|y_2 - y_1|$.

So we can measure across and we can measure up. The trick is to use both at once.

Hiding a right triangle under the segment

From Meera's point, run a line straight across. From Nikhil's point, run a line straight down. They meet at $(6, 1)$.

Those three points form a right-angled triangle, and our slanting segment is its hypotenuse.

The two legs are easy, because each is along a grid line:

• horizontal leg $= |6 - 2| = 4$
• vertical leg $= |4 - 1| = 3$

The Baudhāyana–Pythagoras theorem finishes it

In a right-angled triangle the square on the hypotenuse equals the sum of the squares on the other two sides. That result is stated in Baudhāyana's Sulbasutras, centuries before it reached Greece, which is why it carries both names.

Applying it:

$$MN = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

Five units, so 500 metres apart. The whole method is that one picture: legs from the coordinates, hypotenuse from Pythagoras.

The distance formula

Nothing about that argument depended on the particular numbers. For any two points $A(x_1, y_1)$ and $B(x_2, y_2)$, the horizontal leg is $x_2 - x_1$ and the vertical leg is $y_2 - y_1$, so:

$$\boxed{AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$$

This is the distance formula, and it is the single most used result in the whole of coordinate geometry.

The order of the points does not matter

You may wonder which point to call the first. It makes no difference:

$$(x_2 - x_1)^2 = (x_1 - x_2)^2$$

because squaring destroys the sign. Take the points in whichever order makes the subtraction easier.

Negative coordinates change nothing

The same squaring saves us in the other quadrants. Take $C(-4, -1)$ and $D(-1, 3)$:

$$CD = \sqrt{(-1 - (-4))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$$

Notice $-1 - (-4) = -1 + 4 = 3$. Subtracting a negative adds, and that single step is where most marks are lost in this topic. Write it out rather than doing it in your head; the squaring afterwards will not rescue a wrong difference.

Distance from the origin

The origin is only the point $(0, 0)$, so the formula still applies and simply gets shorter:

$$OP = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}$$

Example. The distance of $P(8, 15)$ from the origin is $\sqrt{64 + 225} = \sqrt{289} = 17$.

Questions ask for this constantly, so the short form is worth knowing on sight.

Leave a surd as a surd

Most distances are not whole numbers. For $(2, 3)$ and $(5, 6)$:

$$\sqrt{3^2 + 3^2} = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt2$$

Write $3\sqrt2$, not $4.24$. Simplify the surd by pulling out any square factor, then stop. A decimal is an approximation, and the exact form is what earns the mark.

Compare the squares, not the roots

Very often a question does not want the distances themselves — only whether two of them are equal, or which is larger. Then the square roots are wasted work.

Because distances are never negative, $AB^2 = AC^2$ already tells you $AB = AC$. So compute the squared distances, which are whole numbers, and compare those.

This one habit removes most of the arithmetic from the harder questions in this chapter.

What kind of triangle do three points make?

Find the three squared distances and read them off.

Are $A(1, 7)$, $B(4, 2)$ and $C(-1, -1)$ the vertices of an isosceles triangle?

$$AB^2 = 3^2 + (-5)^2 = 34, \qquad BC^2 = (-5)^2 + (-3)^2 = 34, \qquad CA^2 = (-2)^2 + (-8)^2 = 68$$

$AB^2 = BC^2$, so $AB = BC$ and the triangle is isosceles.

Look again and there is more: $AB^2 + BC^2 = 34 + 34 = 68 = CA^2$. By the converse of the theorem, the triangle is also right-angled, with the right angle at $B$. Three subtractions and no square root at all.

A reflection does not change a length

Reflect a figure in the y-axis and every x-coordinate changes sign, while the y-coordinates stay put: $A(2, 1)$ becomes $A'(-2, 1)$ and $B(6, 4)$ becomes $B'(-6, 4)$.

The horizontal gap was $|6 - 2| = 4$; now it is $|-6 - (-2)| = 4$. The vertical gap is untouched. Both legs survive, so the hypotenuse does too, and $A'B' = AB = 5$.

A reflection moves a shape but never resizes it. The same holds for a reflection in the x-axis, where the y-coordinates change sign instead.

Finding a missing coordinate

Sometimes the distance is given and a coordinate is not.

Find $x$ if the distance from $P(x, 2)$ to $Q(3, 6)$ is 5.

Square both sides at once to avoid carrying a root:

$$(3 - x)^2 + (6 - 2)^2 = 25 \quad\Rightarrow\quad (3 - x)^2 + 16 = 25 \quad\Rightarrow\quad (3 - x)^2 = 9$$

So $3 - x = 3$ or $3 - x = -3$, giving $x = 0$ or $x = 6$.

Both are genuine answers. There are two points on the line $y = 2$ at distance 5 from $Q$, one either side. Offering only one of them loses a mark.

A point equidistant from two others

Find the point on the x-axis that is the same distance from $A(2, -5)$ and $B(-2, 9)$.

A point on the x-axis is $(x, 0)$. Set the squared distances equal:

$$(2 - x)^2 + 25 = (-2 - x)^2 + 81$$ $$4 - 4x + x^2 + 25 = 4 + 4x + x^2 + 81$$ $$29 - 4x = 85 + 4x \quad\Rightarrow\quad -8x = 56 \quad\Rightarrow\quad x = -7$$

The point is $(-7, 0)$. The $x^2$ terms always cancel, which is why this method leaves a linear equation and is far easier than it first looks.

Slips that cost marks

• Mishandling a double negative in the subtraction, as in $-1 - (-4)$.
• Adding the differences instead of their squares.
• Turning an exact surd into a decimal without being asked.
• Giving only one answer when squaring has produced two.
• Forgetting the square root at the end and quoting $AB^2$ as the distance.

What to write in the exam

Quote the formula before you substitute. Show the two differences on separate lines so a sign slip stays visible and recoverable. Keep surds exact and simplified. And when the question asks what kind of triangle three points form, work with the squared distances throughout — it is quicker, and it removes every rounding error.

Meeting exactly halfway

Two friends live at $A(-3, 2)$ and $B(7, 6)$ on a town map. They want to meet at the point exactly halfway along the straight road between them.

You could measure the distance, halve it, and then try to step that far along the road. There is no need. The halfway point can be written down straight from the coordinates, in one line.

Averaging the two coordinates

Halfway along means halfway across and halfway up. So take the average of the two x-coordinates, and the average of the two y-coordinates:

$$x = \frac{-3 + 7}{2} = 2, \qquad y = \frac{2 + 6}{2} = 4$$

The meeting point is $M(2, 4)$. That is the whole method.

The midpoint formula

For any two points $A(x_1, y_1)$ and $B(x_2, y_2)$, the midpoint of the segment $AB$ is:

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

Two averages, nothing more. Notice how much gentler this is than the distance formula — there is no squaring and no square root, because we are locating a point rather than measuring a length.

Why averaging is the right thing to do

Going from $A$ to $B$, the total sideways shift is $x_2 - x_1$. Half of that shift is $\tfrac{1}{2}(x_2 - x_1)$, so the halfway point sits at

$$x_1 + \frac{x_2 - x_1}{2} = \frac{2x_1 + x_2 - x_1}{2} = \frac{x_1 + x_2}{2}$$

which is exactly the average. The same argument works vertically. So the formula is not a rule to memorise blindly; it is what "half of the way across" simplifies to.

Finding the other end

The formula can be run backwards, and questions often ask for this.

$M(2, -1)$ is the midpoint of $A(-3, 4)$ and $B$. Where is $B$?

The x-coordinates must average to 2, so $\dfrac{-3 + x}{2} = 2$, giving $-3 + x = 4$ and $x = 7$. The y-coordinates must average to $-1$, so $\dfrac{4 + y}{2} = -1$, giving $4 + y = -2$ and $y = -6$.

So $B$ is $(7, -6)$. There is a quicker version worth knowing: double the midpoint and subtract the known end.

$$B = (2 \times 2 - (-3),; 2 \times (-1) - 4) = (7, -6)$$

Cutting a segment into three

Sometimes a segment must be divided into three equal parts. The two dividing points are the points of trisection.

Find them for $A(1, 2)$ and $B(10, 11)$.

The total shift is 9 across and 9 up, so each third is 3 across and 3 up. Starting from $A$:

• first point $P = (1 + 3,; 2 + 3) = (4, 5)$
• second point $Q = (4 + 3,; 5 + 3) = (7, 8)$

You can check these with the midpoint formula alone: $P$ should be the midpoint of $A$ and $Q$, and indeed $\left(\tfrac{1+7}{2}, \tfrac{2+8}{2}\right) = (4, 5)$. Likewise $Q$ is the midpoint of $P$ and $B$.

Do three points lie on one straight line?

Three points are collinear when they all lie on a single straight line. Using only the distance formula, there is a neat test:

Three points are collinear exactly when the two shorter distances add up to the longest one.

If they do, the middle point sits on the segment joining the outer two, with no detour.

Are $A(-2, -3)$, $B(1, 1)$ and $C(4, 5)$ collinear?

$$AB = \sqrt{9 + 16} = 5, \qquad BC = \sqrt{9 + 16} = 5, \qquad AC = \sqrt{36 + 64} = 10$$

$AB + BC = 5 + 5 = 10 = AC$. Yes, they are collinear, and $B$ is the midpoint of $AC$.

The one place where squares do not help

Everywhere else in this chapter we worked with squared distances to dodge the surds. Do not do it here.

The collinearity test is an addition of lengths, and squaring does not respect addition. In the example above $AB^2 + BC^2 = 25 + 25 = 50$, while $AC^2 = 100$. The squares do not add, even though the lengths do.

So for collinearity, keep the surds and write each distance as a multiple of the same root where you can — then the addition is easy to see.

A near miss worth seeing

Are $P(0, 0)$, $Q(3, 4)$ and $R(7, 9)$ collinear?

$$PQ = 5, \qquad QR = \sqrt{16 + 25} = \sqrt{41}, \qquad PR = \sqrt{49 + 81} = \sqrt{130}$$

On a calculator $5 + \sqrt{41} \approx 11.403$ and $\sqrt{130} \approx 11.402$. They are close enough to look equal, but they are not: squaring $5 + \sqrt{41}$ gives $66 + 10\sqrt{41}$, which is not $130$.

Three points can look collinear on a sketch and fail the test. Always check by calculation, never by eye.

Inside, on, or outside a circle

The set of points at a fixed distance $r$ from a centre is a circle of radius $r$. So the distance formula decides where a point sits relative to one.

Show that $(3, 4)$, $(-4, 3)$ and $(0, -5)$ all lie on a circle centred at the origin, and find its radius.

Each squared distance from the origin is $9 + 16 = 25$, $16 + 9 = 25$ and $0 + 25 = 25$. All equal, so all three points are the same distance from the centre — they lie on a circle of radius $\sqrt{25} = 5$.

Now test two more points against that circle by comparing squared distances with $r^2 = 25$:

• $(2, 4)$: $4 + 16 = 20 < 25$, so it lies inside.
• $(5, 1)$: $25 + 1 = 26 > 25$, so it lies outside.

Here the squares are safe again, because we are comparing rather than adding.

Naming a quadrilateral from its corners

Given four points, the side and diagonal lengths name the figure. Work in squared distances and run down a checklist.

A useful extra check: in every one of these, the diagonals bisect each other, so both diagonals share the same midpoint.

A worked quadrilateral

What figure is $E(1, 2)$, $F(4, 1)$, $G(5, 4)$, $H(2, 5)$?

Sides: $EF^2 = 9 + 1 = 10$, $FG^2 = 1 + 9 = 10$, $GH^2 = 9 + 1 = 10$, $HE^2 = 1 + 9 = 10$. All four equal.

Diagonals: $EG^2 = 16 + 4 = 20$ and $FH^2 = 4 + 16 = 20$. Equal as well.

Four equal sides and equal diagonals, so $EFGH$ is a square of side $\sqrt{10}$.

As a final check, the midpoint of $EG$ is $\left(\tfrac{1+5}{2}, \tfrac{2+4}{2}\right) = (3, 3)$ and the midpoint of $FH$ is $\left(\tfrac{4+2}{2}, \tfrac{1+5}{2}\right) = (3, 3)$. The diagonals do bisect each other, as they must.

Working back from the midpoints of a triangle

A harder use of the formula. Suppose $D(3, 2)$, $E(7, 4)$ and $F(5, 8)$ are the midpoints of the sides $BC$, $CA$ and $AB$ of a triangle $ABC$. Where are $A$, $B$ and $C$?

Adding the three midpoint equations in pairs gives a short route:

$$A = E + F - D, \qquad B = F + D - E, \qquad C = D + E - F$$

treating each point as its pair of coordinates. So $A = (7 + 5 - 3,; 4 + 8 - 2) = (9, 10)$, $B = (5 + 3 - 7,; 8 + 2 - 4) = (1, 6)$, and $C = (3 + 7 - 5,; 2 + 4 - 8) = (5, -2)$.

Always check: the midpoint of $B(1, 6)$ and $C(5, -2)$ is $(3, 2) = D$. It works.

Slips that cost marks

• Subtracting the coordinates instead of adding them when finding a midpoint.
• Dividing only one of the two coordinates by 2.
• Using squared distances in the collinearity test, where they do not add.
• Forgetting to check the diagonals, and so calling a rhombus a square.
• Giving a midpoint as a single number rather than as a pair of coordinates.

What to write in the exam

Quote the midpoint formula before substituting, and give the answer as a coordinate pair in brackets. For collinearity, show all three distances and the addition that settles it. For naming a quadrilateral, show all four sides and both diagonals — the sides alone cannot tell a square from a rhombus, and that missing line is where the mark goes.