Orienting Yourself: The Use of Coordinates

Contents hide

1The Coordinate Plane

2Quadrants

3Distance Between Two Points

4Midpoint, Collinearity and Circles

Chapter mind map: how it all connects

1 · The coordinate planeTwo perpendicular number lines, the x-axis and y-axis, meeting at the origin

2 · QuadrantsThe axes cut the plane into four regions, each with its own sign pattern

Orienting Yourself: Coordinates

3 · Distance between two pointsThe Baudhāyana–Pythagoras theorem turns any two points into a right triangle

4 · Midpoint, collinearity, circlesThe distance idea unlocks midpoints, checking straight lines, and points on a circle

What you will learn in this chapter

How to plot any point in a plane using an ordered pair of numbers (x, y)
How to tell which of the four quadrants a point lies in, just from the signs of its coordinates
How to find the exact distance between any two points, using a theorem you already know
How to find the midpoint of a segment, and use it to check if three points lie on one straight line

coordinate axesoriginCartesian planequadrantordered pairdistance formulamidpointcollinear points

1The Coordinate Plane

A single number line can only locate a point along one direction: left-right. But a room, a city, or a computer screen is flat: to pin down an exact spot you need two pieces of information, not one. Those two numbers are what this chapter is about.

Figure 1: A room drawn on a coordinate grid. Every corner and the door can now be described with just two numbers. — **Figure 1**: A room drawn on a coordinate grid. Every corner and the door can now be described with just two numbers.

Learn by heartDefinition 1

The coordinate axes are two number lines drawn perpendicular to each other: one horizontal, called the x-axis, and one vertical, called the y-axis. Their point of intersection is called the origin, written O, with coordinates (0, 0).

Learn by heartDefinition 2

The plane containing the two axes is called the Cartesian plane, the coordinate plane, or the xy-plane. Distances to the right of O or above O are positive; distances to the left of O or below O are negative.

**Figure 2**: Points on the axes: a point on the x-axis always has y = 0, a point on the y-axis always has x = 0.

Learn by heartDefinition 3

The coordinates of a point P are written as an ordered pair (x, y), where x is the distance of P from the y-axis (measured along the x-axis) and y is the distance of P from the x-axis (measured along the y-axis). x is called the x-coordinate (or abscissa) and y the y-coordinate (or ordinate).

Common Mistake

(x, y) and (y, x) are not the same point unless x = y. Always write the x-coordinate first, students lose marks by swapping the order when reading a point off a graph.

Did you know?

Coordinate thinking is far older than Descartes. Cities in the Sindhu-Sarasvatī Civilisation were laid out on a precise North-South/East-West grid over 4,000 years ago. Āryabhaṭa (c. 499 CE) used coordinates to map the sky, and Brahmagupta (c. 628 CE) gave zero and negative numbers their algebraic meaning, without which the origin and the left/below-axis directions we use today would not exist. Fermat and Descartes only formalised the system in the 1600s, in Europe.

2Quadrants

The two axes divide the Cartesian plane into four regions, called quadrants, numbered I to IV starting from the top-right and going anticlockwise.

Quadrant	x-coordinate	y-coordinate	Form	Example
I	positive	positive	(+, +)	P(3, 5)
II	negative	positive	(–, +)	Q(–5, 3)
III	negative	negative	(–, –)	R(–4, –4)
IV	positive	negative	(+, –)	S(3, –5)

**Figure 3**: One point plotted in each quadrant, with its sign pattern.

Exam Tip

A point on the x-axis is always of the form (x, 0), and a point on the y-axis is always of the form (0, y). Neither one lies “in” any quadrant. This is a favourite 1-mark question.

Solved Example

Q. State the quadrant (or axis) for: A(–2, 7), B(0, –6), C(8, –1), D(–3, –3).

A. A → Quadrant II (–, +) · B → on the y-axis (x = 0) · C → Quadrant IV (+, –) · D → Quadrant III (–, –)

3Distance Between Two Points

If two points lie on the same axis, or on a line parallel to an axis, finding the distance between them is simple subtraction.

Formula

Distance

= |x_2 – x_1|

(points on a line parallel to the x-axis)

Distance

= |y_2 – y_1|

(points on a line parallel to the y-axis). The modulus is used because distance is never negative

But what if the segment joining two points is slanting, not parallel to either axis? Look at points A(3, 4) and D(7, 1). Drop a helper point C(3, 1): now AC is vertical, CD is horizontal, and AD is the hypotenuse of a right triangle.

**Figure 4**: AD is the hypotenuse of right triangle ACD, with legs AC = 3 units and CD = 4 units.

AC = 4 – 1 = 3

unitsdifference of y-coordinates of A and C

CD = 7 – 3 = 4

unitsdifference of x-coordinates of C and D

AD = \sqrt{AC^2 + CD^2} = \sqrt{3^2+4^2} = \sqrt{25} = 5

unitsBaudhāyana–Pythagoras theorem on right triangle ACD

Repeating this for any two points $(x_1, y_1)$ and $(x_2, y_2)$ gives the general distance formula, and it makes no difference whether the coordinates are positive or negative, since we are only measuring lengths.

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

…(1)

What each symbol means

d

distance between the two pointssame unit as the coordinates

(x_1,y_1)

coordinates of the first point–

(x_2,y_2)

coordinates of the second point–

Given

Points M(–3, 6) and N(5, 0)

Find

Distance MN

Solution

$MN = \sqrt{(5-(-3))^2 + (0-6)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64+36} = \sqrt{100} = \;$ $10$ units

Exam Tip

Reflecting a figure in an axis flips the signs of one coordinate but never changes any of its side lengths. The distance formula gives the same answer before and after, because it only ever uses squares of differences.

4Midpoint, Collinearity and Circles

The distance formula is the one tool this whole chapter is built on: the next three ideas are all applications of it.

Formula

Midpoint of

(x_1,y_1)

and

(x_2,y_2)

= \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)

The midpoint’s coordinates are simply the average of the two endpoints’ coordinates

**Figure 5**: M(5, 6) is the midpoint of segment ST, since 5 = (2+8)/2 and 6 = (3+9)/2.

Given

M(–7, 1) is the midpoint of A(3, –4) and B(x, y)

Find

Coordinates of B

Solution

$\dfrac{3+x}{2} = -7 \Rightarrow x = -17$ and $\dfrac{-4+y}{2} = 1 \Rightarrow y = 6$ . So B $=\;$ $(-17, 6)$

Formula

Trisection points of

A(x_1,y_1)

and

B(x_2,y_2)

P = A + \tfrac{1}{3}(B-A)

Q = A + \tfrac{2}{3}(B-A)

P and Q split AB into three equal parts, with P closer to A and Q closer to B

Given

A(4, 7) and B(16, –2)

Find

Trisection points P (near A) and Q (near B)

Solution

$P = \left(4+\tfrac{16-4}{3},\ 7+\tfrac{-2-7}{3}\right) = (8, 4)$

$Q = \left(4+\tfrac{2(16-4)}{3},\ 7+\tfrac{2(-2-7)}{3}\right) = \;$ $(12, 1)$

Checking if three points are collinear

Three points A, B, C (with B between A and C) are collinear, meaning they lie on one straight line, exactly when $AB + BC = AC$ . Find all three distances with the distance formula and check the sum. No plotting needed.

Solved Example

Q. Are M(–3, –4), A(0, 0) and G(6, 8) collinear?

$MA = \sqrt{3^2+4^2} = 5$ , $AG = \sqrt{6^2+8^2} = 10$ , $MG = \sqrt{9^2+12^2} = 15$

Since $MA + AG = 5 + 10 = 15 = MG$ , the points are collinear.

A circle is the set of all points at a fixed distance (the radius) from a fixed centre, so the distance formula also tells us whether a point lies on, inside, or outside a circle.

**Figure 6**: A(1,–8), B(–4,7) and C(–7,–4) all lie on circle K (radius √65, centre O). D is inside, E is outside.

Solved Example

Q. Show A(1, –8), B(–4, 7), C(–7, –4) lie on a circle centred at O(0, 0), and find its radius.

$OA = \sqrt{1^2+8^2}=\sqrt{65}$ , $OB=\sqrt{4^2+7^2}=\sqrt{65}$ , $OC=\sqrt{7^2+4^2}=\sqrt{65}$

All three distances from O are equal, so A, B, C lie on one circle of radius $\sqrt{65}$ units.

Memory Trick

Inside → distance < radius. On → distance = radius. Outside → distance > radius.

Just compare the point’s distance from the centre with the radius, and the three cases follow directly.

Every idea in this chapter, quadrants, distance, midpoint, collinearity, circles, comes from just one thing: two perpendicular number lines and the Baudhāyana–Pythagoras theorem.

the single idea to carry out of this chapter

Check yourself before the exam

Can I plot a point and name its quadrant instantly from its signs?
Can I write the distance formula from memory and apply it without a diagram?
Can I find a midpoint, and use it to check three points are collinear?
Can I tell whether a point lies inside, on, or outside a given circle?

Quick Revision: read this the night before the exam

Coordinates (x, y): x from the y-axis, y from the x-axis. (x, 0) is on the x-axis; (0, y) is on the y-axis
Quadrants I–IV: (+,+), (–,+), (–,–), (+,–)
Distance $= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ ; on a line parallel to an axis it is just $|$ difference $|$
Midpoint $= \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
A, B, C are collinear iff $AB+BC=AC$ (B between A and C)
A point is on a circle of radius r centred at O iff its distance from O equals r

Practice Questions

1 markWhat are the coordinates of the origin?
1 markIn which quadrant does the point (–6, –2) lie?
1 markWrite the coordinates of a point that lies on the y-axis, 5 units below the origin.
2 marksFind the distance between P(–5, 7) and Q(–1, 4).
2 marksFind the midpoint of the segment joining (–8, 7) and (6, –3).
3 marksPoint W has x-coordinate –5. H lies on the line through W parallel to the y-axis. Which quadrants can H lie in? Explain.
3 marksCheck whether R(–5, –1), B(–2, –5) and C(4, –12) are collinear.
3 marksPlot A(2, 1), B(–1, 2), C(–2, –1), D(1, –2). Show ABCD is a square and find its area.
4 marksGiven D(–5, 6) and E(0, 9), and circle K of radius √65 centred at the origin, state whether each point lies inside, on, or outside K. Show your working.
5 marksThe midpoints of the sides of triangle ABC are D(5, 1), E(6, 5) and F(0, 3). Find the coordinates of A, B and C.
3 marksFind the trisection points of the segment joining A(2, –2) and B(–7, 4).
3 marksFor quadrilateral RAMP with R(3, 0), A(0, –2), M(–5, –2), P(–5, 2): name two sides that are perpendicular to each other, one side parallel to an axis, and a pair of points that are mirror images of each other in an axis.
4 marksA computer screen is 800 px wide and 600 px high (origin at the bottom-left corner). A circular icon of radius 80 px has centre A(100, 150); another of radius 100 px has centre B(250, 230). Does any part of either circle lie outside the screen? Do the two circles intersect?

Multiple Choice

1 markThe point (0, –7) lies on:
(a) the x-axis(b) the y-axis(c) Quadrant III(d) Quadrant IV
1 markIf x ≠ y, then (x, y) and (y, x):
(a) are always the same point(b) are never the same point(c) lie in the same quadrant always(d) both lie on an axis
1 markThe distance of the point (–8, 6) from the origin is:
(a) 2(b) 10(c) 14(d) 100

Assertion (A): The point (4, 0) lies on the x-axis.
Reason (R): Every point on the x-axis has y-coordinate 0.

Case Study

A city has two main roads crossing at its centre, running North–South and East–West. Every other street runs parallel to these, 200 m apart, with 10 streets in each direction. A street intersection is named (a, b) where a is the a-th street from the N–S direction and b is the b-th street from the E–W direction.

1 markWhat real-world idea does this intersection-naming system represent?
1 markUsing 1 cm = 200 m, what distance on paper represents the real distance between two streets?
2 marksTwo friends live at intersections (2, 5) and (2, 9). What is the actual distance between their homes?

Answer Key

Q1–3(0,0) · Quadrant III · (0, –5)

PQ=\sqrt{4^2+3^2}=5

units

Q5Midpoint

=(-1,2)

Q6H can be (–5, y) for any y ≠ 0; lies in Quadrant II if y > 0, Quadrant III if y < 0

RB=5, BC=\sqrt{85}, RC=\sqrt{202}

. Since

RB+BC\ne RC

, not collinear

Q8All sides = √10, ABCD is a square, area = 10 sq. units

OD=\sqrt{61}<\sqrt{65}

→ inside;

OE=9>\sqrt{65}

→ outside

Q10A(1, 7), B(–1, –1), C(11, 3), taking D, E, F as the midpoints of BC, CA, AB respectively

Q11P(–1, 0), Q(–4, 2)

Q12AM and MP are perpendicular; AM is parallel to the x-axis (MP to the y-axis); M and P are mirror images in the x-axis

Q13Both circles lie fully inside the screen; distance AB = 170, which is between the difference (20) and sum (180) of the radii, so the circles intersect

MCQ 1–3(b) · (b) · (b)

A/RBoth A and R are true, and R is the correct explanation of A

Case StudyThe coordinate system · 1 cm : 200 m · 800 m