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Coordinate Geometry | What are coordinates | Real Life Connections of coordinate geometry
Mathematics | Class 10 | CBSE & SEBA Board
Edunes Online Education
BOARD: CBSE | CLASS: 10 | Mathematics | Chapter 7: coordinate geometry
๐ 6.1 Introduction to the Coordinate Geometry
๐ฏ Chapter 7: Coordinate Geometry โ Why This Chapter Is Not Just Maths
Imagine this ๐
You open Google Maps. You drop a pin ๐.
That pin is not magic. It is Coordinate Geometry.
๐ง Neurological Hook:
The brain remembers ideas better when they connect to movement, space, and location. Coordinate Geometry talks directly to your spatial memory system.
The brain remembers ideas better when they connect to movement, space, and location. Coordinate Geometry talks directly to your spatial memory system.
7.1 ๐ What Is Coordinate Geometry? (Think Before You Define)
Before definitions, ask this thinking question:
๐ง Think:
How does Swiggy know where you live?
How does Uber find your driver?
How does PUBG know where your enemy is hiding?
How does Swiggy know where you live?
How does Uber find your driver?
How does PUBG know where your enemy is hiding?
All of them use the same idea: Every position needs a reference.
In maths, that reference is:
- x-axis โ left / right movement
- y-axis โ up / down movement
๐ Definition (Now It Makes Sense):
Coordinate Geometry is the study of geometry using numbers that describe position with respect to two perpendicular lines called axes.
Coordinate Geometry is the study of geometry using numbers that describe position with respect to two perpendicular lines called axes.
๐ง Memory Trick:
First walk sideways (x), then climb (y). x comes before y โ always.
First walk sideways (x), then climb (y). x comes before y โ always.
๐งญ Understanding Coordinates: (x, y) Is a Journey
Students often see (x, y) as just brackets and commas. Thatโs a mistake.
๐ถ Real-Life Thinking:
You exit your house.
You walk 3 steps right โ then 4 steps up.
Your final position is (3, 4).
You exit your house.
You walk 3 steps right โ then 4 steps up.
Your final position is (3, 4).
๐ How to Think:
Coordinates are not points. Coordinates are instructions to reach a point.
Coordinates are not points. Coordinates are instructions to reach a point.
๐ง Brain Hack:
Say it aloud while solving: โRight/Left first, then Up/Down.โ
Say it aloud while solving: โRight/Left first, then Up/Down.โ
7.2 ๐ Distance Formula โ Measuring Without a Ruler
Big question first:
๐ค Think:
If two mobile towers are on a map, how does the network company know the distance between them without visiting the place?
If two mobile towers are on a map, how does the network company know the distance between them without visiting the place?
They donโt measure on land. They calculate using coordinates.
Distance between two points A(xโ, yโ) and B(xโ, yโ) is:
๐ Distance Formula:
\( AB = \sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2} \)
\( AB = \sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2} \)
๐ง Why This Sticks:
The brain loves patterns. This formula is just Pythagoras wearing coordinates as clothes.
The brain loves patterns. This formula is just Pythagoras wearing coordinates as clothes.
๐ง How to Think While Applying Distance Formula
๐ง Step-by-Step Thinking Pattern:
- Identify Point 1 and Point 2
- Subtract x-values โ square
- Subtract y-values โ square
- Add โ square root
โ
Golden Rule:
Never memorise blindly.
Always imagine a right-angled triangle between the points.
Never memorise blindly.
Always imagine a right-angled triangle between the points.
๐ง Exam Survival Tip:
If you forget the formula, draw axes โ make a right triangle โ apply Pythagoras.
If you forget the formula, draw axes โ make a right triangle โ apply Pythagoras.
๐ฌ Final Punchline
Coordinate Geometry is not about points. It is about position, movement, and distance.
Every time you use GPS, maps, or games โ you are already doing Chapter 7.
๐ง Lock It In:
Maths becomes easy the moment your brain sees it as real life.
Maths becomes easy the moment your brain sees it as real life.
Plot a Point on Cartesian Plane
X: Y:๐ Distance Formula โ Why We Need a Proof
How can we find the distance between two points on a plane
without measuring it physically?
๐ง The brain understands distance best when it can
see a right-angled triangle.
This proof is about creating that triangle.
๐งญ Step 1: Visualise Before You Calculate
Let the two points be: P(xโ, yโ) and Q(xโ, yโ).
We are not interested in formulas yet. First, we ask: How are these points positioned?
๐ From points P and Q, draw perpendiculars to the X-axis:
- PR โ X-axis
- QS โ X-axis
Geometry becomes easy when you force a right angle.
That is the hidden strategy here.
๐ Step 2: Convert Coordinates into Lengths
Now think slowly: coordinates are not numbers โ they are distances from axes.
Along the X-axis:
OR = xโ and OS = xโ
โด RS = OS โ OR = xโ โ xโ
OR = xโ and OS = xโ
โด RS = OS โ OR = xโ โ xโ
Since PT is parallel to RS,
PT = xโ โ xโ
๐ง Horizontal change always comes from
x-coordinates.
๐ Step 3: Understand Vertical Distance
Along the Y-axis:
SQ = yโ and ST = yโ
SQ = yโ and ST = yโ
โด QT = SQ โ ST = yโ โ yโ
๐ง Vertical change always comes from
y-coordinates.
๐ Step 4: The Key Triangle Appears
Observe triangle ฮPTQ.
PT โ QT
โด ฮPTQ is right-angled at T
โด ฮPTQ is right-angled at T
The moment you see a right-angled triangle,
your brain should automatically think:
Pythagoras theorem.
๐งฎ Step 5: Apply Pythagoras Theorem
In ฮPTQ:
PQยฒ = PTยฒ + QTยฒ
PQยฒ = PTยฒ + QTยฒ
PQยฒ = (xโ โ xโ)ยฒ + (yโ โ yโ)ยฒ
โด
\( PQ = \sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2} \)
โ Final Result: Distance Formula
The distance between two points
P(xโ, yโ) and
Q(xโ, yโ) is:
\( PQ = \sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2} \)
\( PQ = \sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2} \)
๐ง Permanent Memory Trigger:
Distance formula = Pythagoras + coordinates
Nothing new. Just smart geometry.
Distance formula = Pythagoras + coordinates
Nothing new. Just smart geometry.
๐ Example 1: Do Three Points Form a Triangle?
Do the points (3, 2), (โ2, โ3) and (2, 3) form a triangle?
If yes, name the type of triangle formed.
๐ง Before calculation, ask your brain one question:
Are these points connected in a straight line or not?
Are these points connected in a straight line or not?
๐งญ Step 1: Label the Points Clearly
Let the given points be:
P(3, 2), Q(โ2, โ3) and R(2, 3)
P(3, 2), Q(โ2, โ3) and R(2, 3)
Geometry problems become easy when you
name points and stick to those names.
๐ Step 2: Decide the Tool โ Distance Formula
To check whether three points form a triangle,
we must compare the lengths of
PQ, QR and PR.
๐ง Three points form a triangle only when
no one side equals or exceeds the sum of the other two.
๐งฎ Step 3: Calculate Distance PQ
\( PQ = \sqrt{{(3 + 2)}^2 + {(2 + 3)}^2} \)
\( = \sqrt{5^2 + 5^2} = \sqrt{50} \)
\( = \sqrt{5^2 + 5^2} = \sqrt{50} \)
Always subtract coordinates carefully:
minus minus becomes plus โ donโt rush.
minus minus becomes plus โ donโt rush.
๐งฎ Step 4: Calculate Distance QR
\( QR = \sqrt{{(- 2 - 2)}^2 + {(- 3 - 3)}^2} \)
\( = \sqrt{(-4)^2 + (-6)^2} = \sqrt{52} \)
\( = \sqrt{(-4)^2 + (-6)^2} = \sqrt{52} \)
๐ง Squaring removes the sign โ distance is always positive.
๐งฎ Step 5: Calculate Distance PR
\( PR = \sqrt{{(3 - 2)}^2 + {(2 - 3)}^2} \)
\( = \sqrt{1^2 + (-1)^2} = \sqrt{2} \)
\( = \sqrt{1^2 + (-1)^2} = \sqrt{2} \)
๐บ Step 6: Do the Points Form a Triangle?
Check triangle condition:
PQ + PR > QR
QR + PR > PQ
PQ + QR > PR
PQ + PR > QR
QR + PR > PQ
PQ + QR > PR
Since the sum of any two sides is greater than the third,
points P, Q and R form a triangle.
๐ง If any two distances exactly add up to the third,
the triangle collapses into a straight line.
๐ Step 7: Identify the Type of Triangle
PQยฒ + PRยฒ = 50 + 2 = 52 = QRยฒ
By the converse of Pythagoras theorem,
the angle opposite the longest side is 90ยฐ.
โด โ P = 90ยฐ
Triangle PQR is a right-angled triangle.
Triangle PQR is a right-angled triangle.
๐ง Final Memory Lock
๐ Exam Brain Formula:
Distance formula โ triangle condition โ square comparison
Geometry is not calculation โ it is pattern recognition.
Distance formula โ triangle condition โ square comparison
Geometry is not calculation โ it is pattern recognition.
๐ Example 2: Proving a Square Using Distance Formula
Show that the points A(1, 7), B(4, 2), C(โ1, โ1) and D(โ4, 4) are the vertices of a square.
๐ง Before calculating anything, pause and ask:
What makes a square a square?
What makes a square a square?
๐งญ Step 1: Decide the Strategy (Most Important Step)
A quadrilateral is a square if:
- All four sides are equal
- Both diagonals are equal
Geometry proofs are not about formulas โ
they are about choosing the correct property.
๐ Step 2: Name the Points Systematically
Let the given points be:
A(1, 7), B(4, 2), C(โ1, โ1), D(โ4, 4)
A(1, 7), B(4, 2), C(โ1, โ1), D(โ4, 4)
๐ง Always label points in order โ
confusion in naming leads to wrong conclusions.
๐ Step 3: Find Length of Side AB
\( AB = \sqrt{{(1 - 4)}^2 + {(7 - 2)}^2} \)
\( = \sqrt{(-3)^2 + 5^2} = \sqrt{34} \)
\( = \sqrt{(-3)^2 + 5^2} = \sqrt{34} \)
๐ Step 4: Find Length of Side BC
\( BC = \sqrt{{(4 + 1)}^2 + {(2 + 1)}^2} \)
\( = \sqrt{(5)^2 + (3)^2} = \sqrt{34} \)
\( = \sqrt{(5)^2 + (3)^2} = \sqrt{34} \)
When two consecutive sides are equal,
the shape is already becoming special.
๐ Step 5: Find Length of Side CD
\( CD = \sqrt{{(- 1 + 4)}^2 + {(- 1 - 4)}^2} \)
\( = \sqrt{(-3)^2 + (-5)^2} = \sqrt{34} \)
\( = \sqrt{(-3)^2 + (-5)^2} = \sqrt{34} \)
๐ Step 6: Find Length of Side DA
\( DA = \sqrt{{(1 + 4)}^2 + {(7 - 4)}^2} \)
\( = \sqrt{5^2 + 3^2} = \sqrt{34} \)
\( = \sqrt{5^2 + 3^2} = \sqrt{34} \)
โด AB = BC = CD = DA
๐ง Step 7: Check the Diagonals
\( AC = \sqrt{{(1 + 1)}^2 + {(7 + 1)}^2} \)
\( = \sqrt{(2)^2 + (8)^2} = \sqrt{68} \)
\( = \sqrt{(2)^2 + (8)^2} = \sqrt{68} \)
\( BD = \sqrt{{(4 + 4)}^2 + {(2 - 4)}^2} \)
\( = \sqrt{8^2 + (-2)^2} = \sqrt{68} \)
\( = \sqrt{8^2 + (-2)^2} = \sqrt{68} \)
โด AC = BD
โ Final Conclusion
Since:
- All four sides are equal
- Both diagonals are equal
๐ Permanent Exam Memory:
Square = equal sides + equal diagonals
Distance formula is just the measuring tape.
Square = equal sides + equal diagonals
Distance formula is just the measuring tape.
Example 3: The figure shows the arrangement of desks in a classroom.
Ashima, Bharti and Camilla are seated at
A(3, 1), B(6, 4) and C(8, 6) respectively.
Do you think they are seated in a line?
Give reasons for your answer.
If three points lie on a straight line (are collinear), then the sum of distances of two consecutive points equals the distance between the extreme points.
Mathematically:
If AB + BC = AC โ A, B, C are collinear
- Whenever coordinates are given, distance formula must ring a bell.
- To check whether points are in a straight line, we must compare distances.
- Choose one point in the middle (here B) and calculate:
- Distance from A to B
- Distance from B to C
- Distance from A to C
Using distance formula:
\( AB = \sqrt{(6 - 3)^2 + (4 - 1)^2} \)
\( = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \)
\( BC = \sqrt{(8 - 6)^2 + (6 - 4)^2} \)
\( = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \)
\( AC = \sqrt{(8 - 3)^2 + (6 - 1)^2} \)
\( = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2} \)
AB + BC = \( 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} \)
AB + BC = AC
This satisfies the condition for collinearity.
โIf the sum of two smaller distances equals the biggest distance โ the points must lie on one straight road.โ
๐ Always remember: Distance Formula โ Add โ Compare โ Decide
Example 4: Find a relation between x and y such that the point (x, y) is equidistant from the points
A(7, 1) and B(3, 5).
A point is equidistant from two fixed points if its distance from both points is equal.
To avoid square roots, we compare squares of distances:
If AP = BP โ APยฒ = BPยฒ
- The word โequidistantโ immediately tells us to use the distance formula.
- Since the point is unknown, assume it as P(x, y).
- Form two distances:
- Distance from P to A
- Distance from P to B
- Equate their squares to remove square roots.
Let P(x, y) be equidistant from A(7, 1) and B(3, 5).
So, AP = BP โ APยฒ = BPยฒ
\( AP^2 = (x - 7)^2 + (y - 1)^2 \)
\( BP^2 = (x - 3)^2 + (y - 5)^2 \)
\( (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2 \)
\( \Rightarrow x^2 - 14x + 49 + y^2 - 2y + 1 \) \( = x^2 - 6x + 9 + y^2 - 10y + 25 \)
Cancelling \( x^2 \) and \( y^2 \) from both sides,
\( \Rightarrow - 14x + 49 - 2y + 1 + 6x - 9 + 10y - 25 = 0 \)
\( \Rightarrow -14x + 6x + 10y - 2y + 16 = 0 \)
\( \Rightarrow -8x + 8y + 16 = 0 \)
Divide throughout by 8:
\( \Rightarrow x - y - 2 = 0 \)
โEquidistant means equal distances โ so square them, equate them, and simplify.โ
๐ Trigger words for your brain: Equidistant โ Distance formula โ Square โ Linear equation
Any point on the y-axis has its x-coordinate equal to 0.
So, a general point on the y-axis is of the form (0, y)
- Notice the phrase โon the y-axisโ โ this immediately fixes one coordinate.
- Since the point must be equidistant from A and B, apply the distance formula.
- To avoid square roots, compare squares of distances.
- You will get a linear equation in y.
Let the required point be P(0, y) on the y-axis.
Given that P is equidistant from A(6, 5) and B(โ4, 3):
AP = BP โ APยฒ = BPยฒ
\( (6 - 0)^2 + (5 - y)^2 = (-4 - 0)^2 + (3 - y)^2 \)
\( \Rightarrow 36 + (25 - 10y + y^2) = 16 + (9 - 6y + y^2) \)
\( \Rightarrow 61 + y^2 - 10y = 25 + y^2 - 6y \)
Subtract yยฒ from both sides and rearrange:
\( \Rightarrow -10y + 6y = 25 - 61 \)
\( \Rightarrow -4y = -36 \)
\( y = 9 \)
\( AP = \sqrt{(6 - 0)^2 + (5 - 9)^2} = \sqrt{36 + 16} = \sqrt{52} \)
\( BP = \sqrt{(-4 - 0)^2 + (3 - 9)^2} = \sqrt{16 + 36} = \sqrt{52} \)
โOn the y-axis โ x = 0. Equidistant โ square the distances and equate.โ
๐ Brain trigger sequence: Axis โ Coordinate fixed โ Distance formula โ Linear equation