Coordinate Geometry — Detailed Notes

Chapter 7 | CBSE Class 10 Mathematics (NCERT)

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Coordinate Geometry — Decision Map

Coordinate Geometry — Formula Strip

Distance & Section on the Plane

7.1 Introduction

To locate a point in a plane you need a pair of axes. The $x$ -coordinate (abscissa) is the distance from the $y$ -axis; the $y$ -coordinate (ordinate) is the distance from the $x$ -axis.

Point on $x$ -axis: $(x, 0)$
Point on $y$ -axis: $(0, y)$

Coordinate geometry studies geometric figures using algebra (and vice versa). This chapter focuses on distance and section formulae.

7.2 Distance Formula

Special cases

On $x$ -axis: $A (4, 0)$ , $B (6, 0)$ $\Rightarrow$ $A B = ∣6 - 4∣ = 2$
On $y$ -axis: $C (0, 3)$ , $D (0, 8)$ $\Rightarrow$ $C D = ∣8 - 3∣ = 5$
Mixed axis points: use Pythagoras, e.g. $A (4, 0)$ , $C (0, 3)$ $\Rightarrow$ $A C = 5$

General formula

For $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ , drop perpendiculars to the axes and form a right triangle. By Pythagoras: $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} .$ Distance is non-negative, so take the positive square root. Order of points does not matter: $P Q = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} .$

Distance from origin $O (0, 0)$ to $P (x, y)$ : $O P = x^{2} + y^{2} .$

Worked ideas

Triangle type (Example 1): Points $(3, 2)$ , $(- 2, - 3)$ , $(2, 3)$ . Compute three distances; triangle inequality holds. Check $P Q^{2} + P R^{2} = Q R^{2}$ $\Rightarrow$ right-angled at $P$ .

Square (Example 2): Show four sides equal and diagonals equal; or four sides equal and one angle $9 0^{\circ}$ via converse of Pythagoras ( $A D^{2} + D C^{2} = A C^{2}$ ).

Better Method / Strategy: For “is it a square?”, either (all sides + equal diagonals) or (all sides + one right angle). The second often needs fewer calculations if one diagonal is already computed.

Collinear seating (Example 3): $A (3, 1)$ , $B (6, 4)$ , $C (8, 6)$ : $A B + B C = 32 + 22 = 52 = A C \Rightarrow collinear .$

Equidistant relation (Example 4): $P (x, y)$ equidistant from $(7, 1)$ and $(3, 5)$ : $A P^{2} = B P^{2} \Rightarrow (x - 7)^{2} + (y - 1)^{2} = (x - 3)^{2} + (y - 5)^{2} \Rightarrow x - y = 2.$ (This is the perpendicular bisector of the segment joining the two points.)

Point on $y$ -axis (Example 5): $P (0, y)$ equidistant from $(6, 5)$ and $(- 4, 3)$ $\Rightarrow$ $y = 9$ , so $(0, 9)$ .

7.3 Section Formula

Motivation

A point $P$ on $A B$ with $P A : P B = 1 : 2$ (relay-tower style) can be found using similar right triangles; this generalises to any ratio.

Internal section formula

If $P (x, y)$ divides the join of $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally in the ratio $m_{1} : m_{2}$ (i.e. $P A : P B = m_{1} : m_{2}$ ), then $P = (\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}}) .$

$k : 1$ form: If the ratio is $k : 1$ , $P = (\frac{k x _{2} + x _{1}}{k + 1}, \frac{k y _{2} + y _{1}}{k + 1}) .$

Mid-point (special case $1 : 1$ )

$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}) .$

Worked ideas

Given ratio (Example 6): Divide $(4, - 3)$ and $(8, 5)$ in $3 : 1$ : $x = \frac{3 \cdot 8 + 1 \cdot 4}{4} = 7, y = \frac{3 \cdot 5 + 1 \cdot ( - 3 )}{4} = 3 \Rightarrow (7, 3) .$

Find ratio (Example 7): Point $(- 4, 6)$ divides $A (- 6, 10)$ and $B (3, - 8)$ . Let ratio be $k : 1$ : $- 4 = \frac{3 k - 6}{k + 1} \Rightarrow k = \frac{2}{7} \Rightarrow ratio 2 : 7.$ Verify with the $y$ -coordinate.

Better Method / Strategy: Prefer $k : 1$ when finding an unknown ratio — one variable instead of two. Always verify the second coordinate.

Trisection (Example 8): Points dividing $A (2, - 2)$ , $B (- 7, 4)$ in three equal parts: ratios $1 : 2$ and $2 : 1$ give $(- 1, 0)$ and $(- 4, 2)$ . (Second point is also mid-point of first trisection point and $B$ .)

Axis intersection (Example 9): $y$ -axis divides join of $(5, - 6)$ and $(- 1, - 4)$ . Let ratio $k : 1$ ; set $x = 0$ : $\frac{- k + 5}{k + 1} = 0 \Rightarrow k = 5 \Rightarrow point (0, - \frac{13}{3}) .$

Parallelogram (Example 10): Vertices $A (6, 1)$ , $B (8, 2)$ , $C (9, 4)$ , $D (p, 3)$ . Diagonals bisect: $mid-point of A C = mid-point of B D \Rightarrow p = 7.$

Note (external division): If $P$ lies on line $A B$ but outside segment $A B$ , division is external — studied in higher classes. Class 10 NCERT focuses on internal division.

Extra Board-Style Patterns

Pattern A — Classify the figure

Goal	Compute	Conclude
Right triangle	Check $a^{2} + b^{2} = c^{2}$	Right-angled opposite $c$
Isosceles	Two sides equal	Base angles equal (geometry)
Equilateral	All three equal	Equilateral
Square	4 sides equal + diagonals equal (or one right angle)	Square
Rhombus (not square)	4 sides equal, diagonals unequal	Rhombus
Rectangle	Opposite sides + equal diagonals, not all sides equal	Often needs more checks
Collinear	$A B + B C = A C$	On a straight line

Pattern B — Equidistant locus

$P (x, y)$ equidistant from $A$ and $B$ $\Rightarrow$ $P A^{2} = P B^{2}$ $\Rightarrow$ linear equation (perpendicular bisector of $A B$ ).

Pattern C — Axis cuts a segment

Use section formula with ratio $k : 1$ .

Hits $y$ -axis $\Rightarrow$ set $x = 0$ , solve $k$ .
Hits $x$ -axis $\Rightarrow$ set $y = 0$ , solve $k$ .
Then substitute $k$ to get the point.

Pattern D — Equal division points

Division	Ratios from $A$ to $B$
Mid-point	$1 : 1$
Trisection	$1 : 2$ and $2 : 1$
Four equal parts	$1 : 3$ , $1 : 1$ , $3 : 1$

Pattern E — Parallelogram / diameter

Parallelogram (order $A, B, C, D$ ): mid-point of $A C$ = mid-point of $B D$ .
Diameter $A B$ with centre $C$ : $C$ is mid-point of $A$ and $B$ .

Pattern F — “ $A P = \frac{m}{n} A B$ on the segment”

If $P$ lies on $A B$ and $A P = \frac{m}{n} A B$ , then $A P : P B = m : (n - m)$ . Example: $A P = \frac{3}{7} A B$ $\Rightarrow$ ratio $3 : 4$ .

Memory Hooks for Section Formula

Think: “ $m_{1}$ pulls toward $B$ ” — the weight $m_{1}$ multiplies coordinates of $B (x_{2}, y_{2})$ .

$x = \frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}$

Check with mid-point: $m_{1} = m_{2} = 1$ recovers the average.

Worked Mini-Set (Exam Pace)

1. Distance $(a, b)$ to $(- a, - b)$ : $(2 a)^{2} + (2 b)^{2} = 2 a^{2} + b^{2}$ .

2. Point on $x$ -axis equidistant from $(2, - 5)$ and $(- 2, 9)$ : let $(x, 0)$ ; $(x - 2)^{2} + 25 = (x + 2)^{2} + 81 \Rightarrow x = - \frac{7}{2} (after simplifying) .$ (Verify arithmetic carefully in exam.)

3. Divide $(- 1, 7)$ and $(4, - 3)$ in $2 : 3$ : $(\frac{2 \cdot 4 + 3 \cdot ( - 1 )}{5}, \frac{2 \cdot ( - 3 ) + 3 \cdot 7}{5}) = (1, 3) .$

Exam-Oriented Analysis

Aspect	Detail
Weight	Typically 3–5 marks
Question types	Distance; classify $△$ /quadrilateral; collinear; section; mid-point; find ratio; missing coordinate
1-mark	Direct distance / mid-point
2–3 mark	Section ratio or figure type
3–4 mark	Combined (trisection + distance, parallelogram + unknown)
Common mistakes	Swapping $m_{1} x_{2}$ with $m_{1} x_{1}$ ; forgetting $P A : P B = m_{1} : m_{2}$ ; using $+$ instead of checking collinearity with largest side; sign errors with negative coordinates
Scoring tips	Write formula once, substitute carefully; for equidistant use squares; state “diagonals bisect” for parallelograms

Scoring tip: In classification questions, compute all needed lengths systematically and state the geometric property used (e.g. “all sides equal and diagonals equal $\Rightarrow$ square”).

Method chooser

Need length / type / collinear? → Distance
Need a point in known ratio? → Section
Need unknown ratio? → k:1 then verify other coordinate
Axis intersection? → Section + x=0 or y=0
Parallelogram vertex missing? → Equal mid-points of diagonals