Coordinate Geometry | Mathematics Class 10

Internal Division of Line Segment

Concept Used: Section Formula (Internal Division)

If a point divides a line segment joining two points $ (x_1, y_1) $ and $ (x_2, y_2) $ internally in the ratio $ m:n $, then its coordinates are:

$$\begin{aligned} x = \dfrac{mx_2 + nx_1}{m+n}, \quad y = \dfrac{my_2 + ny_1}{m+n} \end{aligned}$$

Here, "internal division" means the point lies between the two given points.

Solution Roadmap

Identify coordinates of given points
Identify ratio $ m:n $
Apply section formula
Simplify step-by-step carefully

Solution:

Let the required point be $ P(x, y) $ which divides the line joining $ A(-1,7) $ and $ B(4,-3) $ in the ratio $ 2:3 $.

Comparing with standard form:

$$\begin{aligned} x_1 = -1, \quad y_1 = 7 \\ x_2 = 4, \quad y_2 = -3 \\ m = 2, \quad n = 3 \end{aligned}$$

Using section formula:

$$\begin{aligned} x &= \dfrac{mx_2 + nx_1}{m+n} \end{aligned}$$ $$\begin{aligned} x &= \dfrac{2 \times 4 + 3 \times (-1)}{2+3} \\ &= \dfrac{8 - 3}{5} \\ &= \dfrac{5}{5} \\ &= 1 \end{aligned}$$ $$\begin{aligned} y &= \dfrac{my_2 + ny_1}{m+n} \end{aligned}$$ $$\begin{aligned} y &= \dfrac{2 \times (-3) + 3 \times 7}{5} \\ &= \dfrac{-6 + 21}{5} \\ &= \dfrac{15}{5} \\ &= 3 \end{aligned}$$

Therefore, the required point is $ (1, 3) $.

Exam Significance

Very frequently asked in CBSE Board Exams (2–3 marks guaranteed)
Foundation concept for Coordinate Geometry in Class 11
Used in midpoint, centroid, and vector-based problems
Important for NDA, SSC, Banking and other competitive exams
Common trap: wrong substitution of $ x_1, x_2 $

Trisection of Line Segment

Concept Used: Trisection using Section Formula

Trisection means dividing a line segment into three equal parts.

The two required points divide the line internally in ratios:

First point → $ 1:2 $
Second point → $ 2:1 $

$$\begin{aligned} x = \dfrac{mx_2 + nx_1}{m+n}, \quad y = \dfrac{my_2 + ny_1}{m+n} \end{aligned}$$

Solution Roadmap

Identify endpoints A and B
Apply section formula for ratio 1:2
Apply section formula for ratio 2:1
Simplify carefully step-by-step

Solution:

Let A(4, –1) and B(–2, –3)

$$\begin{aligned} x_1 = 4,\quad y_1 = -1 \\ x_2 = -2,\quad y_2 = -3 \end{aligned}$$

First trisection point (ratio 1 : 2)

$$\begin{aligned} x &= \dfrac{1 \cdot (-2) + 2 \cdot 4}{1+2} \\ &= \dfrac{-2 + 8}{3} \\ &= \dfrac{6}{3} \\ &= 2 \end{aligned}$$ $$\begin{aligned} y &= \dfrac{1 \cdot (-3) + 2 \cdot (-1)}{3} \\ &= \dfrac{-3 - 2}{3} \\ &= \dfrac{-5}{3} \end{aligned}$$

First point = $ (2, -\frac{5}{3}) $

Second trisection point (ratio 2 : 1)

$$\begin{aligned} x &= \dfrac{2 \cdot (-2) + 1 \cdot 4}{3} \\ &= \dfrac{-4 + 4}{3} \\ &= \dfrac{0}{3} \\ &= 0 \end{aligned}$$ $$\begin{aligned} y &= \dfrac{2 \cdot (-3) + 1 \cdot (-1)}{3} \\ &= \dfrac{-6 - 1}{3} \\ &= \dfrac{-7}{3} \end{aligned}$$

Second point = $ (0, -\frac{7}{3}) $

Therefore, the points of trisection are $ (2, -\frac{5}{3}) $ and $ (0, -\frac{7}{3}) $.

Exam Significance

Frequently asked 3–4 mark question in CBSE exams
Tests conceptual clarity of section formula
Common error: mixing ratios (1:2 vs 2:1)
Important for coordinate geometry and vectors (Class 11)
Useful in NDA, SSC, Banking exams

Flag Positions on Ground Grid

Niharika runs $ \frac{1}{4} $th of AD on 2nd line → green flag Preet runs $ \frac{1}{5} $th of AD on 8th line → red flag

Concept Used

Coordinate representation of real-life grid
Distance Formula
Midpoint Formula

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] \[\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)\]

Solution Roadmap

Convert ground into coordinate system
Find coordinates of both flags
Apply distance formula
Find midpoint

Solution:

Step 1: Convert into coordinates

Total length AD = 100 m

Niharika:

$$\begin{aligned} y &= \frac{1}{4} \times 100 = 25 \\ x &= 2 \\ \Rightarrow \text{Coordinates} = (2, 25) \end{aligned}$$

Preet:

$$\begin{aligned} y &= \frac{1}{5} \times 100 = 20 \\ x &= 8 \\ \Rightarrow \text{Coordinates} = (8, 20) \end{aligned}$$

Step 2: Distance between flags

$$\begin{aligned} d &= \sqrt{(8 - 2)^2 + (20 - 25)^2} \\ &= \sqrt{(6)^2 + (-5)^2} \\ &= \sqrt{36 + 25} \\ &= \sqrt{61} \end{aligned}$$

Distance between flags = $ \sqrt{61} $ m

Step 3: Midpoint (Rashmi’s flag)

$$\begin{aligned} x &= \frac{2 + 8}{2} = \frac{10}{2} = 5 \\ y &= \frac{25 + 20}{2} = \frac{45}{2} = 22.5 \end{aligned}$$

Midpoint = $ (5, 22.5) $

Rashmi should place the blue flag on the 5th line at a distance of 22.5 m.

Exam Significance

Very important application-based question (CBSE HOTS)
Tests conversion of real situation → coordinate plane
Combines distance + midpoint (2 concepts in one)
Common mistake: wrong coordinate assignment
Highly relevant for NDA, SSC, and foundation maths

Point dividing line internally

Concept Used: Reverse Section Formula

If a point divides a line internally in ratio $ m:n $, then:

\[x = \frac{mx_2 + nx_1}{m+n}\]

Here, coordinates are known → we find ratio.

Solution Roadmap

Assume ratio = m : n
Substitute given point into section formula
Solve linear equation
Verify using y-coordinate (optional check)

Solution:

Let A(–3, 10), B(6, –8) and P(–1, 6) divides AB in ratio $ m:n $

$$\begin{aligned} x_1 = -3,\quad y_1 = 10 \\ x_2 = 6,\quad y_2 = -8 \end{aligned}$$

Using x-coordinate

$$\begin{aligned} -1 &= \frac{m(6) + n(-3)}{m+n} \end{aligned}$$ Multiply both sides by $ (m+n) $: $$\begin{aligned} -1(m+n) &= 6m - 3n \\ -m - n &= 6m - 3n \end{aligned}$$ Rearranging: $$\begin{aligned} -m - n - 6m + 3n &= 0 \\ -7m + 2n &= 0 \\ 7m &= 2n \\ \frac{m}{n} &= \frac{2}{7} \end{aligned}$$

Ratio = $ 2 : 7 $

Verification using y-coordinate

$$\begin{aligned} 6 &= \frac{m(-8) + n(10)}{m+n} \end{aligned}$$ Substitute $ m:n = 2:7 $: $$\begin{aligned} &= \frac{2(-8) + 7(10)}{9} \\ &= \frac{-16 + 70}{9} \\ &= \frac{54}{9} \\ &= 6 \quad \checkmark \end{aligned}$$

Hence verified, ratio = $ 2:7 $

Exam Significance

Very common CBSE 3-mark question
Tests reverse thinking (finding ratio)
Common mistake: sign handling errors
Always verify using second coordinate
Important for higher maths and vectors

Division by x-axis (y = 0)

Concept Used

Point on x-axis ⇒ $ y = 0 $
Section Formula
Reverse ratio finding

\[y = \frac{my_2 + ny_1}{m+n}\]

Solution Roadmap

Assume ratio = m : n
Use y = 0 condition
Find ratio
Substitute into x-formula

Solution:

Let A(1, –5), B(–4, 5)

$$\begin{aligned} x_1 = 1,\quad y_1 = -5 \\ x_2 = -4,\quad y_2 = 5 \end{aligned}$$

Let the point P divide AB in the ratio $ m:n $

Since P lies on x-axis:

$$y = 0$$ Using section formula: $$\begin{aligned} 0 &= \frac{m(5) + n(-5)}{m+n} \end{aligned}$$ Multiply both sides: $$\begin{aligned} 0 &= 5m - 5n \\ 5m &= 5n \\ m &= n \end{aligned}$$

Ratio = $ 1 : 1 $

Now find coordinates of point

$$\begin{aligned} x &= \frac{m x_2 + n x_1}{m+n} \end{aligned}$$ Substitute $ m = n = 1 $: $$\begin{aligned} x &= \frac{(-4) + 1}{2} \\ &= \frac{-3}{2} \end{aligned}$$ $$y = 0$$

Required point = $ \left(-\frac{3}{2}, 0 \right) $

Exam Significance

Classic CBSE question (3 marks)
Key idea: axis ⇒ coordinate becomes zero
Shortcut: opposite y-values ⇒ midpoint
Common mistake: wrong substitution in formula
Useful in coordinate geometry & vectors

Diagonals of Parallelogram Bisect Each Other

Concept Used: Diagonals of Parallelogram

Diagonals of a parallelogram bisect each other.

So, midpoint of AC = midpoint of BD

\[\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\]

Solution Roadmap

Label points A, B, C, D
Find midpoint of AC
Find midpoint of BD
Equate coordinates
Solve for x and y

Solution:

Let A(1,2), B(4,y), C(x,6), D(3,5)

Step 1: Midpoint of AC

$$\begin{aligned} M_1 &= \left(\frac{1 + x}{2}, \frac{2 + 6}{2}\right) \\ &= \left(\frac{1 + x}{2}, \frac{8}{2}\right) \\ &= \left(\frac{1 + x}{2}, 4\right) \end{aligned}$$

Step 2: Midpoint of BD

$$\begin{aligned} M_2 &= \left(\frac{4 + 3}{2}, \frac{y + 5}{2}\right) \\ &= \left(\frac{7}{2}, \frac{y + 5}{2}\right) \end{aligned}$$

Step 3: Equate midpoints

Since diagonals bisect each other:

$$M_1 = M_2$$

Equating x-coordinates:

$$\begin{aligned} \frac{1 + x}{2} &= \frac{7}{2} \\ 1 + x &= 7 \\ x &= 6 \end{aligned}$$

Equating y-coordinates:

$$\begin{aligned} 4 &= \frac{y + 5}{2} \\ 8 &= y + 5 \\ y &= 3 \end{aligned}$$

Therefore, $ x = 6 $ and $ y = 3 $

Exam Significance

Very important CBSE 4-mark question
Tests understanding of parallelogram properties
Common mistake: wrong midpoint pairing
Can also be solved using vector method (advanced)
Frequently appears in competitive exams

Centre is midpoint of diameter

Concept Used

The centre of a circle is the midpoint of its diameter.

\[\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]

Solution Roadmap

Use midpoint formula
Substitute centre and one endpoint
Solve for unknown coordinates

Solution:

Let coordinates of A be $ (x, y) $ and B = (1, 4)

Given centre = (2, –3)

Using midpoint formula:

$$\begin{aligned} \left(2, -3\right) = \left( \frac{x + 1}{2}, \frac{y + 4}{2} \right) \end{aligned}$$

Equating x-coordinates:

$$\begin{aligned} 2 &= \frac{x + 1}{2} \\ 4 &= x + 1 \\ x &= 3 \end{aligned}$$

Equating y-coordinates:

$$\begin{aligned} -3 &= \frac{y + 4}{2} \\ -6 &= y + 4 \\ y &= -10 \end{aligned}$$

Therefore, coordinates of A = $ (3, -10) $

Exam Significance

Very common CBSE 3-mark question
Direct application of midpoint concept
Common mistake: wrong substitution order
Also appears in circle geometry problems
Useful in coordinate & vector geometry

Point dividing line internally

Concept Used

Internal division of line segment
Ratio conversion from given condition
Section Formula

Solution Roadmap

Convert AP = 3/7 AB into ratio AP : PB
Apply section formula
Substitute values carefully

Solution:

Given A(–2, –2), B(2, –4)

Step 1: Find ratio AP : PB

$$\begin{aligned} AP &= \frac{3}{7} AB \\ PB &= AB - AP \\ &= AB - \frac{3}{7}AB \\ &= \frac{4}{7}AB \end{aligned}$$ $$\begin{aligned} \frac{AP}{PB} = \frac{3/7}{4/7} = \frac{3}{4} \end{aligned}$$

Ratio = $ 3 : 4 $

Step 2: Apply section formula

$$\begin{aligned} x &= \frac{3(2) + 4(-2)}{3+4} \\ &= \frac{6 - 8}{7} \\ &= \frac{-2}{7} \end{aligned}$$ $$\begin{aligned} y &= \frac{3(-4) + 4(-2)}{7} \\ &= \frac{-12 - 8}{7} \\ &= \frac{-20}{7} \end{aligned}$$

Therefore, $ P = \left( -\frac{2}{7}, -\frac{20}{7} \right) $

Exam Significance

Frequently asked CBSE 3-mark question
Tests ratio conversion (very important step)
Common mistake: directly using 3:7 instead of 3:4
Strong foundation for vectors and coordinate geometry
Appears in NDA, SSC, Banking exams

Division into four equal parts

Concept Used

To divide a line into 4 equal parts, we find midpoints repeatedly:
First midpoint → divides into 2 parts Then midpoints of each half → 4 equal parts

Solution Roadmap

Find midpoint P of AB
Find midpoint R of AP
Find midpoint S of PB

Solution:

Let A(–2, 2) and B(2, 8)

Step 1: Midpoint P of AB

$$\begin{aligned} x_P &= \frac{-2 + 2}{2} = 0 \\ y_P &= \frac{2 + 8}{2} = 5 \end{aligned}$$

$ P = (0, 5) $

Step 2: Midpoint R of AP

$$\begin{aligned} x_R &= \frac{-2 + 0}{2} = -1 \\ y_R &= \frac{2 + 5}{2} = \frac{7}{2} \end{aligned}$$

$ R = \left(-1, \frac{7}{2}\right) $

Step 3: Midpoint S of PB

$$\begin{aligned} x_S &= \frac{0 + 2}{2} = 1 \\ y_S &= \frac{5 + 8}{2} = \frac{13}{2} \end{aligned}$$

$ S = \left(1, \frac{13}{2}\right) $

Therefore, the points dividing AB into four equal parts are:
$ \left(-1,\frac{7}{2}\right), (0,5), \left(1,\frac{13}{2}\right) $

Exam Significance

Classic CBSE 4-mark construction-based problem
Tests repeated midpoint concept
Alternative method: section formula (1:3, 2:2, 3:1)
Common mistake: wrong midpoint pairing

Rhombus with diagonals AC and BD

Concept Used

Diagonals of rhombus
Distance Formula
Area = $ \frac{1}{2} \times d_1 \times d_2 $

Solution Roadmap

Find diagonal AC
Find diagonal BD
Apply area formula

Solution:

Let A(3,0), B(4,5), C(–1,4), D(–2,–1)

Step 1: Length of diagonal AC

$$\begin{aligned} AC &= \sqrt{(3 - (-1))^2 + (0 - 4)^2} \\ &= \sqrt{(4)^2 + (-4)^2} \\ &= \sqrt{16 + 16} \\ &= \sqrt{32} \\ &= 4\sqrt{2} \end{aligned}$$

Step 2: Length of diagonal BD

$$\begin{aligned} BD &= \sqrt{(4 - (-2))^2 + (5 - (-1))^2} \\ &= \sqrt{(6)^2 + (6)^2} \\ &= \sqrt{36 + 36} \\ &= \sqrt{72} \\ &= 6\sqrt{2} \end{aligned}$$

Step 3: Area of rhombus

$$\begin{aligned} \text{Area} &= \frac{1}{2} \times AC \times BD \\ &= \frac{1}{2} \times (4\sqrt{2}) \times (6\sqrt{2}) \\ &= \frac{1}{2} \times 24 \times 2 \\ &= 24 \end{aligned}$$

Area of rhombus = 24 square units

Exam Significance

Very important CBSE 4-mark problem
Tests diagonal identification correctly
Common mistake: wrong coordinate pairing
Alternative method: Shoelace formula
Useful in coordinate geometry + mensuration

Chapter 7 — Coordinate Geometry

Concept Used: Section Formula (Internal Division)

Solution Roadmap

Solution:

Exam Significance

Concept Used: Trisection using Section Formula

Solution Roadmap

Solution:

Exam Significance

Concept Used

Solution Roadmap

Solution:

Exam Significance

Concept Used: Reverse Section Formula

Solution Roadmap

Solution:

Exam Significance

Concept Used

Solution Roadmap

Solution:

Exam Significance

Concept Used: Diagonals of Parallelogram

Solution Roadmap

Solution:

Exam Significance

Concept Used

Solution Roadmap

Solution:

Exam Significance

Concept Used

Solution Roadmap

Solution:

Exam Significance

Concept Used

Solution Roadmap

Solution:

Exam Significance

Concept Used

Solution Roadmap

Solution:

Exam Significance

Chapter Complete!

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