Chapter 7: Coordinate Geometry

Overview

This page provides comprehensive Coordinate Geometry Class 10 Notes. Free Class 10 Maths Chapter 7 Coordinate Geometry notes. Covers Distance Formula, Section Formula, Midpoint for CBSE Board Exams.

Distance Formula • Section Formula • Midpoint • Centroid

Exam Weightage & Blueprint

Total: 6 Marks

This chapter falls under Unit III: Coordinate Geometry (6 marks total). As per the latest syllabus, focus is on: Review of concepts of coordinate geometry, graphs of linear equations, Distance formula, and Section formula (internal division).

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Distance, Midpoint
Short Answer	2 or 3	High	Section Formula
Case Study	4	Medium	Real-life applications (positions, maps)

⏰ Last 24-Hour Checklist

Distance Formula: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Distance from Origin: $\sqrt{x^2 + y^2}$.

Section Formula: $(\frac{m_1x_2+m_2x_1}{m_1+m_2}, \frac{m_1y_2+m_2y_1}{m_1+m_2})$.
Mid-point Formula: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
Centroid: $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.

📐 Concepts & Definitions

Coordinate Geometry: The branch of mathematics that links geometry and algebra using coordinates on a Cartesian plane.

Key Components

Distance Formula

Measures length between two points $(x_1, y_1)$ and $(x_2, y_2)$.

Section Formula

Coordinates of a point dividing a segment in ratio $m_1 : m_2$ (or $k:1$).

Points of Trisection

Two points dividing a segment into 3 equal parts (ratios $1:2$ and $2:1$).

Collinear Points (Distance Method): Points $A, B, C$ are collinear if the sum of lengths of any two segments equals the third (e.g. $AB + BC = AC$).

🧮 Important Formulas

1. Distance Formula

$$ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
Distance from Origin $(0, 0)$ to $(x, y)$: $$ D = \sqrt{x^2 + y^2} $$

2. Section Formula (Internal Division)

$$ P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) $$
When using ratio $k:1$: $$ P(x, y) = \left( \frac{kx_2 + x_1}{k + 1}, \frac{ky_2 + y_1}{k + 1} \right) $$

3. Mid-point Formula

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

Figure: Distance Formula (Pythagorean Triangle) & Section Formula (Internal Division m₁ : m₂)

4. Points of Trisection of Line Segment $AB$

The points $P$ and $Q$ which divide segment $AB$ into three equal parts ($AP = PQ = QB$):

$$ P \text{ divides } AB \text{ in ratio } 1:2 \qquad \text{and} \qquad Q \text{ divides } AB \text{ in ratio } 2:1 $$ $$ (\text{Alternatively, } Q \text{ is the midpoint of } PB) $$

5. Centroid of a Triangle

The coordinates of the centroid $G$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are:

$$ G(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$

Solved Examples (Board Marking Scheme)

Q1. Find the distance between the points (2, 3) and (4, 1). (2 Marks)

Step 1: Identify Parameters 0.5 Mark

$(x_1, y_1) = (2, 3)$, $(x_2, y_2) = (4, 1)$

Step 2: Apply Formula 1 Mark

$D = \sqrt{(4-2)^2 + (1-3)^2}$

$D = \sqrt{2^2 + (-2)^2}$

Step 3: Calculate 0.5 Mark

$D = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.

Q2. Find the coordinates of the point which divides the line segment joining the points (4, -3) and (8, 5) in the ratio 3:1 internally. (3 Marks)

Step 1: Identify Parameters 1 Mark

$(x_1, y_1) = (4, -3)$, $(x_2, y_2) = (8, 5)$, $m_1=3, m_2=1$

Step 2: Apply Formula 1.5 Marks

$x = \frac{3(8) + 1(4)}{3+1} = \frac{24+4}{4} = 7$

$y = \frac{3(5) + 1(-3)}{3+1} = \frac{15-3}{4} = 3$

Step 3: Answer 0.5 Mark

The point is (7, 3).

Q3. Find the area of the triangle whose vertices are (1, -1), (-4, 6) and (-3, -5). (3 Marks)

Q3. Find the coordinates of the points of trisection of the line segment joining $(2, -2)$ and $(-7, 4)$. (3 Marks)

Step 1: Understand Trisection Ratios 0.5 Mark

Let $P$ and $Q$ be the points of trisection of $AB$. Then $AP = PQ = QB$.

$P$ divides $AB$ internally in the ratio $1 : 2$, and $Q$ divides $AB$ in the ratio $2 : 1$ (or $Q$ is midpoint of $PB$).

Step 2: Find Coordinates of P ($m_1:m_2 = 1:2$) 1.5 Marks

$x_P = \frac{1(-7) + 2(2)}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1$

$y_P = \frac{1(4) + 2(-2)}{1 + 2} = \frac{4 - 4}{3} = 0 \implies P = (-1, 0)$

Step 3: Find Coordinates of Q (Midpoint of PB) 1 Mark

$x_Q = \frac{-1 + (-7)}{2} = \frac{-8}{2} = -4$

$y_Q = \frac{0 + 4}{2} = 2 \implies Q = (-4, 2)$

Points of trisection are $(-1, 0)$ and $(-4, 2)$.

Previous Year Questions (PYQs)

2023 (1 Mark): Find the distance of the point (3, 4) from the origin.
Ans: $D = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$.

2020 (3 Marks): Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4).
Ans: Let the ratio be k:1 and point on y-axis be (0, y). $0 = \frac{k(-1)+1(5)}{k+1} \Rightarrow -k+5=0 \Rightarrow k=5$. Ratio is 5:1.

2019 (4 Marks): If A(1,2), B(4,y), C(x,6) and D(3,5) are the vertices of a parallelogram taken in order, find x and y.
Ans: Diagonals of a parallelogram bisect each other. Midpoint of AC = Midpoint of BD. $(\frac{1+x}{2}, \frac{2+6}{2}) = (\frac{4+3}{2}, \frac{y+5}{2})$. $1+x=7 \Rightarrow x=6$. $8=y+5 \Rightarrow y=3$.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Sign Errors: Be careful with double negatives like $(x_2 - (-x_1))^2 = (x_2 + x_1)^2$ in distance and section formulas.

Mixing x and y: Don't mix up $x$ and $y$ coordinates when substituting into the section formula.

Equidistant Points: In $PA = PB$, always square both sides ($PA^2 = PB^2$) immediately to avoid square roots.

💡 Scoring Tips

Ratio Shortcut: Always assume unknown division ratio as $k : 1$. If point is on x-axis, set $y=0$. If on y-axis, set $x=0$.

Check for Collinearity: Show $AB + BC = AC$ using Distance Formula.

Parallelogram Vertex: Use equality of midpoints of diagonals instead of distance formula (saves 5 minutes).

Self-Assessment Mock Test (10 Marks)

Q1 (1M): Find the coordinates of the midpoint of the line segment joining the points $(-2, 3)$ and $(4, -5)$.

Q2 (2M): Find the point on the x-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.

Q3 (3M): Determine if the points $(1, 5)$, $(2, 3)$, and $(-2, -11)$ are collinear using the distance formula.

Q4 (4M): If $A(1, 2), B(4, y), C(x, 6)$ and $D(3, 5)$ are vertices of a parallelogram taken in order, find the values of $x$ and $y$.

📝 More Solved Board Questions

Q3. Find the coordinates of the point equidistant from the points A(1, 2), B(3, -4), and C(5, -6). 3 Marks

Sol. Let the point be $P(x, y)$. Since it is equidistant, $PA = PB = PC$.

$PA^2 = PB^2 \Rightarrow (x-1)^2 + (y-2)^2 = (x-3)^2 + (y+4)^2$

$x^2-2x+1 + y^2-4y+4 = x^2-6x+9 + y^2+8y+16$

$4x - 12y = 20 \Rightarrow x - 3y = 5$ ...(1)

Similarly $PB^2 = PC^2$ gives another equation. Solve (1) and (2) to get $(x, y)$.

Answer: (11, 2)

Q4. Find the ratio in which the line $2x + y - 4 = 0$ divides the line segment joining A(2, -2) and B(3, 7). 4 Marks

Sol. Let the ratio be $k:1$.

Point $P = \left( \frac{3k+2}{k+1}, \frac{7k-2}{k+1} \right)$.

Substitute $P$ in the line equation $2x + y - 4 = 0$:

$2\left( \frac{3k+2}{k+1} \right) + \left( \frac{7k-2}{k+1} \right) - 4 = 0$

$6k + 4 + 7k - 2 - 4k - 4 = 0 \Rightarrow 9k = 2 \Rightarrow k = \frac{2}{9}$.

Ratio = 2:9

🎯 Board Pattern (2018–2025): Questions involving "equidistant points" or "points on x-axis/y-axis" are very common. Remember: Point on x-axis is $(x, 0)$ and point on y-axis is $(0, y)$.

📋 Board Revision Checklist

✅ Distance between $(x_1, y_1)$ and $(x_2, y_2)$: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
✅ Distance from origin $(0, 0)$: $\sqrt{x^2 + y^2}$
✅ Section Formula (Internal): $\left( \frac{m_1x_2 + m_2x_1}{m_1+m_2}, \frac{m_1y_2 + m_2y_1}{m_1+m_2} \right)$
✅ Midpoint Formula: $\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$
✅ Centroid of Triangle: $\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$
✅ Collinearity: Show that $AB + BC = AC$ (Distance Method)
✅ Deleted Topic: Area of a triangle formula is NOT in 2025-26 syllabus
✅ Point on x-axis: $(x, 0)$; Point on y-axis: $(0, y)$

💡 Exam Tip:
When calculating ratio, always assume it as $k:1$. It reduces the number of variables to one, making the calculation much faster.

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. The distance of the point $P(2, 3)$ from the x-axis is:

2. The distance between the points $(0, 5)$ and $(-5, 0)$ is:

3. If the midpoint of $(4, 0)$ and $(0, y)$ is $(2, 3)$, then $y$ is:

4. The ratio in which the x-axis divides the join of $(2, -3)$ and $(5, 6)$ is:

5. The coordinates of the centroid of a triangle with vertices $(0, 6), (8, 12)$ and $(8, 0)$ are: