Class 10 Maths Ch 7: Coordinate Geometry Notes

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)

CBSE 2026–27 Syllabus Status: Coordinate Geometry

Unit III: Coordinate Geometry • 6 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

Deleted / Not in Board Exam:

⏰ 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)

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

3. Mid-point Formula

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

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 ($AP = PQ = QB$):

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

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:

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

⚡ High-Yield Conceptual Rules (Board MCQs & A-R):

Perpendicular Distances from Axes (Classic MCQ Trap):
• Distance of $P(x, y)$ from $x$-axis $= |y|$ (absolute value of ordinate).
• Distance of $P(x, y)$ from $y$-axis $= |x|$ (absolute value of abscissa).
Division by Coordinate Axes:
• If divided by $x$-axis, the intersection point is $(x, 0) \implies$ equate $y$-coordinate to 0.
• If divided by $y$-axis, the intersection point is $(0, y) \implies$ equate $x$-coordinate to 0.
Geometric Quadrilateral Checklist:
• Parallelogram: Midpoint of diagonal $AC$ = Midpoint of diagonal $BD$.
• Rectangle: Parallelogram with equal diagonals ($AC = BD$).
• Rhombus: All 4 sides equal, but diagonals unequal ($AC \neq BD$).
• Square: All 4 sides equal AND diagonals equal ($AC = BD$).
Collinearity Condition: Points $A, B, C$ are collinear if $AB + BC = AC$ (or $AB + AC = BC$).

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

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

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

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

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 $(0, y)$.

CBSE Section E • 4 Marks

Case Study: Sports Day Field & Relay Baton Handover

Scenario: On an inter-school annual sports day, rectangular track lines are marked with chalk powder at intervals of 1 meter each. 100 flower pots have been placed at a distance of 1 m from each other along the boundary AD. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag at point G. Preet runs 1/5th the distance AD on the 8th line and posts a red flag at point R.

Q1: Find the coordinates of the green flag G posted by Niharika. 1 Mark

Q2: Find the coordinates of the red flag R posted by Preet. 1 Mark

Q3: Find the distance between both flags. (OR) If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, at what coordinates should she post her flag? 2 Marks

📋 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): $( \frac{m_1x_2 + m_2x_1}{m_1+m_2}, \frac{m_1y_2 + m_2y_1}{m_1+m_2} )$
✅ Midpoint Formula: $( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} )$
✅ Centroid of Triangle: $( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} )$
✅ Distance from x-axis $= |y|$; Distance from y-axis $= |x|$
✅ Divided by x-axis $\implies y = 0$; Divided by y-axis $\implies x = 0$
✅ Parallelogram verification: midpoints of diagonals $AC$ and $BD$ coincide
✅ Collinearity: Show that $AB + BC = AC$ (Distance Method)
✅ Excluded Topic: The area-of-a-triangle formula is not included in the 2026-27 CBSE Class 10 Mathematics 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:

Frequently Asked Questions: Class 10 Maths Chapter 7

What is the weightage of Chapter 7 Coordinate Geometry in CBSE Class 10?

Coordinate Geometry carries 6 marks in Unit 3: Coordinate Geometry, usually comprising two 1-mark questions and one 2-mark or 4-mark case study problem.

Is the Area of Triangle formula included in the Class 10 syllabus?

No, the area of triangle formula in coordinate geometry has been removed from the rationalized CBSE syllabus. The active topics are the Distance Formula and the Section Formula (internal division only).

What is the Section Formula for internal division?

The coordinates of point P(x, y) dividing line segment AB in ratio m1:m2 internally are ((m1*x2 + m2*x1)/(m1 + m2), (m1*y2 + m2*y1)/(m1 + m2)). The midpoint formula is ((x1 + x2)/2, (y1 + y2)/2).

How do you prove three points are collinear using the distance formula?

Points A, B, and C are collinear if the distance between any two points equals the sum of the distances between the other pairs (for instance, AB + BC = AC).

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Chapter 7: Coordinate Geometry

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