Coordinate Geometry Class 10 Notes | Chapter 7 Maths

Exam Weightage & Blueprint

Total: 6 Marks

This chapter is part of the Coordinate Geometry Unit. It's a very important and high-scoring chapter.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Distance, Midpoint
Short Answer	2 or 3	High	Section Formula, Area of Triangle
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: $(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$.
Mid-point Formula: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
Area of Triangle: $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.

Concepts & Definitions ★★★★★

Coordinate Geometry: The study of geometry using a coordinate system.

Key Components

Distance Formula

Finds the length of a line segment.

Section Formula

Divides a line segment in a given ratio.

Area of a Triangle

Calculates the area of a triangle from its vertices.

Collinear Points: Three points are collinear if the sum of the lengths of any two line segments is equal to the length of the third line segment. Or, the area of the triangle formed by them is zero.

Important Formulas 🔥🔥🔥

1. Distance Formula

$$ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

2. Section Formula

$$ P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) $$

3. Mid-point Formula

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

4. Area of a Triangle

$$ Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $$

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)

Step 1: Identify Parameters 0.5 Mark

$(x_1, y_1) = (1, -1)$, $(x_2, y_2) = (-4, 6)$, $(x_3, y_3) = (-3, -5)$

Step 2: Apply Formula 1.5 Marks

$Area = \frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|$

$Area = \frac{1}{2} |1(11) + (-4)(-4) + (-3)(-7)|$

Step 3: Solve 1 Mark

$Area = \frac{1}{2} |11 + 16 + 21| = \frac{1}{2} |48| = 24$ sq. units.

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 negative coordinates in the distance and section formulas.

Mixing x and y: Don't mix up x and y coordinates when applying formulas.

Area Formula: The area formula is long. Practice it well. Remember the absolute value for the final answer.

Scoring Tips 🏆

Draw a Diagram: A rough sketch of the points can help you visualize the problem.

Check for Collinearity: To check if 3 points are collinear, the easiest method is to show the area of the triangle is zero.

Parallelogram Properties: Remember the properties of parallelograms, rhombuses, squares, and rectangles.

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 value of 'a' so that the point (3, a) lies on the line represented by 2x - 3y = 5.

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

Q4 (4M): Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

Chapter 7: Coordinate Geometry