CBSE Class 10 Maths — Coordinate Geometry

Chapter Overview & Weightage

Coordinate Geometry is one of the most scoring chapters in Class 10 Maths. Expect 6 marks in board exams — typically a 2-mark question and a 4-mark question.

Year	Marks	Question Types
2024	6	1 SA (2m) + 1 LA (4m)
2023	6	MCQ + SA + LA
2022	6	2 SA (3m each)

This chapter has predictable question patterns. The 4-mark question usually asks you to find a point dividing a line segment in a given ratio, or prove three points are collinear, or find the area of a triangle. Practice all three templates.

Key Concepts You Must Know

1. Distance Formula Distance between points $(x_1, y_1)$ and $(x_2, y_2)$ :

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

2. Section Formula Point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in ratio $m:n$ internally:

\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)

3. Midpoint Formula (special case of section formula with $m = n = 1$ ):

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

4. Area of Triangle Triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ :

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

Collinearity condition: Area = 0 means the three points are collinear.

Important Formulas

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

Use when: finding distance between two points, checking if a triangle is equilateral/isosceles, finding radius.

P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)

Use when: a point divides a line in ratio $m:n$ . Remember: $m$ is associated with $(x_2, y_2)$ , $n$ with $(x_1, y_1)$ .

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

Use when: finding area of triangle or checking collinearity (area = 0 means collinear).

Solved Previous Year Questions

PYQ 1 — CBSE 2024 (2 marks)

Q: Find the distance between A(3, 4) and B(−2, 1).

Solution:

d = \sqrt{(-2-3)^2 + (1-4)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}

\boxed{d = \sqrt{34} \approx 5.83 \text{ units}}

PYQ 2 — CBSE 2023 (4 marks)

Q: Find the coordinates of the point P which divides the line segment joining A(4, −3) and B(8, 5) in the ratio 3:1 internally.

Solution: Using section formula with $m:n = 3:1$ , $A(4, -3)$ , $B(8, 5)$ :

P = \left(\frac{3 \times 8 + 1 \times 4}{3+1}, \frac{3 \times 5 + 1 \times (-3)}{3+1}\right)

= \left(\frac{24+4}{4}, \frac{15-3}{4}\right) = \left(\frac{28}{4}, \frac{12}{4}\right) = \boxed{(7, 3)}

PYQ 3 — CBSE 2022 (4 marks)

Q: Show that the points A(1, 7), B(4, 2), C(−1, −1), D(−4, 4) are vertices of a square.

Solution (strategy): For a square: all four sides equal AND diagonals equal.

$AB = \sqrt{(4-1)^2 + (2-7)^2} = \sqrt{9+25} = \sqrt{34}$

$BC = \sqrt{(-1-4)^2 + (-1-2)^2} = \sqrt{25+9} = \sqrt{34}$

$CD = \sqrt{(-4+1)^2 + (4+1)^2} = \sqrt{9+25} = \sqrt{34}$

$DA = \sqrt{(1+4)^2 + (7-4)^2} = \sqrt{25+9} = \sqrt{34}$

All sides equal. Now check diagonals:

$AC = \sqrt{(-1-1)^2 + (-1-7)^2} = \sqrt{4+64} = \sqrt{68}$

$BD = \sqrt{(-4-4)^2 + (4-2)^2} = \sqrt{64+4} = \sqrt{68}$

All sides equal and diagonals equal → ABCD is a square. ✓

Difficulty Distribution

Difficulty	%	Topics
Easy	40%	Distance formula, midpoint
Medium	40%	Section formula, collinearity
Hard	20%	Area of triangle applications, finding unknown points

Expert Strategy

For distance formula questions: Be careful with negative coordinates. $(-2-3)^2 = (-5)^2 = 25$ , NOT $(-5)^2 = -25$ . Always square after subtraction.

For section formula: Write the formula first, substitute carefully. The most common error is swapping $m$ and $n$ — $m$ goes with the second point $(x_2, y_2)$ .

For quadrilateral type questions: Proving square/rhombus/parallelogram requires both side calculations AND diagonal calculations. Just showing equal sides isn’t enough (a rhombus also has equal sides but unequal diagonals).

For 4-mark questions, show every step of calculation. Even if the final answer is wrong, you’ll get partial credit for correct formula application and correct substitution.

Common Traps

Trap 1: Forgetting the absolute value in area formula. Area $= \frac{1}{2}|\ldots|$ . Without the modulus, you may get a negative “area.” Area is always positive — take the absolute value.

Trap 2: Saying points are collinear when area is non-zero. If area $\neq 0$ , the points form a triangle — they are NOT collinear. The collinearity condition is area $= 0$ .

Trap 3: Using the wrong section formula for external division. Internal division uses $(mx_2 + nx_1)/(m+n)$ . External division uses $(mx_2 - nx_1)/(m-n)$ . Class 10 CBSE mostly tests internal division only — don’t confuse them.