What is the section formula?

If a point P divides the line joining (x₁, y₁) and (x₂, y₂) in the ratio m : n internally, then P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)). For the midpoint, m = n, giving ((x₁ + x₂)/2, (y₁ + y₂)/2).

How do you check if three points are collinear using coordinates?

Three points are collinear if the sum of the distances between two pairs equals the distance of the third pair (they lie on one line), or if the area of the triangle they form is zero.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry — Exercise Solutions

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

How to Use This Page

Complete NCERT Solutions for CBSE Class 10 Maths Chapter 7 — Coordinate Geometry: the distance formula, the section formula, the midpoint, and collinearity.

Key Concepts (Quick Recap)

Distance: d = √[(x₂ − x₁)² + (y₂ − y₁)²].
Section (ratio m:n): P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)).
Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2).

Exercise 7.1 — Distance Formula

Q1. Find the distance between the points (2, 3) and (4, 1). d = √[(4 − 2)² + (1 − 3)²] = √[4 + 4] = √8 = 2√2 units.

Q2. Find the distance between (−5, 7) and (−1, 3). d = √[(−1 + 5)² + (3 − 7)²] = √[16 + 16] = √32 = 4√2 units.

Q3. Find the distance of the point (6, 8) from the origin. d = √[6² + 8²] = √[36 + 64] = √100 = 10 units.

Q4. Show that the points (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square. Sides: (1,7)-(4,2): √(9+25)=√34; (4,2)-(−1,−1): √(25+9)=√34; (−1,−1)-(−4,4): √(9+25)=√34; (−4,4)-(1,7): √(25+9)=√34. All four sides = √34. Diagonals: (1,7)-(−1,−1): √(4+64)=√68; (4,2)-(−4,4): √(64+4)=√68. Equal sides and equal diagonals → square.

Q5. Find a point on the x-axis equidistant from (2, −5) and (−2, 9). Point (x, 0). (x − 2)² + 25 = (x + 2)² + 81 → −4x + 29 = 4x + 85 → −8x = 56 → x = −7. Point: (−7, 0).

Exercise 7.2 — Section Formula

Q1. Find the coordinates of the point dividing the segment joining (4, −3) and (8, 5) in the ratio 3 : 1 internally. P = ((3·8 + 1·4)/4, (3·5 + 1·(−3))/4) = (28/4, 12/4) = (7, 3).

Q2. Find the midpoint of the segment joining (3, 4) and (−1, 2). M = ((3 + (−1))/2, (4 + 2)/2) = (1, 3).

Q3. In what ratio does the point (−1, 6) divide the segment joining (−3, 10) and (6, −8)? Let the ratio be k : 1. x-coordinate: −1 = (6k + (−3))/(k + 1) → −(k + 1) = 6k − 3 → −k − 1 = 6k − 3 → 7k = 2 → k = 2/7. Ratio = 2 : 7.

Q4. Find the coordinates of the points of trisection of the segment joining (4, −1) and (−2, −3). Trisection points divide in 1:2 and 2:1. P₁ (1:2) = ((1·(−2) + 2·4)/3, (1·(−3) + 2·(−1))/3) = (6/3, −5/3) = (2, −5/3). P₂ (2:1) = ((2·(−2) + 1·4)/3, (2·(−3) + 1·(−1))/3) = (0/3, −7/3) = (0, −7/3).

Collinearity

Q. Show that the points A(1, 5), B(2, 3) and C(−2, −11) are not collinear. AB = √[(2−1)² + (3−5)²] = √5. BC = √[(−2−2)² + (−11−3)²] = √(16 + 196) = √212. AC = √[(−2−1)² + (−11−5)²] = √(9 + 256) = √265. Since AB + BC ≠ AC, the points are not collinear.

Common Mistakes

Sign slips inside the squares — (x₂ − x₁) can be negative, but its square is positive.
In the section formula, swapping which point is (x₁,y₁) vs (x₂,y₂), or swapping m and n.
Assuming a figure is a square from equal sides alone — also check the diagonals.

Exam Tips

To find a ratio, use the section formula on one coordinate and solve for k.
Midpoint is just the section formula with ratio 1 : 1.
For "type of quadrilateral," compute all four sides and both diagonals.

Frequently Asked Questions

The distance between two points (x₁, y₁) and (x₂, y₂) is d = √[(x₂ − x₁)² + (y₂ − y₁)²]. It comes directly from the Pythagoras theorem applied to the horizontal and vertical gaps.