How to Solve: Coordinate Geometry (Distance & Section Formula) – Complete Guide

How to Solve: Coordinate Geometry (Distance & Section Formula) – Complete Guide

For SSC / Bank / Railway Exams

Introduction

"Mastering distance and section formula lets you crack 3–5 marks in every SSC, Bank, or Railway exam—marks that decide whether you clear the cutoff or not. Imagine two points on a map: how far apart are they? Where’s the midpoint? These formulas answer it in seconds."

What You Need To Know First

1. Coordinate Plane Basics: Know how to plot points (x, y) on a graph.
2. Pythagoras’ Theorem: Used to derive the distance formula.
3. Ratio Concept: Understand parts (m:n) for the section formula.

Key Vocabulary

Formulas To Know

1. Distance Formula

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

Variables: - ( (x_1, y_1) ): First point. - ( (x_2, y_2) ): Second point. - ( d ): Distance between them.

MEMORISE THIS – Not given in most exam sheets.

2. Section Formula (Internal Division)

Formula: [ P = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) ]

Variables: - ( (x_1, y_1) ): First point. - ( (x_2, y_2) ): Second point. - ( m:n ): Ratio in which point ( P ) divides the line. - ( P ): Coordinates of the dividing point.

MEMORISE THIS – Usually not provided.

Special Case (Midpoint): If ( m = n = 1 ), the formula simplifies to: [ P = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) ]

Step-by-Step Method

For Distance Between Two Points

1. Label the points: Write down ( (x_1, y_1) ) and ( (x_2, y_2) ).
2. Subtract x-coordinates: ( x_2 - x_1 ).
3. Subtract y-coordinates: ( y_2 - y_1 ).
4. Square both results: ( (x_2 - x_1)^2 ) and ( (y_2 - y_1)^2 ).
5. Add the squares: ( (x_2 - x_1)^2 + (y_2 - y_1)^2 ).
6. Take the square root: ( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ).

For Section Formula (Internal Division)

1. Label the points: ( (x_1, y_1) ) and ( (x_2, y_2) ).
2. Write the ratio: ( m:n ).
3. Apply x-coordinate formula: ( \frac{m x_2 + n x_1}{m + n} ).
4. Apply y-coordinate formula: ( \frac{m y_2 + n y_1}{m + n} ).
5. Simplify: Write the final coordinates as ( (x, y) ).

Worked Examples

Example 1 – Basic Distance

Question: Find the distance between ( A(2, 3) ) and ( B(5, 7) ).

Solution: 1. ( x_1 = 2, y_1 = 3 ) 2. ( x_2 = 5, y_2 = 7 ) 3. ( x_2 - x_1 = 5 - 2 = 3 ) 4. ( y_2 - y_1 = 7 - 3 = 4 ) 5. ( 3^2 + 4^2 = 9 + 16 = 25 ) 6. ( \sqrt{25} = 5 )

Answer: Distance = 5 units.

What we did and why: We plugged the coordinates into the distance formula. The numbers were simple, so we could calculate mentally. Always subtract first, then square.

Example 2 – Medium Section Formula

Question: Find the point that divides the line joining ( A(1, 2) ) and ( B(4, 6) ) in the ratio 2:1.

Solution: 1. ( x_1 = 1, y_1 = 2 ) 2. ( x_2 = 4, y_2 = 6 ) 3. Ratio ( m:n = 2:1 ) 4. ( x = \frac{2 \times 4 + 1 \times 1}{2 + 1} = \frac{8 + 1}{3} = \frac{9}{3} = 3 ) 5. ( y = \frac{2 \times 6 + 1 \times 2}{2 + 1} = \frac{12 + 2}{3} = \frac{14}{3} )

Answer: The point is ( \left( 3, \frac{14}{3} \right) ).

What we did and why: We used the section formula. The ratio was 2:1, so we multiplied ( x_2 ) by 2 and ( x_1 ) by 1, then divided by the sum of the ratio (3). Same for y-coordinates.

Example 3 – Exam-Style (Disguised)

Question: Three points ( A(1, 1) ), ( B(3, 3) ), and ( C(5, k) ) are collinear. Find ( k ).

Solution: 1. If points are collinear, the area formed by them is zero. 2. Alternatively, the section formula can be used: ( B ) divides ( AC ) in some ratio. 3. Let ( B ) divide ( AC ) in ratio ( m:n ). 4. Using section formula for x-coordinate:
[ 3 = \frac{m \times 5 + n \times 1}{m + n} ] 5. Simplify:
[ 3(m + n) = 5m + n ]
[ 3m + 3n = 5m + n ]
[ 2n = 2m ]
[ m:n = 1:1 ] 6. Now, use y-coordinate:
[ 3 = \frac{1 \times k + 1 \times 1}{1 + 1} ]
[ 3 = \frac{k + 1}{2} ]
[ 6 = k + 1 ]
[ k = 5 ]

Answer: ( k = 5 ).

What we did and why: We used collinearity to set up the section formula. Since ( B ) lies between ( A ) and ( C ), it divides ( AC ) in some ratio. We found the ratio (1:1) and then solved for ( k ).

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up! For distance, subtract x’s, subtract y’s, square both, add, then square root. For section formula, multiply x₂ by m, x₁ by n, add them, divide by m+n. Same for y. Midpoint? Just average the x’s and y’s. Watch out for negative numbers—they mess up signs. If points are collinear, the section formula will give the same ratio for x and y. Practice 2–3 problems tonight, and you’ll own this topic tomorrow. Go crush it!