Coordinate Geometry Basics for Competitive Exams: Lines, Slopes and Circles

Coordinate geometry is the bridge between algebra and geometry. Instead of working with abstract shapes, you work with points on a grid — and every geometric property becomes a calculation using coordinates.

For competitive exams, coordinate geometry is a high-reward topic. The concepts are limited and well-defined: distance between points, midpoint of a segment, slope of a line, equations of lines and circles, and area of triangles using coordinates. Once you have these formulas memorized and can apply them without hesitation, every coordinate geometry question becomes a direct substitution problem.

In SSC CGL, coordinate geometry contributes 2–4 questions in Tier 1 and 3–5 in Tier 2. In CAT, it appears in 1–2 questions within geometry sets. In IBPS PO, it appears occasionally in Mains.

Part 1: The Coordinate System — Foundation Concepts

Axes and Quadrants

Exam trap: Point on x-axis has y = 0. Point on y-axis has x = 0.

Part 2: Distance Formula

Between Two Points

d = √((x₂ − x₁)² + (y₂ − y₁)²)

Worked Example 1

Distance between A(3, 4) and B(7, 1)?

d = √((7−3)² + (1−4)²) = √(16 + 9) = √25 = 5 units

Worked Example 2

Distance from P(5, −3) to origin?

d = √(5² + (−3)²) = √(25 + 9) = √34 units

Shortcut — Distance from origin: d = √(x² + y²)

Checking Shape from Points

Worked Example 3:
A(0,0), B(4,0), C(4,3) — right triangle?

AB = 4, BC = 3, AC = √(16+9) = 5
AC² = AB² + BC²? → 25 = 16 + 9 ✓ Right triangle (3-4-5 triplet)

Part 3: Midpoint Formula

M = ((x₁+x₂)/2, (y₁+y₂)/2)

Worked Example 4

Midpoint of A(2,6) and B(8,4)?

M = ((2+8)/2, (6+4)/2) = (5, 5)

Section Formula — Internal Division

P divides A(x₁,y₁) and B(x₂,y₂) in ratio m:n:

P = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))

Memory trick: m part uses B's coordinates, n part uses A's coordinates.

Worked Example 5

P divides A(1,2) and B(7,8) in ratio 2:1.

P = ((2×7 + 1×1)/3, (2×8 + 1×2)/3) = (15/3, 18/3) = (5, 6)

Worked Example 6 — Finding the Ratio

In what ratio does P(4,5) divide A(2,3) and B(6,7)?

Let ratio = k:1
(6k + 2)/(k+1) = 4 → 6k + 2 = 4k + 4 → k = 1

Ratio = 1:1 (midpoint)

Part 4: Slope of a Line

m = (y₂ − y₁) / (x₂ − x₁)

Slope Properties

Parallel and Perpendicular Lines

Parallel: m₁ = m₂

Perpendicular: m₁ × m₂ = −1

Worked Example 7

Line 1: (1,2) to (3,6). Line 2: (0,4) to (2,5). Relationship?

m₁ = (6−2)/(3−1) = 2
m₂ = (5−4)/(2−0) = 1/2

m₁ × m₂ = 2 × 1/2 = 1 ≠ −1, m₁ ≠ m₂ → Neither

Worked Example 8

Slope perpendicular to line through (2,3) and (5,9)?

Slope of line = (9−3)/(5−2) = 2
Perpendicular slope = −1/2

Part 5: Equations of Lines

Three Forms

Form 1 — Slope-Intercept:
y = mx + c (m = slope, c = y-intercept)

Form 2 — Point-Slope:
y − y₁ = m(x − x₁)

Form 3 — Standard:
ax + by + c = 0

Worked Example 9 — Equation from Two Points

Line through A(2,3) and B(5,9)?

Slope = (9−3)/(5−2) = 2
y − 3 = 2(x − 2)
y = 2x − 1

Worked Example 10 — Reading Slope

Slope and y-intercept of 3x + 4y − 12 = 0?

4y = −3x + 12 → y = −3x/4 + 3

Slope = −3/4, y-intercept = 3

Intercepts Shortcut

For ax + by + c = 0:

• x-intercept = −c/a
• y-intercept = −c/b

Worked Example 11:
Intercepts of 2x + 3y − 6 = 0?

x-intercept: x = 6/2 = 3
y-intercept: y = 6/3 = 2

Distance from Point to Line

d = |ax₁ + by₁ + c| / √(a² + b²)

Worked Example 12:
Distance from (3,4) to line 3x + 4y − 5 = 0?

d = |3(3) + 4(4) − 5| / √(9+16) = |9+16−5| / 5 = 20/5 = 4 units

Part 6: Area of Triangle Using Coordinates

Area = (1/2) × |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|

Worked Example 13

Triangle: A(1,2), B(4,6), C(7,2)?

= (1/2) × |1(6−2) + 4(2−2) + 7(2−6)|
= (1/2) × |4 + 0 − 28|
= (1/2) × 24 = 12 sq units

Collinearity Check

Area = 0 means three points are collinear.

Worked Example 14:
A(1,1), B(2,2), C(3,3) — collinear?

= (1/2) × |1(2−3) + 2(3−1) + 3(1−2)|
= (1/2) × |−1 + 4 − 3| = 0 ✓ Collinear

Part 7: Circle Equations

Standard Form

(x − h)² + (y − k)² = r²

Center = (h, k), Radius = r

Center at origin: x² + y² = r²

General Form

x² + y² + 2gx + 2fy + c = 0

• Center = (−g, −f)
• Radius = √(g² + f² − c)

Worked Example 15 — Center and Radius

x² + y² − 6x + 4y − 3 = 0?

2g = −6 → g = −3, 2f = 4 → f = 2, c = −3

Center = (3, −2)
Radius = √(9 + 4 + 3) = √16 = 4 units

Worked Example 16 — Writing Circle Equation

Center (2, −3), radius 5?

(x−2)² + (y+3)² = 25
x² + y² − 4x + 6y − 12 = 0

Position of Point Relative to Circle

Substitute (x₁, y₁) into circle equation:

• Result < 0 → Inside
• Result = 0 → On the circle
• Result > 0 → Outside

Worked Example 17:
Is (1,2) inside x² + y² − 4x − 6y + 9 = 0?

1 + 4 − 4 − 12 + 9 = −2 < 0 → Inside

Part 8: Centroid, Circumcenter, and Incenter

Centroid

Centroid = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

Worked Example 18:
Centroid of (2,4), (6,2), (4,6)?

= ((2+6+4)/3, (4+2+6)/3) = (4, 4)

Circumcenter

Equidistant from all three vertices. For a right triangle — midpoint of hypotenuse.

Incenter

= ((a×x₁ + b×x₂ + c×x₃)/(a+b+c), (a×y₁ + b×y₂ + c×y₃)/(a+b+c))

Where a, b, c are side lengths opposite to vertices A, B, C.

Part 9: Exam Traps and Speed Tips

Speed Tip — Pythagorean Triplets:
Before computing √ in distance formula, check: 3-4-5, 5-12-13, 8-15-17, 7-24-25, 6-8-10, 9-12-15

Quick Reference Formula Sheet

Exam-Wise Strategy

2-Week Practice Plan