Coordinate Geometry in SSC CGL: What Is Asked
SSC CGL Tier 1 usually has 1 question and Tier 2 has 2 to 3 from coordinate geometry. Nearly all of them use eight formulas: distance, section, midpoint, centroid, area of a triangle, slope, equation of a line and distance from a point to a line. Learn these and the axis-intercept shortcut, and you can solve most questions in under 40 seconds.
Key takeaway: Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]. Slope = (y₂ − y₁)/(x₂ − x₁). Parallel lines: equal slopes; perpendicular lines: m₁ × m₂ = −1. Area of triangle formed by ax + by = c with the axes = c²/(2|ab|). Point-to-line distance = |ax₁ + by₁ + c| / √(a² + b²).
Every worked answer on this page was computed and checked in Python. Pair this note with our algebra formulas for SSC CGL and mensuration formulas for SSC CGL.
Coordinate Geometry Formulas List (One Table)
| Topic | Formula |
|---|---|
| Distance between (x₁, y₁) and (x₂, y₂) | √[(x₂ − x₁)² + (y₂ − y₁)²] |
| Distance from origin | √(x² + y²) |
| Midpoint | ((x₁ + x₂)/2, (y₁ + y₂)/2) |
| Section formula (internal, ratio m : n) | ((m x₂ + n x₁)/(m + n), (m y₂ + n y₁)/(m + n)) |
| Section formula (external, ratio m : n) | ((m x₂ − n x₁)/(m − n), (m y₂ − n y₁)/(m − n)) |
| Centroid of triangle | ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3) |
| Area of triangle | ½ |
| Collinear points | Area of triangle = 0 |
| Slope of line through two points | m = (y₂ − y₁)/(x₂ − x₁) |
| Slope of ax + by + c = 0 | m = −a/b |
| Slope-intercept form | y = mx + c |
| Point-slope form | y − y₁ = m(x − x₁) |
| Intercept form | x/a + y/b = 1 |
| Parallel lines | m₁ = m₂ |
| Perpendicular lines | m₁ × m₂ = −1 |
| Angle between lines | tan θ = |
| Distance from (x₁, y₁) to ax + by + c = 0 | |
| Distance between parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 |
Quadrants and Reflection of a Point (Quick Table)
| Point (x, y) reflected in... | Image |
|---|---|
| x-axis | (x, −y) |
| y-axis | (−x, y) |
| Origin | (−x, −y) |
| Line y = x | (y, x) |
| Line y = −x | (−y, −x) |
Signs in quadrants: I (+, +), II (−, +), III (−, −), IV (+, −).
Example: the reflection of (3, −5) in the x-axis is (3, 5); in the y-axis (−3, −5); in the origin (−3, 5); in y = x it is (−5, 3).
Shortcut: Triangle Made by a Line with the Axes
For the line ax + by = c: x-intercept = c/a, y-intercept = c/b, and
Area with the axes = c² / (2|ab|).
Example: 2x + 3y = 12 cuts the axes at (6, 0) and (0, 4). Area = 12²/(2 × 2 × 3) = 144/12 = 12 sq units (check: ½ × 6 × 4 = 12).
Solved Coordinate Geometry Questions (SSC CGL Pattern)
- Distance between (3, −4) and (−5, 2)? √[(−8)² + 6²] = √100 = 10 units.
- Point dividing the join of (2, 3) and (8, 9) internally in 1 : 2? ((1×8 + 2×2)/3, (1×9 + 2×3)/3) = (4, 5).
- External division: point dividing (1, 2) and (4, 5) externally in 2 : 1? ((2×4 − 1×1)/1, (2×5 − 1×2)/1) = (7, 8).
- Centroid of (1, 2), (3, 4), (5, 0)? (9/3, 6/3) = (3, 2).
- Area of triangle with vertices (1, 2), (4, 6), (7, 2)? ½ |1(6 − 2) + 4(2 − 2) + 7(2 − 6)| = ½ |4 + 0 − 28| = 12 sq units.
- Collinear: find k if (1, 2), (3, k), (5, 10) are collinear. Area = 0 gives k = 6 (the middle point is the midpoint).
- Slope of the line through (2, 3) and (6, 11)? 8/4 = 2. Its equation: y − 3 = 2(x − 2), i.e. y = 2x − 1.
- Perpendicular: find k if 2x + 3y = 5 and kx − 4y = 1 are perpendicular. Slopes −2/3 and k/4; (−2/3)(k/4) = −1 gives k = 6.
- Point to line: distance of (3, 4) from 3x + 4y − 5 = 0 = |9 + 16 − 5|/5 = 4 units.
- Parallel lines: distance between 3x + 4y + 7 = 0 and 3x + 4y − 8 = 0 = |7 − (−8)|/5 = 3 units.
- Midpoint of (−2, 5) and (6, −1)? (2, 2).
Common Mistakes in Coordinate Geometry
- Forgetting the ½ and the modulus in the area formula (area is never negative).
- Mixing up internal (m + n) and external (m − n) section formulas.
- Taking the slope of ax + by + c = 0 as a/b instead of −a/b.
- In point-to-line distance, not writing the line as ax + by + c = 0 first (move everything to one side).
- Using m₁ = −m₂ for perpendicular lines. The rule is m₁ × m₂ = −1.
Coordinate Geometry Formulas FAQs
What is the distance formula in coordinate geometry?
The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]. From the origin it is √(x² + y²).
How do you find the area of a triangle from its vertices?
Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. If the area is 0, the three points are collinear.
What is the condition for two lines to be perpendicular?
The product of their slopes is −1 (m₁ × m₂ = −1). For parallel lines the slopes are equal.
How many coordinate geometry questions come in SSC CGL?
Usually about 1 in Tier 1 and 2 to 3 in Tier 2, often mixed with straight-line and triangle-area questions.
Practice Next
- Apply these formulas in the SSC CGL Coordinate Geometry Mock Test 03 and SSC CPO Quant Mock Test 01.
- New SSC CGL practice: SSC CGL Tier 1 Full Mock Test 06.
- New SSC CGL tests (Tier 1 level): SSC CGL Geometry Mock Test 01 and SSC CGL Tier 1 Full Mock Test 05.
- Mock tests: SSC CGL Tier 1 Mock Test: Geometry and Coordinate Geometry, SSC CGL Tier 1 Geometry Practice Set (30 Questions) and SSC Geometry and Mensuration Mock Test.
- Related notes: algebra formulas for SSC CGL, mensuration formulas for SSC CGL, trigonometry formulas for SSC CGL and height and distance for SSC CGL, plus geometry formulas for SSC CGL.
- Exam details: SSC CGL 2026.
- Browse all study notes.