Coordinate Geometry Formulas and Solved Examples

Coordinate geometry converts geometric problems into algebra by assigning every point a pair of numbers on a plane. For placement aptitude rounds, that means most questions reduce to plugging coordinates into 5 or 6 standard formulas and simplifying.

This article covers every formula that appears in campus placement evaluation tests, verifies each one with worked arithmetic, and ends with multi-step problems of the type you’ll see in TCS NQT, AMCAT, and Cocubes quantitative sections.

Core Formulas: Distance, Midpoint, and Section

Distance Formula

The distance between two points (x1, y1) and (x2, y2) is:

• d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

This is the Pythagorean theorem applied to horizontal and vertical differences. According to the NCERT Class 11 textbook on Straight Lines, the derivation follows directly from constructing a right triangle between the two points.

Worked example:

• Points: (2, 3) and (7, 15)
• Horizontal difference: 7 - 2 = 5
• Vertical difference: 15 - 3 = 12
• Distance: sqrt(5^2 + 12^2) = sqrt(25 + 144) = sqrt(169) = 13

The pair (5, 12, 13) is a Pythagorean triplet. Recognising these saves time in timed tests.

Midpoint Formula

The midpoint of a segment joining (x1, y1) and (x2, y2) is:

• M = ((x1 + x2)/2, (y1 + y2)/2)

Worked example:

• Points: (4, 6) and (10, 2)
• Midpoint: ((4 + 10)/2, (6 + 2)/2) = (7, 4)

Section Formula (Internal Division)

A point dividing the segment from (x1, y1) to (x2, y2) in ratio m:n internally:

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

Worked example:

• Points: (3, 5) and (11, 8), ratio 5:2 internally
• x-coordinate: (5*11 + 2*3)/(5 + 2) = (55 + 6)/7 = 61/7
• y-coordinate: (5*8 + 2*5)/(5 + 2) = (40 + 10)/7 = 50/7
• Result: (61/7, 50/7)

Section Formula (External Division)

For external division in ratio m:n:

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

Worked example (same points, ratio 5:2 externally):

• x-coordinate: (5*11 - 2*3)/(5 - 2) = (55 - 6)/3 = 49/3
• y-coordinate: (5*8 - 2*5)/(5 - 2) = (40 - 10)/3 = 30/3 = 10
• Result: (49/3, 10)

Slope and Conditions for Parallel and Perpendicular Lines

Slope of a Line

The slope of a line through (x1, y1) and (x2, y2) is:

• m = (y2 - y1)/(x2 - x1)

Slope is undefined when x2 = x1 (vertical line). A horizontal line has slope 0.

Parallel and Perpendicular Conditions

• Two lines are parallel when m1 = m2
• Two lines are perpendicular when m1 * m2 = -1

Quick check: if one line has slope 3/4, a perpendicular line has slope -4/3. Multiply: (3/4) * (-4/3) = -1. Confirmed.

Angle Between Two Lines

The acute angle between two lines with slopes m1 and m2:

• tan(theta) = |(m1 - m2)/(1 + m1*m2)|

This formula breaks down (gives undefined) when 1 + m1*m2 = 0, which is exactly the perpendicular case. That’s a useful consistency check in tests.

Equations of a Line

There are 3 standard forms you’ll use in aptitude rounds. Each fits a different “given” scenario. As covered in Khan Academy’s analytic geometry module, converting between forms is a frequent exam task.

Slope-Intercept Form

• y = mx + c

Use when you know the slope and y-intercept directly.

Point-Slope Form

• y - y1 = m(x - x1)

Use when you know the slope and one point on the line.

Worked example:

• Slope: 2/3, point: (2, 3)
• Equation: y - 3 = (2/3)(x - 2)
• Multiply through by 3: 3y - 9 = 2x - 4
• Rearranged: 2x - 3y + 5 = 0

Two-Point Form

Given points (x1, y1) and (x2, y2):

• (y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)

This is point-slope form with the slope substituted in.

Perpendicular Line Example

• Given line: 3x + 2y + 4 = 0
• Slope of given line: -3/2 (rearrange to y = (-3/2)x - 2)
• Perpendicular slope: 2/3 (negative reciprocal)
• Through point (2, 3): y - 3 = (2/3)(x - 2)
• Simplify: 3y - 9 = 2x - 4, so 2x - 3y + 5 = 0

Verification: slopes are -3/2 and 2/3. Product: (-3/2)(2/3) = -1. Perpendicularity confirmed.

Area of a Triangle from Coordinates

Given vertices (x1, y1), (x2, y2), (x3, y3):

• Area = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

The absolute value ensures area is non-negative. If the result is zero, the three points are collinear.

Worked example:

• Vertices: (1, 2), (4, 6), (7, 2)
• Calculation: (1/2)|1*(6 - 2) + 4*(2 - 2) + 7*(2 - 6)|
• Expand: (1/2)|1*4 + 4*0 + 7*(-4)|
• Simplify: (1/2)|4 + 0 - 28| = (1/2)*|-24| = (1/2)*24 = 12
• Area: 12 square units

Collinearity Check

• Points: (1, 1), (3, 3), (5, 5)
• Area: (1/2)|1*(3 - 5) + 3*(5 - 1) + 5*(1 - 3)|
• Expand: (1/2)|1*(-2) + 3*4 + 5*(-2)| = (1/2)|-2 + 12 - 10| = (1/2)*0 = 0
• Area is zero, so the points are collinear (they all lie on y = x)

Worked Examples for Placement Aptitude

These are multi-step problems combining 2 or more formulas. The difficulty level matches what appears in time and work aptitude questions sections of campus drives.

Problem 1: Triangle Median Length

• Given: Triangle with vertices A(0, 0), B(6, 0), C(3, 6)
• Find: Length of the median from C to side AB
• Step 1: Midpoint of AB = ((0+6)/2, (0+0)/2) = (3, 0)
• Step 2: Distance from C(3, 6) to midpoint (3, 0) = sqrt((3-3)^2 + (6-0)^2) = sqrt(0 + 36) = 6
• Answer: The median from C has length 6 units

Problem 2: Line Through Intersection

• Given: Find the equation of a line through the point (1, 2) and parallel to 4x - 3y + 7 = 0
• Step 1: Slope of 4x - 3y + 7 = 0 is 4/3 (rearrange: y = (4/3)x + 7/3)
• Step 2: Parallel line has the same slope: 4/3
• Step 3: Point-slope form: y - 2 = (4/3)(x - 1)
• Step 4: Multiply by 3: 3y - 6 = 4x - 4
• Step 5: Rearrange: 4x - 3y + 2 = 0
• Answer: 4x - 3y + 2 = 0

Problem 3: Perpendicular Distance

• Given: Find the perpendicular distance from point (3, 4) to line 3x + 4y - 5 = 0
• Formula: d = |ax1 + by1 + c|/sqrt(a^2 + b^2)
• Substitution: d = |3*3 + 4*4 - 5|/sqrt(9 + 16) = |9 + 16 - 5|/sqrt(25) = |20|/5 = 4
• Answer: 4 units

The formula pattern across all three problems is consistent: identify what’s given, pick the right formula, substitute, simplify. For placement preparation books that cover these topics in depth, R.S. Aggarwal’s Quantitative Aptitude dedicates 2 chapters to coordinate geometry with 100-plus practice problems at this difficulty level.

Bridge to Analytical Thinking

Every problem above follows the same structure: translate a geometric question into coordinates, apply an algebraic formula, verify the arithmetic. That systematic breakdown is exactly how structured learning works in any technical domain. If you found the step-by-step pattern in the perpendicular-distance problem (Problem 3 above) satisfying, TinkerLLM applies the same decomposition approach to building with large language models. It costs ₹299 to start, and the problems scale from single-formula to multi-step just like the progression in this article.