Coordinate geometry is where algebra meets geometry. Instead of drawing shapes and eyeballing, we use numbers — every point has an address (x, y), and we can calculate distances, areas, and slopes with formulas. It’s one of the most formula-driven topics in aptitude, which is great news because formulas mean guaranteed marks once we memorize them.
The Coordinate Plane
Quick refresher: the plane is divided into 4 quadrants by two axes.
- Q1: x > 0, y > 0 (both positive)
- Q2: x < 0, y > 0 (x negative, y positive)
- Q3: x < 0, y < 0 (both negative)
- Q4: x > 0, y < 0 (x positive, y negative)
Points on axes have one coordinate as zero. The origin is (0, 0).
Key Formulas
Distance Formula
The distance between two points (x₁, y₁) and (x₂, y₂) is:
d = √[(x₂ − x₁)² + (y₂ − y₁)²]
In simple language: it’s just the Pythagoras theorem applied to coordinates. The horizontal difference is one leg, the vertical difference is the other, and the distance is the hypotenuse.
Special cases:
- Distance from origin: d = √(x² + y²)
- Horizontal line (y₁ = y₂): d = |x₂ − x₁|
- Vertical line (x₁ = x₂): d = |y₂ − y₁|
Midpoint and Section Formula
Midpoint Formula
The midpoint of (x₁, y₁) and (x₂, y₂) is simply the average of the coordinates:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Section Formula (Internal Division)
If point P divides the line joining A(x₁, y₁) and B(x₂, y₂) in the ratio m:n internally:
P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))
Memory trick: The point closer to B gets the weight of m (associated with B’s coordinates), and the point closer to A gets the weight of n.
The midpoint is just a special case where m:n = 1:1.
Centroid of a Triangle
The centroid (intersection of medians) of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is:
G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)
In simple language: just average all three x-coordinates and all three y-coordinates.
Slope of a Line
The slope tells us how steep a line is — how much it rises (or falls) for every unit it moves right.
Slope m = (y₂ − y₁)/(x₂ − x₁) = rise / run
- Positive slope: line goes up from left to right (↗)
- Negative slope: line goes down from left to right (↘)
- Zero slope: horizontal line (m = 0)
- Undefined slope: vertical line (division by zero — the run is 0)
Parallel and Perpendicular Lines
- Parallel lines have the same slope: m₁ = m₂
- Perpendicular lines have slopes that multiply to −1: m₁ × m₂ = −1
In simple language: for perpendicular lines, flip the fraction and change the sign. If one line has slope 2/3, the perpendicular line has slope −3/2.
Equation of a Line
There are several forms — use whichever fits the given information:
1. Slope-intercept form: y = mx + c
- m = slope, c = y-intercept (where the line crosses the y-axis)
- Best when we know the slope and y-intercept
2. Point-slope form: y − y₁ = m(x − x₁)
- Best when we know the slope and one point
3. Two-point form: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁)
- Best when we know two points on the line
4. Intercept form: x/a + y/b = 1
- a = x-intercept, b = y-intercept
- Best when we know both intercepts
Area of Triangle from Coordinates
Given three vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):
Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
The absolute value is important — area can’t be negative! If the formula gives us zero, the three points are collinear (they lie on the same line).
Collinearity check: Three points are collinear if and only if the area of the triangle they form is zero.
Shortcut Methods and Tricks
Trick 1: Quick distance check Before applying the distance formula, check if the differences form a Pythagorean triplet. E.g., if Δx = 3 and Δy = 4, the distance is immediately 5.
Trick 2: Vertical/horizontal sides If a triangle or quadrilateral has any vertical or horizontal sides, calculate those distances by simple subtraction — no need for the full formula.
Trick 3: Area using the “shoelace” pattern For the triangle area formula, arrange the coordinates in a column and cross-multiply diagonally:
Write the vertices in order and repeat the first:
x₁ y₁
x₂ y₂
x₃ y₃
x₁ y₁Sum of (down-right diagonals) − Sum of (down-left diagonals), divide by 2, take absolute value.
Trick 4: When given the equation ax + by + c = 0
- Slope = −a/b
- x-intercept = −c/a
- y-intercept = −c/b
Worked Examples
Example 1: Distance
Find the distance between (3, −2) and (−1, 1).
d = √[(−1 − 3)² + (1 − (−2))²] d = √[(−4)² + (3)²] d = √[16 + 9] d = √25 = 5 units
(We spotted the 3-4-5 triplet! Δx = 4, Δy = 3.)
Example 2: Section Formula
Find the point that divides the line joining A(2, 3) and B(8, −1) in the ratio 1:3.
P = ((1×8 + 3×2)/(1+3), (1×(−1) + 3×3)/(1+3)) P = ((8 + 6)/4, (−1 + 9)/4) P = (14/4, 8/4) P = (3.5, 2)
Example 3: Equation of Line
Find the equation of a line passing through (2, 5) with slope 3.
Using point-slope form: y − 5 = 3(x − 2) y − 5 = 3x − 6 y = 3x − 1
In general form: 3x − y − 1 = 0
Example 4: Area of Triangle
Find the area of the triangle with vertices A(1, 2), B(4, 6), C(7, 2).
Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| = ½ |1(6 − 2) + 4(2 − 2) + 7(2 − 6)| = ½ |1(4) + 4(0) + 7(−4)| = ½ |4 + 0 − 28| = ½ |−24| = ½ × 24 = 12 sq units
Example 5: Perpendicular Lines
Line L₁ passes through (1, 3) and (4, 9). Find the slope of a line perpendicular to L₁.
Slope of L₁ = (9 − 3)/(4 − 1) = 6/3 = 2
For perpendicular: m₁ × m₂ = −1 2 × m₂ = −1 m₂ = −1/2
Common Exam Variations
- Collinearity check — show that area = 0, or show that slope between any two pairs is the same
- Finding the fourth vertex of a parallelogram (use midpoint property — diagonals bisect each other)
- Circumcenter/Incenter — usually avoided in aptitude, but centroid is commonly tested
- Line through origin — always has the form y = mx (no c term, since it passes through (0,0))
- Distance of a point from a line — d = |ax₁ + by₁ + c| / √(a² + b²) for line ax + by + c = 0
Practice Problems
Q1: Show that the points (1, 1), (3, 5), and (5, 9) are collinear.
Q2: Find the equation of the line passing through (−2, 3) and (4, −1). Also find the x-intercept and y-intercept.
Q3: The vertices of a triangle are P(0, 0), Q(8, 0), and R(4, 6). Find the area of the triangle and the coordinates of the centroid.
Answers
A1: Area = ½|1(5−9) + 3(9−1) + 5(1−5)| = ½|−4 + 24 − 20| = ½|0| = 0. Since the area is zero, the points are collinear. (Alternatively: slope of AB = (5−1)/(3−1) = 2, slope of BC = (9−5)/(5−3) = 2. Same slope, so collinear.)
A2: Slope = (−1−3)/(4−(−2)) = −4/6 = −2/3. Using point-slope: y − 3 = −2/3(x + 2) → 3y − 9 = −2x − 4 → 2x + 3y − 5 = 0. x-intercept: set y = 0 → x = 5/2 = 2.5. y-intercept: set x = 0 → y = 5/3 ≈ 1.67.
A3: Area = ½|0(0−6) + 8(6−0) + 4(0−0)| = ½|0 + 48 + 0| = 24 sq units. Centroid = ((0+8+4)/3, (0+0+6)/3) = (4, 2).