Coordinate Geometry Calculator

Distance AB
10
√(6² + 8²) = √100
Midpoint
(4, 7)
((1 + 7) ÷ 2, (3 + 11) ÷ 2)
Slope
4/3
(11 − 3) ÷ (7 − 1)
Line AB
4x − 3y + 5 = 0
y = 4/3x + 5/3
Point dividing AB in 2 : 1
(5, 25/3)
Area of triangle ABC
24

Coordinate geometry formulas

Coordinate geometry describes points by their (x, y) coordinates, so lengths, midpoints and areas can be calculated with formulas instead of measured. This calculator covers the Class 10 NCERT formulas: the distance formula, the section formula, the midpoint and the area of a triangle, plus the slope and equation of the line through two points.

Formulas

AB = √((x₂ − x₁)² + (y₂ − y₁)²) · P = ((mx₂ + nx₁) ÷ (m + n), (my₂ + ny₁) ÷ (m + n)) · Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| · slope = (y₂ − y₁) ÷ (x₂ − x₁)

The section formula gives the point P that divides AB internally in the ratio m : n; the midpoint is the case m = n = 1. Points are collinear when the triangle they form has zero area.

Good to know

  • The distance of a point (x, y) from the origin is √(x² + y²).
  • Parallel lines have equal slopes; perpendicular lines have slopes whose product is −1.
  • A vertical line has an undefined slope and the equation x = constant.
  • The centroid of a triangle is ((x₁ + x₂ + x₃) ÷ 3, (y₁ + y₂ + y₃) ÷ 3).
  • To check whether three points are collinear, find the area of the triangle they make: it is 0 if they're on one line.

Worked example: A(1, 3) and B(7, 11)

  1. Distance AB = √((7 − 1)² + (11 − 3)²) = √(36 + 64) = √100 = 10
  2. Midpoint = ((1 + 7) ÷ 2, (3 + 11) ÷ 2) = (4, 7)
  3. Slope = (11 − 3) ÷ (7 − 1) = 8/6 = 4/3; line AB: 4x − 3y + 5 = 0
  4. Point dividing AB in 2 : 1 = ((2 × 7 + 1 × 1) ÷ 3, (2 × 11 + 1 × 3) ÷ 3) = (5, 25/3)
  5. With C(1, 11): area = ½ |1(11 − 11) + 7(11 − 3) + 1(3 − 11)| = ½ × 48 = 24

Distances between common points

Pythagorean-triple distances that give whole numbers.

ABDistance AB
(0, 0)(3, 4)5
(1, 3)(7, 11)10
(2, -1)(7, 11)13
(-3, 2)(5, 17)17
(0, 0)(8, 15)17
(-1, -1)(6, 23)25

Distance = √(Δx² + Δy²).

Frequently Asked Questions

AB = √((x₂ − x₁)² + (y₂ − y₁)²). The distance between (1, 3) and (7, 11) is √(36 + 64) = 10.
The point dividing the line joining (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁) ÷ (m + n), (my₂ + ny₁) ÷ (m + n)).
Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. For (1, 3), (7, 11) and (1, 11), the area is 24.
Find the slope m = (y₂ − y₁) ÷ (x₂ − x₁), then use y − y₁ = m(x − x₁). Through (1, 3) and (7, 11): y − 3 = 4/3 (x − 1), or 4x − 3y + 5 = 0.