On the coordinate plane, two points already fix their distance: the horizontal and vertical gaps make a right triangle whose hypotenuse is that distance, by Pythagoras.
The midpoint is just the average of the two points' coordinates on each axis.
A line, in turn, is set entirely by its slope and y-intercept.
Drag B to watch the distance and its triangle change, then adjust the slope and intercept to see the line pivot and slide.
Distance & Midpoint of Two Points
On the coordinate plane the distance uses Pythagoras, the midpoint is the average of coordinates.
4
3
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Distance and midpoint between A(1,1) and B
💡 Intuition for Distance & Midpoint
①Distance = hypotenuse of a right triangle (Pythagoras)
②Midpoint = average of the two coordinates
③Drawing the Δx, Δy right triangle makes it clear
Distance & Midpoint Formulas
Distance Between Two Points
d = √((x2 - x1)² + (y2 - y1)²)
Coordinate version of Pythagoras
Midpoint
M = (x1 + x22, y1 + y22)
Arithmetic mean of each coordinate
Line Equation — Slope & y-intercept
2
-1
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y = mx + b: a line set by slope and intercept
Slope-Intercept Form
y = mx + b
m: slope (Δy/Δx), b: y-intercept
Point + Slope
y - y1 = m(x - x1)
Line through (x₁, y₁) with slope m
Other Line Forms & Point-Line Distance
General Form
ax + by + c = 0
Covers every line (including vertical)
Distance from Point to Line
d = |ax1 + by1 + c|√(a² + b²)
From (x₁, y₁) to ax + by + c = 0
📐 Two Lines
①Parallel: m₁ = m₂
②Perpendicular: m₁ · m₂ = -1
③Identical: same slope and intercept
Work It Out
Example 1
Find the point P that divides A(1, 2) and B(7, 8) internally in the ratio 2 : 1.
1
Apply the internal-division formula with m : n = 2 : 1.
P = ( (2·7 + 1·1)/3, (2·8 + 1·2)/3 )
2
Compute.
P = (15/3, 18/3) = (5, 6)
▸ P(5, 6)
In ratio m : n, multiply the far endpoint by m and the near one by n.
Example 2
For A(−1, 3) and B(3, −1), find the length of segment AB and its midpoint M.
1
Use the distance and midpoint formulas.
AB = √((3 − (−1))² + (−1 − 3)²) = √(16 + 16)
2
Simplify.
AB = √32 = 4√2, M = ((−1 + 3)/2, (3 + (−1))/2) = (1, 1)
▸ AB = 4√2, midpoint M(1, 1)
For distance and midpoint, just substitute the coordinates.
Wrap-up
Distance Formula
d = √((x2 - x1)² + (y2 - y1)²)
Core tool on the coordinate plane
Point-to-Line Distance
d = |ax0 + by0 + c|√(a² + b²)
Shortest distance to the foot of perpendicular
Grade-10 school exam type
Find the distance from the point (2, 3) to the line 3x − 4y + 1 = 0.
①1
②2
③3
④4
⑤5
▸ ① 1
1
Use the point-to-line distance formula.
d = |a·x₁ + b·y₁ + c| / √(a² + b²)
2
Substitute the values.
d = |3·2 − 4·3 + 1| / √(9 + 16) = |−5| / 5 = 1
🎯 Exam Points
①Distance: square root of sum of squared differences
②Midpoint: special case of internal division (1:1)
③Slope = tanθ — angle of inclination
④Point-line distance: absolute value top, square root bottom