On the coordinate plane, knowing the coordinates of two points is enough to work out the distance between them: draw a right-angled triangle whose sides are the horizontal and vertical differences, and its hypotenuse is the distance, so Pythagoras' theorem does the rest.
The midpoint, halfway between the two points, is found by averaging the x-coordinates and averaging the y-coordinates.
A straight line is fixed by just two numbers: its gradient, which says how steep it is, and its y-intercept, where it crosses the y-axis.
In the first picture, moving point B changes the right-angled triangle, the distance and the midpoint together; in the second, changing the gradient and the intercept turns the line or shifts it up and down.
Distance and midpoint of two points
On the coordinate plane, the distance between two points comes from Pythagoras' theorem, and their midpoint from averaging the coordinates.
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The distance from the fixed point A(1, 1) to the moving point B found with a right-angled triangle, with the midpoint M marked
💡 Seeing distance and midpoint
①The distance between two points is the hypotenuse of a right-angled triangle with sides Δx (the horizontal difference) and Δy (the vertical difference)
②So the distance is √(Δx² + Δy²)
③The midpoint has as its coordinates the averages of the two x-coordinates and of the two y-coordinates
④In the picture A stays fixed at (1, 1) and only B moves
Distance and midpoint formulas
Distance between two points
d = √((x2 − x1)² + (y2 − y1)²)
Pythagoras' theorem written in coordinates
Coordinates of the midpoint
M = (x1 + x22, y1 + y22)
Average the x-coordinates and the y-coordinates separately
Equation of a line: gradient and y-intercept
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The line y = mx + b as the gradient m and y-intercept b change, with a small right-angled triangle showing y changing by m when x increases by 1
Gradient-intercept form
y = mx + b
m is the gradient (change in y when x increases by 1), b is the y-intercept
Line through a point with a given gradient
y − y1 = m(x − x1)
The line through (x₁, y₁) with gradient m
General form and the distance from a point to a line
General form of a line
ax + by + c = 0
Covers every line, including vertical lines x = k
Distance from a point to a line
d = |ax1 + by1 + c|√(a² + b²)
From the point (x₁, y₁) to the line ax + by + c = 0
📐 How two lines can be related
①Two lines with the same gradient and different y-intercepts are parallel
②If both the gradient and the y-intercept are the same, the lines coincide
③If the product of the gradients is −1, that is m₁m₂ = −1, the lines are perpendicular
Worked examples and unit test
Example 1
Find the coordinates of the point P that divides the segment joining A(1, 2) and B(7, 8) internally in the ratio 2 : 1.
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When AP : PB = m : n, the point is P = ((m·x₂ + n·x₁)/(m + n), (m·y₂ + n·y₁)/(m + n)). Put m : n = 2 : 1.
P = ( (2·7 + 1·1)/3, (2·8 + 1·2)/3 )
2
Work it out.
P = (15/3, 18/3) = (5, 6)
▸ P(5, 6)
The first number of the ratio, m, multiplies B's coordinates and the second, n, multiplies A's: they cross over. Checking that P is nearer to B helps catch mistakes.
Example 2
For A(−1, 3) and B(3, −1), find the length of segment AB and the coordinates of 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, M(1, 1)
The horizontal and vertical differences are both 4, so AB is as long as the diagonal of a square with side 4.
Summary
Distance between two points
d = √((x2 − x1)² + (y2 − y1)²)
The most basic tool on the coordinate plane
Distance from a point to a line
d = |ax0 + by0 + c|√(a² + b²)
The length of the perpendicular from the point to the line, the shortest distance
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)
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Substitute the values.
d = |3·2 − 4·3 + 1| / √(9 + 16) = |−5| / 5 = 1
🎯 Points that often come up
①The distance formula squares the coordinate differences, adds them and takes the square root
②The midpoint divides the segment internally in the ratio 1 : 1
③For the angle θ a line makes with the positive x-axis, its gradient is m = tan θ
④In the point-to-line distance, do not drop the absolute value on top or the square root underneath
⑤The perpendicularity condition m₁m₂ = −1 comes up often