01 · Exact distance
Find the distance between (−1, 2) and (5, 5).
Hint
The changes are 6 and 3.
Worked solution
d = √(6² + 3²) = √45 = 3√5
Understand · explore · practise
Distance and area in coordinate geometry
Calculate exact distances, shortest perpendicular distances and triangle or quadrilateral areas. Combine line intersections with geometry and explore triangle altitudes.
Before you startPythagoras, surds, line equations and perpendicular gradients
01 / Distance
Pythagoras turns coordinate differences into a length.
The horizontal and vertical changes make the two perpendicular sides of a right triangle. The segment between the points is its hypotenuse.
d = √[(x₂ − x₁)² + (y₂ − y₁)²]
Between A(−2, −1) and B(4, 2), the changes are 6 and 3. Hence d = √(36 + 9) = √45 = 3√5.
Squared lengths are enough when you only need to compare lengths: equal positive squares give equal lengths. Keep the square root exact unless a decimal is requested.
A parameter may need an absolute value
Between (2a, a) and (−a, 3a), d = √(9a² + 4a²) = √(13a²) = |a|√13. Do not write a√13 unless a ≥ 0 is known. At a = 0 the points coincide and the distance is zero.
Pythagoras on a coordinate gridDistance
A(−2,−1) and B(4,2) differ by 6 horizontally and 3 vertically. The diagonal length is √45 = 3√5.
02 / Unknown points
Square a known distance to form an equation.
If one coordinate is unknown, insert it into the distance formula and square both sides. Two positions may have the same distance from the fixed point.
Distance from (−2, 1) to (x, 5) is √41:
(x + 2)² + 16 = 41
(x + 2)² = 25
x = 3 or −7
If a point lies on a line, substitute that line’s equation for its y-coordinate before solving. Keep every candidate that satisfies the original distance and line.
A point on a line at a fixed distanceWorked example
P lies on y = 2x − 1
OP = √13
Use x² + y² = 13.
x² + (2x − 1)² = 13
5x² − 4x − 12 = 0
Expand and collect terms.
(5x + 6)(x − 2) = 0
Find each corresponding y-value.
P = (2, 3) or (−6/5, −17/5)
Their squared distances are both 13.
03 / Shortest distance
The shortest route to a line is perpendicular.
For a point P and a line, construct the perpendicular through P, find its intersection H with the line, then calculate PH. This is the perpendicular distance.
For two parallel lines, choose any convenient point on one and apply the same method to the other. A vertical gap between sloping lines is generally longer than their shortest separation.
Distance between parallel linesWorked example
y = 2x + 1 and y = 2x + 6
Choose P = (0, 1) on the first
Both lines have gradient 2.
Perpendicular: y = −x/2 + 1
Find its intersection with the second line.
−x/2 + 1 = 2x + 6
H = (−2, 2)
The displacement from P is (−2, 1).
PH = √(4 + 1) = √5
The y-intercepts differ by 5, but the perpendicular separation is √5.
04 / Triangle area
The height must be perpendicular to your base.
Area = ½ × base × perpendicular height. When the base is horizontal, the height is the absolute difference between the apex’s y-coordinate and the base’s y-coordinate.
Here A(−2, 0) and B(4, 0) give a base of 6. Moving C sideways changes the side lengths but not the height or area. Try putting C beyond the end of the base, or below the axis.
C = (u, v)
Area = ½ × 6 × |v| = 3|v|
At v = 0 the points are collinear and the area is zero: the triangle has collapsed. A negative coordinate never gives a negative area.
Fixed base: 6 unitsMove the apex
C = (1, 3). Height = |3| = 3. Area = ½ × 6 × 3 = 9 square units. Move C sideways: the area stays the same.
Watch base and perpendicular height determine area
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 / Find vertices first
Line equations can supply the missing base and height.
A diagram may define a triangle using intersections rather than giving coordinates directly. Find the vertices first, then choose a convenient base.
For a vertical base, use the difference in y-coordinates as its length and the horizontal distance from the opposite vertex as its perpendicular height.
Two lines and the x-axisWorked example
y = 2x and x + y = 9
They meet at A; the second meets the x-axis at B
Let O be the origin.
3x = 9 ⇒ A = (3, 6)
At y = 0: B = (9, 0)
OB is a horizontal base of length 9.
Area OAB = ½ × 9 × 6 = 27
The height is 6, not the sloping length OA.
06 / Other areas
Choose a method that keeps the working short.
With a sloping base, find a perpendicular height using line equations. Alternatively, enclose the triangle in an axis-aligned rectangle and subtract the outside triangles.
A useful optional formula for A(x₁,y₁), B(x₂,y₂), C(x₃,y₃) is:
Area = ½ |(x₂ − x₁)(y₃ − y₁)
− (y₂ − y₁)(x₃ − x₁)|
This is the area formula obtained by subtracting coordinate triangles. The absolute value makes the answer independent of clockwise or anticlockwise ordering.
For a quadrilateral, split it along a diagonal into two non-overlapping triangles. Use a specialised rectangle or trapezium formula only after checking that the shape has the required properties.
A trapezium
A(0,0), B(6,0), C(4,3), D(1,3) have parallel bases AB = 6 and DC = 3, separated by height 3. Its area is ½(6 + 3) × 3 = 27/2 square units. Splitting along AC gives triangle areas 9 and 9/2, with the same total.
An oblique triangleWorked example
A = (−2, 1), B = (4, 3), C = (1, 6)
The two displacements from A are (6, 2) and (3, 5).
Area = ½ |6 · 5 − 2 · 3|
= 12
A coordinate-area calculation.
AB: y = (x + 5)/3
Altitude through C: y = −3x + 9
For a second method, find the perpendicular foot.
H = (11/5, 12/5)
AB = 2√10, CH = 6√10/5
½ × AB × CH = 12, confirming the same area.
07 / Altitudes
Three perpendicular constructions meet at one point.
An altitude passes through a triangle’s vertex and is perpendicular to the opposite side, extended if necessary. Their common intersection is called the orthocentre.
For A(0,0), B(4,6), C(8,0), the altitude through B is x = 4. BC has gradient −3/2, so the altitude through A is y = 2x/3. They meet at H(4,8/3).
AB has gradient 3/2, so the third altitude through C is y = −2(x − 8)/3. Substituting x = 4 gives y = 8/3 again: all three meet at H.
Generalise without assuming every side has a finite slope
Take O(0,0), U(p,q), V(r,0), with q ≠ 0 and r ≠ 0 so the triangle is non-degenerate. The altitude through U is x = p. An equation of the altitude through O is qy = (r − p)x, giving H = (p, p(r − p)/q).
The altitude through V has equation px + qy = pr. Inserting H gives p² + p(r − p) = pr, so it also passes through H. These equations still work when p = 0 or p = r; then one of the other sides is vertical and the corresponding altitude is horizontal.
Three altitudes, one intersectionExtension
A(0,0), B(4,6), C(8,0). The three altitudes meet at H(4,8/3), inside the triangle.
08 / Your turn
Keep lengths and areas positive and exact.
Use a rough sketch to decide which distance is a perpendicular height.
02 · Unknown coordinate
The distance from (2, −1) to (x, 3) is 5. Find both x values.
Hint
(x − 2)² + 4² = 25.
Worked solution
(x − 2)² = 9
x = 5 or −1
03 · A point on a line
P lies on y = x + 3 and is 5√2 units from A(−1, 2). Find P.
Hint
Both coordinate differences equal x + 1.
Worked solution
(x + 1)² + (x + 1)² = 50
(x + 1)² = 25
x = 4 or −6
P = (4, 7) or (−6, −3).
04 · Shortest separation
Find the distance between y = 3x − 2 and y = 3x + 8.
Hint
Use P(0,−2) and its perpendicular y = −x/3 − 2.
Worked solution
−x/3 − 2 = 3x + 8
H = (−3, −1)
PH = √(9 + 1) = √10
05 · A base above the axis
A(−3,2), B(5,2), C(1,−4) form a triangle. Find its area.
Hint
The height is a difference in coordinates, not just |−4|.
Worked solution
Base = 8
Height = |−4 − 2| = 6
Area = ½ × 8 × 6 = 24
06 · A sloping triangle
Find the area of A(1,1), B(7,3), C(3,6).
Hint
Use displacements (6,2) and (2,5), or construct a perpendicular height.
Worked solution
Area = ½ |6 · 5 − 2 · 2|
= 13
07 · A parameter sign
Find the distance from (2a,a) to (−a,3a), then evaluate it for a = −2.
Hint
The square root of a² is |a|.
Worked solution
d = √(9a² + 4a²) = |a|√13
At a = −2: d = 2√13
08 · An altitude intersection
A(0,0), B(3,4), C(9,0) form a triangle. Find where its three altitudes meet.
Hint
The altitude through B is x = 3. Find the altitude through A from the gradient of BC.
Worked solution
mBC = −4/6 = −2/3
Altitude from A: y = 3x/2
At x = 3: H = (3, 9/2)
AB has gradient 4/3. Its perpendicular through C is y = −3(x − 9)/4, which also gives y = 9/2 at x = 3. H lies outside this obtuse triangle, which is valid.
09 / Recap
Choose lengths that match the geometry.
- Use Pythagoras on coordinate differences.
- Keep exact surds, and use |a| when taking √a².
- Shortest distances to lines are perpendicular.
- Area uses a perpendicular height, even if its foot is outside the base segment.
- Find intersection coordinates before calculating lengths or areas.
- Check a third altitude if you claim that all three are concurrent.
Section 1 of 9 · Distance