ACARA v9 CONTENT DESCRIPTION “find the gradient of a line segment, the midpoint of the line interval and the distance between 2 distinct points on the Cartesian plane”
Builds on: Real Numbers: Rational and Irrational (AC9M9N01). This unit builds on the real number system, square roots, and Pythagoras' theorem. The coordinate methods here set up straight-line graphs and the geometry you will meet in measurement and space later in Year 9.
Three questions about two points
Once two points are pinned to the Cartesian plane, three questions almost always follow: how steep is the line joining them, where is its exact middle, and how far apart are they. Year 9 answers all three with short formulas that are really just careful bookkeeping of horizontal and vertical change. The beauty is that every one of them reduces to the same two raw ingredients: the change in x and the change in y between the points.
Two points, two differences
Every formula in this unit starts from the change in x and the change in y between the points.
P(1, 2) and Q(4, 8) sit on the plane. Reveal the horizontal and vertical differences between them.
Gradient: rise over run
Gradient measures steepness as the ratio of vertical change to horizontal change, often remembered as rise over run. For two points, you subtract the y-values to get the rise, subtract the x-values in the same order to get the run, and divide. Joining one comma two to four comma eight, the rise is eight minus two, which is six, and the run is four minus one, which is three, so the gradient is six over three, equal to two. The order matters only in that you must be consistent: subtract the coordinates of the same point first in both differences. A positive gradient rises to the right, a negative gradient falls, and a horizontal line has a gradient of zero because its rise is nothing at all, as for two comma three joined to six comma three.
Gradient as rise over run
Gradient is the vertical change divided by the horizontal change, never the other way up.
One segment climbs steeply, the other is flat. Reveal the gradient of each as rise over run.
Midpoint: averaging the coordinates
The midpoint is gentler: it is simply the average of the two x-coordinates paired with the average of the two y-coordinates. To average, add and halve. The midpoint of one comma two and four comma eight is the average of the x-values, two and a half, with the average of the y-values, five, giving the point two and a half comma five. This always lands exactly halfway along the segment, and because it is just averaging, it never produces anything more exotic than a fraction.
Midpoint by averaging
The midpoint is the average of the two x-coordinates paired with the average of the two y-coordinates.
The segment from P(1, 2) to Q(4, 8). Reveal its exact middle by averaging the coordinates.
Distance: Pythagoras on the grid
Distance is where the plane reveals its hidden right angle. Between two points, the horizontal and vertical changes are the two short sides of a right-angled triangle, and the segment itself is the hypotenuse, so Pythagoras applies directly. Square the change in x, square the change in y, add them, and take the square root. From three comma four down to the origin, the changes are three and four, so the distance squared is nine plus sixteen, which is twenty-five, and the distance is the square root of twenty-five, exactly five. This is the familiar three, four, five right triangle hiding on the grid.
Distance is the hypotenuse
The horizontal and vertical changes are the legs of a right triangle, so Pythagoras gives the distance.
From (0, 0) to (3, 4) the legs are 3 and 4. Reveal the distance along the blue hypotenuse.
When the distance is irrational
Distances are not always whole numbers, and this is where earlier work on real numbers pays off. Joining one comma two to four comma eight, the distance squared is nine plus thirty-six, which is forty-five, so the distance is the square root of forty-five. That value is irrational; it simplifies to three times the square root of five, and is approximately six point seven one, but it is never exactly six point seven one. Leaving the answer as the square root of forty-five, or as three root five, is the exact form; the decimal is only an approximation, and it is important not to write an equals sign between them. Similarly the distance from minus two comma three to four comma minus three is the square root of seventy-two, or six root two, about eight point four nine.
An irrational distance
Many distances are irrational; keep the exact surd form and treat the decimal only as an approximation.
From (1, 2) to (4, 8) the legs are 3 and 6. Reveal the distance, which turns out to be irrational.
Watching signs and order
A few habits keep this reliable. Always compute the change in x and the change in y once, carefully, and reuse them for all three calculations, since gradient, midpoint and distance all feed on the same differences. Watch signs when points sit in negative regions, because a change such as minus one to two is a rise of three, not one. Keep gradient as rise over run, never the other way up, and remember that distance can never be negative because it comes from a square root of a sum of squares.
Sign care with negatives
When points sit in negative regions, the sign of each difference decides the gradient; distance stays positive.
From (−2, 3) to (4, −3), watch the sign of each difference. Reveal the gradient and distance.
Three views of the same two points
With those differences computed once and handled cleanly, the three formulas stop feeling like separate rules and start looking like three views of the same pair of points. Gradient reads their slope, the midpoint marks their balance point, and distance measures the straight line between them, all from one pair of subtractions.
Quick self-check
1. Find the gradient of the segment joining (1, 2) and (4, 8).
2. What is the gradient of the segment joining (2, 3) and (6, 3)?
3. Find the midpoint of (1, 2) and (4, 8).
4. Find the distance between (0, 0) and (3, 4).
5. The distance between (1, 2) and (4, 8) is √45. Which is correct?