Using pronumerals in place of numbers

Prove general geometric results with coordinates by placing a figure cleverly and using pronumerals for its vertices.

Why pronumerals?

A result checked with numbers is true for one figure only. Using pronumerals such as $A(2a, 0)$ proves it for every figure of that type at once.

Example 1: the midpoint of the hypotenuse●●●

Prove that the midpoint of the hypotenuse of a right-angled triangle is the same distance from all three vertices.

Let $O(0, 0)$, $A(2a, 0)$ and $B(0, 2b)$.

Put the right angle at the origin and the two shorter sides along the axes.

The midpoint of $AB$ is $M(a, b)$.

$MO = \sqrt{a^2 + b^2}$ and $MA = \sqrt{(2a - a)^2 + (0 - b)^2} = \sqrt{a^2 + b^2}$.

$MB = \sqrt{(0 - a)^2 + (2b - b)^2} = \sqrt{a^2 + b^2}$. All three distances are equal.

Check the result on any right-angled triangle

Solid: the hypotenuse AB (the other two sides lie on the axes). Dashed: the circle with centre M through O. Change a and b: the circle always passes through O, A and B.

Try this

Why does this mean the hypotenuse is a diameter of the circle through all three vertices?

Example 2: the midpoint theorem●●●

Prove that the interval joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Let $P(2a, 2b)$, $Q(2c, 2d)$ and $R(2e, 2f)$. The midpoints of $PQ$ and $PR$ are $M(a + c, b + d)$ and $N(a + e, b + f)$.

$m_{MN} = \dfrac{f - d}{e - c}$ and $m_{QR} = \dfrac{2f - 2d}{2e - 2c} = \dfrac{f - d}{e - c}$, so $MN \parallel QR$.

$MN = \sqrt{(e - c)^2 + (f - d)^2}$ and $QR = 2\sqrt{(e - c)^2 + (f - d)^2}$, so $MN = \tfrac12 QR$.

Flashcards

Why use 2a instead of a?

Midpoints come out without fractions.

Best placement

A vertex at the origin, a side along an axis.

Practice

The midpoint of $(2a, 0)$ and $(0, 2b)$ is:
The distance from $(0, 0)$ to $(a, b)$ is:
The gradient of the interval from $(2a, 0)$ to $(0, 2b)$ is:
A good placement for a general rectangle is:
In that rectangle, the diagonals both have midpoint:
Why does a numerical example not prove a general result?
The product of the gradients from $(a, b)$ to $(2a, 0)$ and from $(0, 0)$ to $(a, b)$ is $-1$ exactly when:
With $a = 3$ and $b = 4$ in Example 1, find $MO$.