Quadrilaterals with coordinates

In plane geom­e­try we proved prop­er­ties of par­al­lel­o­grams, rec­tan­gles, rhom­buses and squares. With coor­di­nates we can turn those prop­er­ties into cal­cu­la­tions: given four points, we can decide exactly what kind of quadri­lat­eral they make, find a miss­ing ver­tex, or locate a point that is the same dis­tance from three oth­ers. This les­son com­bines the dis­tance for­mula and the mid­point for­mula to do all of that.

The tests#

Take the ver­tices $A$, $B$, $C$, $D$ in order round the fig­ure.

A rhom­bus whose diag­o­nals are unequal is not a square; a par­al­lel­o­gram whose diag­o­nals are unequal is not a rec­tan­gle. Com­pare squared dis­tances to avoid square roots.

Exam­ple 1#

Show that $A(1, 2)$, $B(4, 3)$, $C(6, 6)$, $D(3, 5)$ form a par­al­lel­o­gram. Is it a rec­tan­gle or a rhom­bus?

$$\displaystyle \text{Midpoint of } AC = \left(\frac{1 + 6}{2}, \frac{2 + 6}{2}\right) = \left(\frac{7}{2}, 4\right), \qquad \text{Midpoint of } BD = \left(\frac{4 + 3}{2}, \frac{3 + 5}{2}\right) = \left(\frac{7}{2}, 4\right)$$

The diag­o­nals bisect each other, so $ABCD$ is a par­al­lel­o­gram.

Sides: $AB^2 = 9 + 1 = 10$ and $BC^2 = 4 + 9 = 13$. Adja­cent sides are unequal, so it is not a rhom­bus.

Diag­o­nals: $AC^2 = 25 + 16 = 41$ and $BD^2 = 1 + 4 = 5$. They are unequal, so it is not a rec­tan­gle either.

Exam­ple 2#

Show that $A(2, 1)$, $B(6, 4)$, $C(3, 8)$, $D(-1, 5)$ are the ver­tices of a square.

$$AB^2 = 16 + 9 = 25, \quad BC^2 = 9 + 16 = 25, \quad CD^2 = 16 + 9 = 25, \quad DA^2 = 9 + 16 = 25$$

All sides are $5$.

$$AC^2 = 1 + 49 = 50, \qquad BD^2 = 49 + 1 = 50$$

The diag­o­nals are equal too, so $ABCD$ is a square.

Exam­ple 3#

Show that $A(3, 0)$, $B(6, 4)$, $C(3, 8)$, $D(0, 4)$ form a rhom­bus, and find its area.

$$AB^2 = 9 + 16 = 25, \quad BC^2 = 9 + 16 = 25, \quad CD^2 = 9 + 16 = 25, \quad DA^2 = 9 + 16 = 25$$

All sides are equal, so it is a rhom­bus. The diag­o­nals are $AC = 8$ (ver­ti­cal) and $BD = 6$ (hor­i­zon­tal), which are unequal, so it is not a square.

The area of a rhom­bus is half the prod­uct of its diag­o­nals: $\displaystyle \frac{1}{2} \times 8 \times 6 = 24$ square units.

Exam­ple 4#

Show that $A(0, 0)$, $B(4, 2)$, $C(3, 4)$, $D(-1, 2)$ form a rec­tan­gle.

The mid­points of $AC$ and $BD$ are both $\displaystyle \left(\frac{3}{2}, 2\right)$, so it is a par­al­lel­o­gram. The diag­o­nals are $AC = \sqrt{9 + 16} = 5$ and $BD = \sqrt{25 + 0} = 5$, which are equal, so it is a rec­tan­gle. (Its sides, $\sqrt{20}$ and $\sqrt{5}$, are unequal, so it is not a square.)

Find­ing a miss­ing ver­tex#

In par­al­lel­o­gram $ABCD$, the diag­o­nals share a mid­point, so $\displaystyle \frac{A + C}{2} = \frac{B + D}{2}$ coor­di­nate by coor­di­nate. Hence

$$D = A + C - B$$

Exam­ple 5#

Three ver­tices of par­al­lel­o­gram $ABCD$ are $A(-1, 0)$, $B(3, 1)$ and $C(2, 2)$. Find $D$.

$$D = (-1 + 2 - 3, \; 0 + 2 - 1) = (-2, 1)$$

Check: the mid­point of $AC$ is $\displaystyle \left(\frac{1}{2}, 1\right)$ and the mid­point of $BD$ is $\displaystyle \left(\frac{3 - 2}{2}, \frac{1 + 1}{2}\right) = \left(\frac{1}{2}, 1\right)$.

A point equidis­tant from three points#

The point equidis­tant from the three ver­tices of a tri­an­gle is its cir­cum­cen­tre, the cen­tre of the cir­cle through all three ver­tices. To find it, let it be $P(x, y)$ and set two pairs of squared dis­tances equal.

Exam­ple 6#

Find the point equidis­tant from $A(1, 1)$, $B(7, 1)$ and $C(1, 9)$, and the radius of the cir­cle through them.

From $PA^2 = PB^2$:

$$(x - 1)^2 + (y - 1)^2 = (x - 7)^2 + (y - 1)^2 \implies -2x + 1 = -14x + 49 \implies x = 4$$

From $PA^2 = PC^2$:

$$(x - 1)^2 + (y - 1)^2 = (x - 1)^2 + (y - 9)^2 \implies -2y + 1 = -18y + 81 \implies y = 5$$

So $P = (4, 5)$, and the radius is $PA = \sqrt{9 + 16} = 5$. (Tri­an­gle $ABC$ is right-angled at $A$, and $P$ is the mid­point of the hypotenuse $BC$, as the angle-in-a-semi­cir­cle the­o­rem pre­dicts.)

Prac­tice#

1. Do $(2, 3)$, $(5, 4)$, $(6, 7)$, $(3, 6)$, taken in order, form a par­al­lel­o­gram? Is it a rhom­bus? A square?
2. Three ver­tices of par­al­lel­o­gram $ABCD$ are $A(1, 2)$, $B(5, 3)$ and $C(6, 7)$. Find $D$.
3. Show that $(-1, 0)$, $(3, 3)$, $(6, -1)$, $(2, -4)$ are the ver­tices of a square.
4. Show that $(2, -1)$, $(5, 3)$, $(2, 7)$, $(-1, 3)$ form a rhom­bus, and find its area.
5. Show that $(1, 0)$, $(5, 2)$, $(4, 4)$, $(0, 2)$ form a rec­tan­gle.
6. Find the point equidis­tant from $(0, 0)$, $(6, 0)$ and $(0, 8)$.
7. The points $(1, 2)$, $(4, y)$, $(x, 6)$ and $(3, 5)$, taken in order, are the ver­tices of a par­al­lel­o­gram. Find $x$ and $y$.

Answers#