Verifying Geometric Properties

A sketch can make a shape look like a square or a right triangle, but looks can fool you. With coordinates you can prove it, using just three tools: slope, length and midpoint. This page shows how to choose the right tool, plan a multi-step check, and finish with a clear concluding statement.

Key ideas

Three tools, three questions

To show that …	use …	and check that …
two segments are parallel	slope	the slopes are equal
two segments are perpendicular	slope	the slopes are negative reciprocals (product $-1$)
two segments are equal in length	length formula	the lengths are equal
one segment is half another	length formula	one length is half the other
two segments bisect each other	midpoint formula	they have the same midpoint

Slopes of parallel and perpendicular lines are covered on parallel and perpendicular lines.

Classifying triangles

• Right triangle: two sides are perpendicular (slopes are negative reciprocals). You can also use the Pythagorean theorem on the side lengths.
• Isosceles: exactly two sides have equal length (if all three are equal, it’s equilateral).
• Equilateral: all three sides have equal length.
• Scalene: no two sides are equal.

Classifying quadrilaterals

Each special quadrilateral has a property you can test:

Shape	One way to verify it
Parallelogram	both pairs of opposite sides are parallel (or: the diagonals bisect each other)
Rectangle	a parallelogram with a right angle (adjacent sides perpendicular)
Rhombus	all four sides are equal
Square	all four sides are equal and there’s a right angle

Name the vertices in order around the shape: in $ABCD$, the sides are $AB$, $BC$, $CD$ and $DA$, and the diagonals are $AC$ and $BD$.

The midsegment of a triangle

A midsegment joins the midpoints of two sides of a triangle. It is always parallel to the third side and half as long. Example 3 verifies this for one triangle.

Planning and concluding

1. Plan: What property do you need? Which tool tests it (slope, length or midpoint)? Which segments?
2. Calculate: show each calculation clearly, labelled with the segment it’s for.
3. Conclude: finish with a sentence that states the result and the reason, like “Since $m_{AB} \times m_{BC} = -1$, $AB \perp BC$, so triangle $ABC$ is a right triangle with the right angle at $B$.”

A quick sketch on a grid helps with the plan, and graphing or dynamic geometry software is a good way to check, but the calculations are the proof.

Worked examples

Example 1: A right triangle, two ways

Show that the triangle with vertices $A(-3, 1)$, $B(1, 3)$ and $C(2, 1)$ is a right triangle.

Plan. A sketch suggests the right angle is at $B$. Check whether $AB$ and $BC$ are perpendicular.

Solution (slopes).

$$m_{AB} = \frac{3 - 1}{1 - (-3)} = \frac{2}{4} = \frac{1}{2} \qquad m_{BC} = \frac{1 - 3}{2 - 1} = \frac{-2}{1} = -2$$

$\dfrac{1}{2} \times (-2) = -1$, so the slopes are negative reciprocals and $AB \perp BC$.

Solution (lengths).

$$AB = \sqrt{4^2 + 2^2} = \sqrt{20} \qquad BC = \sqrt{1^2 + (-2)^2} = \sqrt{5} \qquad AC = \sqrt{5^2 + 0^2} = 5$$

$AB^2 + BC^2 = 20 + 5 = 25 = AC^2$, so the Pythagorean theorem holds.

Conclusion. Triangle $ABC$ is a right triangle with the right angle at $B$.

Example 2: A rhombus that isn’t a square

Show that $PQRS$ with $P(-3, 0)$, $Q(1, 3)$, $R(6, 3)$ and $S(2, 0)$ is a rhombus. Is it a square?

Plan. Rhombus: check that all four sides are equal. Square: also check for a right angle.

Solution.

$$\begin{aligned} PQ &= \sqrt{(1 - (-3))^2 + (3 - 0)^2} = \sqrt{16 + 9} = 5 \\ QR &= 6 - 1 = 5 && \text{horizontal} \\ RS &= \sqrt{(2 - 6)^2 + (0 - 3)^2} = \sqrt{16 + 9} = 5 \\ SP &= 2 - (-3) = 5 && \text{horizontal} \end{aligned}$$

All four sides are $5$, so $PQRS$ is a rhombus.

For a right angle at $Q$, check $PQ$ and $QR$: $m_{PQ} = \dfrac{3}{4}$ and $m_{QR} = 0$. These are not negative reciprocals, so the angle at $Q$ is not a right angle.

Conclusion. $PQRS$ is a rhombus (all sides equal $5$), but it is not a square, because adjacent sides $PQ$ and $QR$ are not perpendicular.

Bonus: the diagonals of a rhombus are perpendicular. Here $m_{PR} = \dfrac{3}{9} = \dfrac{1}{3}$ and $m_{QS} = \dfrac{-3}{1} = -3$, and $\dfrac{1}{3} \times (-3) = -1$. ✓

Example 3: The midsegment of a triangle

Triangle $ABC$ has vertices $A(-2, 6)$, $B(-4, -2)$ and $C(6, 2)$. $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$. Verify that $MN$ is parallel to $BC$ and half its length.

Plan. Find $M$ and $N$ (midpoint). Parallel: compare slopes of $MN$ and $BC$. Half: compare lengths.

Solution.

$$M = \left( \frac{-2 + (-4)}{2},\ \frac{6 + (-2)}{2} \right) = (-3, 2) \qquad N = \left( \frac{-2 + 6}{2},\ \frac{6 + 2}{2} \right) = (2, 4)$$

Slopes:

$$m_{MN} = \frac{4 - 2}{2 - (-3)} = \frac{2}{5} \qquad m_{BC} = \frac{2 - (-2)}{6 - (-4)} = \frac{4}{10} = \frac{2}{5}$$

Lengths:

$$MN = \sqrt{5^2 + 2^2} = \sqrt{29} \qquad BC = \sqrt{10^2 + 4^2} = \sqrt{116}$$

Is $\sqrt{116}$ twice $\sqrt{29}$? Yes: $2\sqrt{29} = \sqrt{4 \times 29} = \sqrt{116}$. (Or compare decimals: $\sqrt{29} \approx 5.385$ and $\sqrt{116} \approx 10.770$.)

Conclusion. $m_{MN} = m_{BC} = \dfrac{2}{5}$, so $MN \parallel BC$, and $MN = \sqrt{29}$ is half of $BC = 2\sqrt{29}$. The midsegment is parallel to the third side and half its length.

The midsegment $MN$ is parallel to $BC$ and half as long.

Example 4: The diagonals of a rectangle bisect each other

Verify that $JKLM$ with $J(-2, 1)$, $K(2, 5)$, $L(4, 3)$ and $M(0, -1)$ is a rectangle, and that its diagonals bisect each other.

Plan. Rectangle: opposite sides parallel and adjacent sides perpendicular (slopes). Bisect: diagonals $JL$ and $KM$ have the same midpoint.

Solution. Slopes of the sides:

$$m_{JK} = \frac{5 - 1}{2 - (-2)} = 1 \qquad m_{KL} = \frac{3 - 5}{4 - 2} = -1 \qquad m_{LM} = \frac{-1 - 3}{0 - 4} = 1 \qquad m_{MJ} = \frac{1 - (-1)}{-2 - 0} = -1$$

Opposite sides have equal slopes ($JK \parallel LM$ and $KL \parallel MJ$), so $JKLM$ is a parallelogram. Adjacent sides have slopes $1$ and $-1$, whose product is $-1$, so the corners are right angles. $JKLM$ is a rectangle.

Midpoints of the diagonals:

$$\text{Midpoint of } JL = \left( \frac{-2 + 4}{2},\ \frac{1 + 3}{2} \right) = (1, 2) \qquad \text{Midpoint of } KM = \left( \frac{2 + 0}{2},\ \frac{5 + (-1)}{2} \right) = (1, 2)$$

Conclusion. $JKLM$ is a rectangle, and both diagonals have midpoint $(1, 2)$, so the diagonals bisect each other. (They’re also equal in length: $JL = \sqrt{6^2 + 2^2} = \sqrt{40}$ and $KM = \sqrt{(-2)^2 + (-6)^2} = \sqrt{40}$.)

Common mistakes

Trusting the sketch. A shape that looks square might have sides of $\sqrt{10}$ and $\sqrt{9}$. Always calculate; the sketch is only for planning.

Checking too little. Four equal sides makes a rhombus, not necessarily a square. One pair of parallel sides doesn’t make a parallelogram. Check every condition the shape needs (see the tables above).

Using the wrong tool. Equal slopes show parallel, not equal length. Equal lengths show equal, not parallel. Matching the property to the tool is half the problem.

Mixing up sides and diagonals. In $ABCD$, the sides go around in order ($AB$, $BC$, $CD$, $DA$) and the diagonals cross ($AC$, $BD$). Comparing $AC$ with $AB$ as if both were sides leads to wrong conclusions. A labelled sketch prevents this.

Comparing radicals carelessly. To check that $\sqrt{116}$ is twice $\sqrt{29}$, don’t compare $116$ with $2 \times 29 = 58$. Square the factor: $2\sqrt{29} = \sqrt{4 \times 29} = \sqrt{116}$. Or compare squared lengths: $116 = 4 \times 29$.

No concluding statement. The calculations aren’t the answer by themselves. Finish with a sentence that names the property and the reason, such as “Since the slopes are equal, $MN \parallel BC$.”

Practice

1. (Warm-up) Line 1 passes through $(1, 2)$ and $(4, 8)$. Line 2 passes through $(0, 5)$ and $(6, 2)$. Are the lines parallel, perpendicular, or neither?

Solution

$m_1 = \dfrac{8 - 2}{4 - 1} = \dfrac{6}{3} = 2$ and $m_2 = \dfrac{2 - 5}{6 - 0} = \dfrac{-3}{6} = -\dfrac{1}{2}$.

$2 \times \left(-\dfrac{1}{2}\right) = -1$, so the lines are perpendicular.

2. (Warm-up) Show that the triangle with vertices $O(0, 0)$, $P(6, 0)$ and $Q(3, 5)$ is isosceles.

Solution

$$OP = 6 \qquad PQ = \sqrt{(3 - 6)^2 + (5 - 0)^2} = \sqrt{34} \qquad OQ = \sqrt{3^2 + 5^2} = \sqrt{34}$$

Since $PQ = OQ = \sqrt{34}$, the triangle is isosceles.

3. (Core) Show that the triangle with vertices $R(-4, -1)$, $S(2, 2)$ and $T(4, -2)$ is a right triangle. Where is the right angle?

Solution

$$m_{RS} = \frac{2 - (-1)}{2 - (-4)} = \frac{3}{6} = \frac{1}{2} \qquad m_{ST} = \frac{-2 - 2}{4 - 2} = \frac{-4}{2} = -2$$

$\dfrac{1}{2} \times (-2) = -1$, so $RS \perp ST$. Triangle $RST$ is a right triangle with the right angle at $S$.

4. (Core) Show that $ABCD$ with $A(-5, -1)$, $B(1, 1)$, $C(3, 5)$ and $D(-3, 3)$ is a parallelogram. Is it a rectangle?

Solution

$$m_{AB} = \frac{1 - (-1)}{1 - (-5)} = \frac{2}{6} = \frac{1}{3} \qquad m_{DC} = \frac{5 - 3}{3 - (-3)} = \frac{2}{6} = \frac{1}{3}$$$$m_{BC} = \frac{5 - 1}{3 - 1} = \frac{4}{2} = 2 \qquad m_{AD} = \frac{3 - (-1)}{-3 - (-5)} = \frac{4}{2} = 2$$

Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram.

Adjacent sides $AB$ and $BC$: $\dfrac{1}{3} \times 2 = \dfrac{2}{3}$, not $-1$. So they’re not perpendicular, and $ABCD$ is not a rectangle.

(Another check for the parallelogram: the midpoints of diagonals $AC$ and $BD$ are both $(-1, 2)$.)

5. (Core) Show that $EFGH$ with $E(1, 1)$, $F(4, 2)$, $G(3, 5)$ and $H(0, 4)$ is a square.

Solution

Sides:

$$\begin{aligned} EF &= \sqrt{3^2 + 1^2} = \sqrt{10} & FG &= \sqrt{(-1)^2 + 3^2} = \sqrt{10} \\ GH &= \sqrt{(-3)^2 + (-1)^2} = \sqrt{10} & HE &= \sqrt{1^2 + (-3)^2} = \sqrt{10} \end{aligned}$$

All four sides are equal, so $EFGH$ is a rhombus.

Right angle: $m_{EF} = \dfrac{1}{3}$ and $m_{FG} = \dfrac{3}{-1} = -3$. Their product is $-1$, so $EF \perp FG$.

Conclusion: $EFGH$ has four equal sides and a right angle, so it is a square.

6. (Core) Triangle $PQR$ has vertices $P(1, 7)$, $Q(-5, -1)$ and $R(7, 1)$. Verify that the segment joining the midpoints of $PQ$ and $PR$ is parallel to $QR$ and half its length.

Solution

Midpoints: $PQ$ gives $D = \left(\dfrac{1 - 5}{2}, \dfrac{7 - 1}{2}\right) = (-2, 3)$, and $PR$ gives $E = \left(\dfrac{1 + 7}{2}, \dfrac{7 + 1}{2}\right) = (4, 4)$.

Slopes: $m_{DE} = \dfrac{4 - 3}{4 - (-2)} = \dfrac{1}{6}$ and $m_{QR} = \dfrac{1 - (-1)}{7 - (-5)} = \dfrac{2}{12} = \dfrac{1}{6}$.

Lengths: $DE = \sqrt{6^2 + 1^2} = \sqrt{37}$ and $QR = \sqrt{12^2 + 2^2} = \sqrt{148} = \sqrt{4 \times 37} = 2\sqrt{37}$.

Conclusion: the slopes are equal, so $DE \parallel QR$, and $DE = \sqrt{37}$ is half of $QR = 2\sqrt{37}$.

7. (Core) Verify that $ABCD$ with $A(-1, -2)$, $B(5, 1)$, $C(3, 5)$ and $D(-3, 2)$ is a rectangle, and that its diagonals bisect each other and are equal in length.

Solution

Slopes of the sides:

$$m_{AB} = \frac{3}{6} = \frac{1}{2} \qquad m_{BC} = \frac{4}{-2} = -2 \qquad m_{CD} = \frac{-3}{-6} = \frac{1}{2} \qquad m_{DA} = \frac{-4}{2} = -2$$

Opposite sides are parallel, and $\dfrac{1}{2} \times (-2) = -1$, so adjacent sides are perpendicular. $ABCD$ is a rectangle.

Midpoints of the diagonals:

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

Same midpoint, so the diagonals bisect each other.

Lengths: $AC = \sqrt{4^2 + 7^2} = \sqrt{65}$ and $BD = \sqrt{(-8)^2 + 1^2} = \sqrt{65}$. The diagonals are equal.

8. (Challenge) Quadrilateral $ABCD$ has vertices $A(-4, 2)$, $B(2, 6)$, $C(6, 0)$ and $D(0, -4)$. Join the midpoints of its sides in order to form quadrilateral $EFGH$. Classify $EFGH$ as precisely as you can.

Solution

Midpoints: $E$ (of $AB$) $= (-1, 4)$, $F$ (of $BC$) $= (4, 3)$, $G$ (of $CD$) $= (3, -2)$, $H$ (of $DA$) $= (-2, -1)$.

Slopes:

$$m_{EF} = \frac{3 - 4}{4 - (-1)} = -\frac{1}{5} \qquad m_{FG} = \frac{-2 - 3}{3 - 4} = 5 \qquad m_{GH} = \frac{-1 - (-2)}{-2 - 3} = -\frac{1}{5} \qquad m_{HE} = \frac{4 - (-1)}{-1 - (-2)} = 5$$

Opposite sides are parallel, and $-\dfrac{1}{5} \times 5 = -1$, so adjacent sides are perpendicular. $EFGH$ is at least a rectangle.

Lengths:

$$EF = \sqrt{5^2 + (-1)^2} = \sqrt{26} \quad FG = \sqrt{(-1)^2 + (-5)^2} = \sqrt{26} \quad GH = \sqrt{(-5)^2 + 1^2} = \sqrt{26} \quad HE = \sqrt{1^2 + 5^2} = \sqrt{26}$$

All four sides are equal too.

Conclusion: $EFGH$ has four equal sides and four right angles, so it is a square.

(Joining the midpoints of any quadrilateral always gives a parallelogram. This one is a square because the diagonals of $ABCD$ happen to be perpendicular and equal in length.)

9. (Challenge) Triangle $ABC$ has vertices $A(-3, -1)$, $B(5, 3)$ and $C(-1, 5)$. Show that the triangle is isosceles, and that the median from $C$ is perpendicular to $AB$.

Solution

Isosceles:

$$AC = \sqrt{(-1 - (-3))^2 + (5 - (-1))^2} = \sqrt{4 + 36} = \sqrt{40} \qquad BC = \sqrt{(-1 - 5)^2 + (5 - 3)^2} = \sqrt{36 + 4} = \sqrt{40}$$

$AC = BC$, so triangle $ABC$ is isosceles.

Median from $C$: it goes to the midpoint of $AB$, $\left(\dfrac{-3 + 5}{2}, \dfrac{-1 + 3}{2}\right) = (1, 1)$. Its slope is $\dfrac{1 - 5}{1 - (-1)} = \dfrac{-4}{2} = -2$.

Slope of $AB$: $\dfrac{3 - (-1)}{5 - (-3)} = \dfrac{4}{8} = \dfrac{1}{2}$.

Conclusion: $-2 \times \dfrac{1}{2} = -1$, so the median from $C$ is perpendicular to $AB$. Since it also passes through the midpoint of $AB$, it is the right bisector of $AB$.