Lesson 10 ~45 min Linear Relationships C · Path +95 XP

Proving Facts About Figures

Put a figure on the Cartesian plane and every geometric claim about it becomes a calculation. The formulas do the work; choosing the right property, and choosing the cheapest test for it, is what turns the numbers into a proof.

Today's hook: A quadrilateral has vertices $(1, 1)$, $(4, 5)$, $(9, 5)$ and $(6, 1)$. Is it a rhombus? You can see that it might be. Seeing is not proving, and the difference is four applications of one formula.

0/5QUESTS

Think First

warm-up

What is the difference between a rhombus and a parallelogram? Now: what would you have to calculate, from four vertices, to be certain which one you had?

Record your answer in your workbook.

The Big Idea

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Every quadrilateral type has a defining test in terms of sides, diagonals or angles. Choose the test that is sufficient and cheapest to run, do the arithmetic, then state the conclusion naming the property you used.

$$\text{compute} \to \text{property} \to \text{conclusion}$$

Two families of test exist for most shapes: one about the sides and one about the diagonals. Either proves the result, so pick whichever the given coordinates make easier, and do not run both.

$\text{compute} \to \text{property} \to \text{conclusion}$

Sufficient, not merely consistent

Two equal sides is true of a rhombus but does not prove one. Ask whether your test could hold for a different shape.

Label the vertices in order

$ABCD$ means the sides are $AB$, $BC$, $CD$, $DA$ and the diagonals are $AC$ and $BD$. Order is information.

Name the property in the conclusion

The marks are for the reason, not the numbers. Say which property you used and what it establishes.

What You'll Master

objectives

Know

the coordinate tests for a parallelogram, rhombus, rectangle and square, in terms of both sides and diagonals
the definitions of a median, an altitude and a perpendicular bisector, and how each is constructed from coordinates
that a test must be sufficient for the conclusion, not merely consistent with it

Understand

why some conditions characterise a shape while others merely fail to rule it out
why the order of the vertex labels determines which segments are sides and which are diagonals
why choosing between the side test and the diagonal test is a matter of arithmetic convenience only

Can Do

prove that a given quadrilateral is a parallelogram, rhombus, rectangle or square
find the equation of a median, an altitude or a perpendicular bisector in a triangle
set out a coordinate proof so that each conclusion names the property that justifies it

Words You Need

vocabulary

RhombusA parallelogram with all four sides equal in length. Its diagonals are perpendicular.

RectangleA parallelogram with a right angle. Its diagonals are equal in length.

SquareA quadrilateral that is both a rhombus and a rectangle.

MedianIn a triangle, the interval from a vertex to the midpoint of the opposite side.

AltitudeIn a triangle, the interval from a vertex perpendicular to the opposite side.

What a Coordinate Proof Looks Like

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A coordinate proof has the same shape every time.

Compute. Apply the formulas to whichever sides, diagonals or angles the chosen test needs, keeping values exact.

Interpret. Say in words what the numbers mean: these two sides are equal, these two gradients are the same, these diagonals share a midpoint.

Conclude. State the result and name the property that gets you there: a quadrilateral with four equal sides is a rhombus.

The middle step is the one most often skipped and the one carrying the marks. Numbers on a page do not say what they are for, and a marker cannot award credit for an inference you did not write down.

The Tests Worth Knowing

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Each shape has more than one sufficient test. These are the ones that use the formulas you have.

Parallelogram: both pairs of opposite sides parallel (equal gradients), or both pairs of opposite sides equal in length, or the diagonals bisect each other (equal midpoints).

Rhombus: all four sides equal in length, or a parallelogram whose diagonals are perpendicular.

Rectangle: a parallelogram with one right angle (a gradient product of $-1$), or a parallelogram whose diagonals are equal in length.

Square: both a rhombus and a rectangle, so four equal sides together with one right angle is enough.

Any one test from a line proves that shape. Running two is not more convincing; it is the same conclusion twice.

Choosing the Cheapest Test

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Since the tests are equivalent, choose by arithmetic.

Diagonal tests need two calculations; side tests need four. So if the diagonals give whole-number answers, prefer them.

But the side test proves more when it works. Four equal sides gives a rhombus directly, whereas the diagonal route requires you to establish a parallelogram first and then show the diagonals are perpendicular, which is two stages.

A quick look at the coordinates usually decides it. If the differences are small whole numbers, sides are painless. If the vertices are spread awkwardly, midpoints and lengths of the two diagonals may be quicker.

There is no wrong choice among sufficient tests, so spend the ten seconds choosing rather than the two minutes regretting.

Sufficient Is Not the Same as Consistent

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This is where marks are most often lost, and the error is invisible to a student who has not been warned.

Showing that $AB = CD$ does not prove a parallelogram. A kite, or an ordinary irregular quadrilateral, can happen to have two equal sides. The fact is consistent with a parallelogram and proves nothing.

Showing that one pair of opposite sides is both parallel and equal does prove it, and that is a genuinely different statement.

The test to apply to your own reasoning: could a shape that is not the one I am claiming also pass my test? If yes, the test is not sufficient and the proof is incomplete.

Two equal diagonals do not prove a rectangle either, for the same reason: an isosceles trapezium has them too. Equal diagonals in a parallelogram do prove a rectangle, and the word parallelogram is doing the work.

Lines Inside a Triangle

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Three named lines come up repeatedly, and each is built from formulas you already have.

A median joins a vertex to the midpoint of the opposite side. Find the midpoint, then find the line through it and the vertex, using Lesson 7.

An altitude runs from a vertex perpendicular to the opposite side. Find the gradient of that side, take the negative reciprocal, then use point–gradient form at the vertex.

A perpendicular bisector of a side passes through its midpoint at right angles to it, so it needs both the midpoint and the negative reciprocal gradient. Note that it usually does not pass through any vertex.

Read the word carefully. A median and an altitude both start at a vertex and both end on the opposite side, but they are the same line only when the triangle happens to be isosceles at that vertex.

Watch Me Solve It · 3 examples

Watch Me Solve It · Proving a rhombus

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PROBLEM

Show that the quadrilateral $ABCD$ with $A(1, 1)$, $B(4, 5)$, $C(9, 5)$ and $D(6, 1)$ is a rhombus.

1

Choose a test

All four coordinate differences are small whole numbers, so the four-equal-sides test will be quick and proves the result in one stage.
2

Find all four sides

$AB = \sqrt{3^2 + 4^2} = 5$

$BC = \sqrt{5^2 + 0^2} = 5$

$CD = \sqrt{(-3)^2 + (-4)^2} = 5$

$DA = \sqrt{(-5)^2 + 0^2} = 5$

Take the sides in order around the quadrilateral, so that $AB$, $BC$, $CD$ and $DA$ are genuinely the sides and not diagonals.
3

Interpret and conclude

All four sides are equal in length. A quadrilateral with four equal sides is a rhombus, so $ABCD$ is a rhombus.
4

Check it is not also a square

$m_{AB} = \tfrac{4}{3}, \qquad m_{BC} = 0$

The product is $0$, not $-1$, so there is no right angle at $B$ and the rhombus is not a square. This was not asked, but it is what a fuller classification would add.

AnswerAll four sides equal $5$, so $ABCD$ is a rhombus

Watch Me Solve It · Proving a square

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PROBLEM

Show that $PQRS$ with $P(0, 0)$, $Q(4, 2)$, $R(2, 6)$ and $S(-2, 4)$ is a square.

1

Find the four side lengths

$PQ = \sqrt{16 + 4} = \sqrt{20}$

$QR = \sqrt{4 + 16} = \sqrt{20}$

$RS = \sqrt{16 + 4} = \sqrt{20}$

$SP = \sqrt{4 + 16} = \sqrt{20}$

All four are equal, so $PQRS$ is a rhombus.
2

Test one angle with gradients

$m_{PQ} = \frac{2 - 0}{4 - 0} = \frac{1}{2}$

$m_{QR} = \frac{6 - 2}{2 - 4} = -2$

These two sides meet at $Q$, so their gradients decide the angle there.
3

Interpret the product

$\tfrac{1}{2} \times (-2) = -1$

The product is $-1$, so $PQ$ and $QR$ are perpendicular and there is a right angle at $Q$.
4

Conclude

A rhombus with a right angle is a square, so $PQRS$ is a square. Only one right angle needs to be shown, because the equal sides force the rest.

AnswerFour sides of $\sqrt{20}$ and a right angle at $Q$, so $PQRS$ is a square

Watch Me Solve It · An altitude of a triangle

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PROBLEM

The triangle $ABC$ has $A(2, 7)$, $B(-1, 1)$ and $C(5, 3)$. Find the equation of the altitude from $A$.

1

Identify what an altitude needs

The altitude from $A$ passes through $A$ and is perpendicular to the opposite side $BC$. So it needs the gradient of $BC$ and the point $A$.
2

Find the gradient of the opposite side

$m_{BC} = \frac{3 - 1}{5 - (-1)} = \frac{2}{6} = \frac{1}{3}$

The altitude does not pass through $B$ or $C$, but its direction comes from them.
3

Take the negative reciprocal

$m = -3$

Check: $\tfrac{1}{3} \times (-3) = -1$.
4

Use point-gradient form at A and check

$y - 7 = -3(x - 2)$

$y = -3x + 13$

$x = 2: \ -6 + 13 = 7$

The point $A(2, 7)$ satisfies the equation, so the altitude passes through the right vertex with the right direction.

Answer$y = -3x + 13$

Brain Trainer · 5 problems

Brain Trainer · Test, then name

5 problems

Five items. For each, state the property you would test before doing any arithmetic.

1 What single calculation proves that a quadrilateral is a parallelogram?

Compare the midpoints of the two diagonals; equal midpoints means they bisect each other.Equal diagonal midpoints
2 Does showing $AB = CD$ prove that $ABCD$ is a parallelogram?

A kite or an irregular quadrilateral can also have two equal sides.No, it is not sufficient
3 A parallelogram has equal diagonals. What does that prove?

Equal diagonals in a parallelogram characterise a rectangle.It is a rectangle
4 In triangle $ABC$, what two things does the median from $A$ need?

The midpoint of the opposite side, and the vertex itself.Midpoint of $BC$, and $A$
5 In triangle $ABC$ with $m_{BC} = \tfrac{2}{5}$, what is the gradient of the altitude from $A$?

Perpendicular to $BC$, so take the negative reciprocal.$-\tfrac{5}{2}$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

Which segments are the diagonals

+10 XP

In the quadrilateral $ABCD$, the diagonals are:

MC2

A sufficient test

+10 XP

Which single finding proves that a quadrilateral is a parallelogram?

MC3

Naming the shape

+10 XP

A quadrilateral has four equal sides and one right angle. It is:

MC4

Median or altitude

+10 XP

In triangle $ABC$, the line from $A$ perpendicular to $BC$ is called:

MC5

Ruling a shape out

+10 XP

A parallelogram is found to have diagonals of lengths $\sqrt{50}$ and $\sqrt{18}$. It follows that the parallelogram is:

Short Answer · 3 questions

Prove and classify

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SHORT ANSWER

The quadrilateral $ABCD$ has vertices $A(1, 0)$, $B(5, 2)$, $C(1, 10)$ and $D(-3, 8)$. Prove that it is a parallelogram, determine with justification whether it is also a rectangle, and state whether it is a square.

Write your working in your book.

Three lines from one vertex

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SHORT ANSWER

The triangle $PQR$ has $P(-2, 4)$, $Q(6, 0)$ and $R(2, -4)$. Find the equation of the median from $P$ and the equation of the altitude from $P$, and state with justification whether they are the same line.

Write your working in your book.

Diagnose an incomplete proof

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SHORT ANSWER

A student writes: "$ABCD$ has $AB = CD = \sqrt{20}$, so it is a parallelogram." Explain why this argument is not valid, state what additional single calculation would make it valid, and describe a quadrilateral that satisfies the student's condition without being a parallelogram.

Write your working in your book.

Stretch Challenge · Proving it for every rhombus at once

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CHALLENGE

Place a rhombus with one vertex at the origin, so its vertices are $O(0, 0)$, $A(a, b)$, $C(a + c, b + d)$ and $B(c, d)$, where $OA$ and $OB$ are two adjacent sides. Use the fact that all sides are equal to show that the diagonals $OC$ and $AB$ are always perpendicular. Identify the case your argument does not cover, and deal with it.

Quick Review

recap

Compute, interpret, conclude

Numbers alone earn little. Say what the values mean, then name the property that turns that meaning into the result being claimed.

Every shape has side tests and diagonal tests

Parallelogram: bisecting diagonals, or both pairs of opposite sides equal or parallel. Rhombus: four equal sides. Rectangle: a parallelogram with equal diagonals or one right angle.

Ask whether another shape could pass

Two equal sides fits a kite. Equal diagonals fit an isosceles trapezium. A test is only a proof if nothing else can satisfy it.

Median, altitude, perpendicular bisector

A median goes to a midpoint, an altitude goes at a right angle, and a perpendicular bisector does both but through the midpoint rather than a vertex.

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