Coordinate Proofs and Polygon Measurement

Watch the video explainer

Learning goals

Choose a coordinate placement and a method that preserve the given geometry
Justify quadrilateral classifications with exact coordinate evidence and valid geometric tests
Prove midsegment relationships and derive centroid coordinates using earlier segment results
Calculate polygon perimeter and area from an ordered boundary and geometric decomposition

Choose the method before calculating

If a problem gives three side lengths and asks for congruence, an earlier triangle test may finish it immediately. If it gives vertices and asks whether opposite sides are parallel, slope or midpoint calculations may give the missing evidence. Coordinates are a method choice, not an instruction to calculate every length and slope.

Direction: use slopes, with vertical and horizontal cases checked separately.
Length: compare squared distances when only equality matters.
Bisection: compare midpoints of the named segments.
Area: find perpendicular heights and split the region into familiar pieces.

The previous three lessons justify these coordinate tools. Here we use them to prove geometric statements.

Place the figure without adding a condition

We may choose the origin and turn the axes without changing the figure’s lengths or angles. Put a known side on an axis when that removes unnecessary variables. Do not put an arbitrary triangle’s third vertex on the other axis: that would quietly assume a right angle.

Worked example 1 Choose axes for a rectangle proof

Problem. Prove with coordinates that the diagonals of every nondegenerate rectangle are congruent. Compare an axis-aligned placement with a turned, shifted placement. Keep both side lengths arbitrary.

Solution. For a rectangle of width 7 and height 4, place $A$ at the origin, $B$ at $(7,0)$ , $C$ at $(7,4)$ , and $D$ at $(0,4)$ in boundary order. The diagonal squares are

\begin{aligned} AC^2&=7^2+4^2\\ AC^2&=65\\ BD^2&=(-7)^2+4^2\\ BD^2&=65. \end{aligned}

Axis placement uses the origin and two side directions. The drawing illustrates the 7 by 4 instance; the proof keeps a and b arbitrary.

The coordinate zeros remove subtraction steps. For the general rectangle, take positive side lengths $a,b$ and place

\begin{aligned} A&=(0,0)\\B&=(a,0)\\C&=(a,b)\\D&=(0,b). \end{aligned}

The same calculation gives

\begin{aligned} AC^2&=a^2+b^2\\ BD^2&=(-a)^2+b^2\\ BD^2&=a^2+b^2. \end{aligned}

Both diagonals have positive lengths, so equal squares give equal lengths. Nothing depended on the sample side lengths. Every rectangle can be placed this way by choosing a vertex as origin and its perpendicular sides as axis directions.

The same 7 by 4 rectangle after a rigid turn and shift. Its geometric properties are unchanged, but no side follows an axis.

In the turned placement, all four vertices may have two nonzero coordinates. Distance still works, but it requires differences of coordinates and equations recording the right angles. The convenient placement keeps those conditions visible in the coordinates themselves.

Check your understanding

To represent every nondegenerate triangle with a horizontal base, which placement avoids an added right-angle condition?

A A at (0,0), B at (a,0), C at (0,c), with a,c positive B A at (0,0), B at (a,0), C at (b,c), with a,c positive and b real C A at (0,0), B at (a,a), C at (2a,0), with a positive D A at (0,0), B at (a,0), C at (a,a), with a positive

A single numerical rectangle checks that instance. Keeping $a,b$ arbitrary, with their required positivity, is what makes the argument a proof for every rectangle. A special placement is useful only if it represents every figure named in the claim.

Prove a classification or disprove a claim

Worked example 2 Use the diagonals to choose a short test

Problem. The vertices of a convex quadrilateral, in boundary order, are $A(-1,2)$ , $B(4,4)$ , $C(9,2)$ , and $D(4,0)$ . Determine its most specific classification among parallelogram, rectangle, rhombus, and square. Give exact evidence.

The named diagonals suggest a midpoint test. No equal-side or right-angle marks are assumed.

Solution. The diagonals have the same midpoint:

\begin{aligned} M_{AC}&=(4,2)\\M_{BD}&=(4,2). \end{aligned}

Thus they bisect each other, proving a parallelogram. Adjacent side squares are

\begin{aligned} AB^2&=5^2+2^2\\AB^2&=29\\ BC^2&=5^2+(-2)^2\\BC^2&=29. \end{aligned}

A parallelogram with congruent adjacent sides is a rhombus. Its diagonals have lengths 10 and 4, which are unequal. A rectangle would have congruent diagonals, so this rhombus is neither a rectangle nor a square.

Our definitions are inclusive: a square is both a rectangle and a rhombus, and every parallelogram is also a trapezoid because it has at least one parallel side pair. A classification should not exclude a stronger family unless the evidence rules it out.

To disprove a universal claim, give one figure meeting its hypotheses and violating its conclusion. For example, the claim that equal diagonals force a rectangle fails for the following convex quadrilateral.

This unequal-base quadrilateral has congruent diagonals. Its coordinates, rather than its appearance, verify the counterexample.

Its vertices are $A(-6,0)$ , $B(6,0)$ , $C(3,4)$ , and $D(-3,4)$ in boundary order. Both diagonal squares are 97: each diagonal has horizontal gap 9 and vertical gap 4. But opposite sides $AB$ and $CD$ have lengths 12 and 6. A rectangle has equal opposite sides, so this is not a rectangle. Equal diagonals become a rectangle test after a parallelogram has been established.

Check your understanding

A convex quadrilateral has diagonals with common midpoint (-2,3). Their lengths are 8 and 12. Which conclusion follows?

A It is a square B It is a rectangle C It is a parallelogram that is not a rectangle D It cannot be a parallelogram

A triangle midsegment in coordinates

Coordinate placement can make the midsegment theorem a comparison of two horizontal lengths.

Worked example 3 Keep the third vertex free

Problem. In a nondegenerate triangle $ABC$ , let $M$ be the midpoint of $AC$ and $N$ the midpoint of $BC$ . Prove that $MN$ is parallel to $AB$ and has half its length, using a placement that includes obtuse triangles.

Solution. First take $A(0,0)$ , $B(14,0)$ , and $C(4,10)$ . Averaging gives $M(2,5)$ and $N(9,5)$ . Both points have height 5, so $MN$ is horizontal, parallel to $AB$ . Its length is 7, half of 14.

M is the midpoint of AC and N is the midpoint of BC. Their common height and horizontal separation are visible.

The triangle midsegment by coordinates#

Now place any nondegenerate triangle with

\begin{aligned} A&=(0,0)\\B&=(2a,0)\\C&=(2b,2c), \end{aligned}

where $a,c$ are positive and $b$ is any real number. Choose the positive vertical direction toward $C$ . Allowing $b$ outside the interval from 0 to $a$ includes triangles whose third vertex lies beyond a base endpoint.

The midpoint formula gives

\begin{aligned} M&=(b,c)\\N&=(a+b,c). \end{aligned}

Their line is horizontal at positive height $c$ , distinct from the horizontal line $AB$ . The length of $MN$ is $a$ , while $AB$ has length $2a$ . This repeats the numerical argument without restricting the side lengths or the third vertex’s horizontal position.

Check your understanding

In the general midsegment placement, take a=5, b=-2, and c=3. What are the midsegment endpoints M and N?

A (-2,3) and (3,3) B (-2,3) and (5,3) C (-4,6) and (6,6) D (2,3) and (7,3)

Find the centroid from all three medians

A median joins a vertex to the opposite side’s midpoint. The earlier centroid theorem says that the vertex-to-centroid portion is twice the centroid-to-midpoint portion. Coordinates also show why all three medians give one common point.

Worked example 4 Average three vertices for a reason

Problem. Find the centroid of the triangle with vertices $A(-3,1)$ , $B(9,-5)$ , and $C(6,10)$ . Then derive the coordinate rule for a general nondegenerate triangle and verify that each median gives the same point.

Solution. The midpoint of $BC$ is $D(7.5,2.5)$ . Two thirds of the coordinate changes from $A$ toward $D$ gives

\begin{aligned} x_G&=-3+\tfrac23(10.5)\\x_G&=4\\ y_G&=1+\tfrac23(1.5)\\y_G&=2. \end{aligned}

The midpoint of $CA$ is $E(1.5,5.5)$ . From $B$ , the changes to $E$ are $-7.5$ and $10.5$ . Two thirds of these changes gives $(9-5,-5+7)$ , again $(4,2)$ .

The midpoint of $AB$ is $F(3,-2)$ . From $C$ , the changes to $F$ are $-3$ and $-12$ . Two thirds gives $(6-2,10-8)$ , again $(4,2)$ .

D, E, and F are the opposite side midpoints. The three computed two-to-one points coincide at G.

The centroid coordinates#

For arbitrary noncollinear vertices $A,B,C$ , first find midpoint $D$ of $BC$ . The point two thirds from $A$ toward $D$ has first coordinate

\begin{aligned} x_D&=\frac{x_B+x_C}{2}\\ x_G&=\frac{x_A+2x_D}{3}\\ x_G&=\frac{x_A+x_B+x_C}{3}. \end{aligned}

Applying the same steps to heights gives

y_G=\frac{y_A+y_B+y_C}{3}.

Starting from $B$ instead puts $x_B$ outside the midpoint sum and $x_A,x_C$ inside it. The resulting total is unchanged. Starting from $C$ also gives the same total, and the heights agree in each case. Thus the three internal two-to-one division points coincide. They lie on their respective medians, proving concurrence as well as the coordinate rule. The fraction two thirds lies strictly between zero and one, so these are internal points, just as in the numerical triangle.

Check your understanding

What is the centroid of the triangle with vertices (-4,7), (8,-2), and (5,4)?

A (9,9) B (3,9/2) C (9/2,3) D (3,3)

Measure the boundary and the region separately

For a simple polygon, list its vertices in boundary order. Perimeter adds only outside edge lengths. Area measures the inside, so a decomposition needs pieces with no overlapping interiors and no missing region. A diagonal used for the split is not an outside edge.

Worked example 5 Split an irregular pentagon

Problem. A simple pentagon has vertices $A(0,0)$ , $B(9,0)$ , $C(9,4)$ , $D(5,7)$ , and $E(0,4)$ in boundary order. Find its exact perimeter and area in coordinate units.

The dashed cut EC separates the lower rectangle from the upper triangle. DF measures the upper triangle height.

Solution. The horizontal and vertical sides have lengths 9, 4, and 4. For the other two sides,

\begin{aligned} CD^2&=4^2+3^2\\CD&=5\\ DE^2&=5^2+3^2\\DE&=\sqrt{34}. \end{aligned}

Adding only the boundary gives

\begin{aligned} P&=9+4+5+\sqrt{34}+4\\P&=22+\sqrt{34}\text{ units}. \end{aligned}

Cut along $EC$ . Below it is a rectangle of width 9 and height 4. Above it is a triangle with base 9 and perpendicular height 3, shown by $DF$ . Their interiors do not overlap and together fill the pentagon.

\begin{aligned} A_{\text{rectangle}}&=36\\A_{\text{triangle}}&=\tfrac12(9)(3)\\ A_{\text{pentagon}}&=\frac{99}{2}\text{ square units}. \end{aligned}

The split segment $EC$ contributes to the area argument, but not to the perimeter sum.

The six named vertices follow the given boundary. No area split or computed area is shown.

Check your understanding

A simple hexagon has boundary vertices (0,0), (8,0), (8,2), (3,2), (3,6), and (0,6). What is its area?

A 48 square units B 28 square units C 32 square units D 20 square units

Coordinate Proofs and Polygon Measurement

Learning goals

Choose the method before calculating

Place the figure without adding a condition

Prove a classification or disprove a claim

A triangle midsegment in coordinates

Find the centroid from all three medians

Measure the boundary and the region separately

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress