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Coordinate Proofs and Polygon Measurement: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A placement worth keeping

You want to prove a statement about the diagonals of every nondegenerate parallelogram. Describe a coordinate placement with one side on the horizontal axis, using the fewest free quantities you need. State the conditions on those quantities. Explain why forcing the adjacent side onto the vertical axis would restrict the claim.
1. Hint 1
  A vertex can become the origin without changing the figure.
2. Hint 2
  Keep an arbitrary horizontal and nonzero vertical change for an adjacent side.
Answer

One valid placement is A(0,0), B(a,0), D(b,c), C(a+b,c), with a,c positive and b real. Forcing b to zero restricts it to rectangles.

Full solution

Choose a vertex A as origin and the positive horizontal direction along AB.
Write its positive length as a.

Choose the positive vertical direction toward D.
Its height c is positive, while its horizontal position b can be any real number.
The parallel and congruent opposite sides then place C at (a+b,c).

Every nondegenerate parallelogram can be placed this way.
Placing AD on the vertical axis would require b to be zero and make the adjacent sides perpendicular, leaving only rectangles.

Answer

One valid placement is A(0,0), B(a,0), D(b,c), C(a+b,c), with a,c positive and b real. Forcing b to zero restricts it to rectangles.

Key idea

Coordinate choices may simplify known geometry but must preserve every allowed shape.
Problem 2 A diagonal midpoint test

Convex quadrilateral ABCD has vertices A(-7,0), B(1,2), C(5,8), and D(-3,6) in boundary order. Determine whether it is a parallelogram, using exact coordinate evidence.

A diagonal midpoint test. Named vertices and segments show only the givens.

Text description of this figure

A diagonal midpoint test. Named vertices and segments show only the givens. The drawing uses equal coordinate scales and labels only the named points.
1. Hint 1
  Look for a short test involving both diagonals.
2. Hint 2
  Compare the coordinate averages for AC and BD.
Answer

It is a parallelogram; both diagonal midpoints are (-1,4).

Full solution

The midpoint of AC is
$M_{AC}=(-1,4).$

The midpoint of BD is
$M_{BD}=(-1,4).$

Thus both diagonals bisect each other.
The diagonal-bisection converse proves a parallelogram.

Answer

It is a parallelogram; both diagonal midpoints are (-1,4).

Key idea

A common diagonal midpoint establishes both required bisections.
Problem 3 Classify a tilted quadrilateral

The boundary vertices of convex ABCD are A(-3,0), B(5,4), C(2,10), and D(-6,6). Determine its most specific classification among parallelogram, rectangle, rhombus, and square. Justify any exclusions.

A tilted quadrilateral. Named vertices and segments show only the givens.

Text description of this figure

A tilted quadrilateral. Named vertices and segments show only the givens. The drawing uses equal coordinate scales and labels only the named points.
1. Hint 1
  First establish a parallelogram rather than assuming one from the sketch.
2. Hint 2
  Check an adjacent angle and adjacent squared side lengths.
Answer

ABCD is a rectangle that is neither a rhombus nor a square.

Full solution

The two diagonal midpoints are both $(-1/2,5)$ .
Hence ABCD is a parallelogram.

Adjacent slopes are
$\begin{gathered}m_{AB}=1/2\\m_{BC}=-2.\end{gathered}$
Their product is -1, so the parallelogram has a right angle and is a rectangle.

The adjacent squared side lengths are
$\begin{gathered}AB^2=80\\BC^2=45.\end{gathered}$
They differ, excluding a rhombus and a square.

Answer

ABCD is a rectangle that is neither a rhombus nor a square.

Key idea

Classification requires the right geometric test and evidence for any stronger family.
Problem 4 A midsegment beyond the base endpoint

Triangle ABC has A(-2,1), B(10,1), and C(-6,9). M is the midpoint of AC, and N is the midpoint of BC. Find M and N, an equation of line MN, and the length MN. Verify its relationship to AB.

A midsegment beyond the base endpoint. Named vertices and segments show only the givens.

Text description of this figure

A midsegment beyond the base endpoint. Named vertices and segments show only the givens. The drawing uses equal coordinate scales and labels only the named points.
1. Hint 1
  Midpoints average coordinates even when the third vertex lies beyond a base endpoint.
2. Hint 2
  Compare the common heights and the horizontal gaps.
Answer

$\begin{gathered}M=(-4,5)\\N=(2,5)\\MN=6\end{gathered}$
Line MN is y=5, parallel to AB and half its length.

Full solution

The midpoint calculations give
$\begin{gathered}M=(-4,5)\\N=(2,5).\end{gathered}$

Their common height gives line $y=5$ .
The base lies on $y=1$ , so the two lines are distinct and parallel.

The midsegment length is
$MN=2-(-4).$
It is 6 units.
The base length is
$AB=10-(-2).$
It is 12 units, twice MN.

The computed M and N share height 5; their horizontal segment lies above the base.

Answer

$\begin{gathered}M=(-4,5)\\N=(2,5)\\MN=6\end{gathered}$
Line MN is y=5, parallel to AB and half its length.

Key idea

A horizontal base makes direction and length comparisons separate and exact.
Problem 5 A centroid constraint

A triangle has vertices A(-8,4), B(7,-5), and C(t,10), where t is real. Its centroid lies on the line y=2x-3. Find t and the centroid. Verify that your result gives a nondegenerate triangle.
1. Hint 1
  The fixed vertex heights already determine the centroid height.
2. Hint 2
  Use that height in the line equation, then recover the third first coordinate.
Answer

$\begin{gathered}t=10\\G=(3,3).\end{gathered}$
The recovered triangle is nondegenerate.

Full solution

The centroid height is
$y_G=(4-5+10)/3.$
It is 3.

Substitution into the required line gives
$\begin{aligned}3&=2x_G-3\\x_G&=3.\end{aligned}$

The first-coordinate average must therefore satisfy
$\begin{aligned}(-8+7+t)/3&=3\\t&=10.\end{aligned}$
The centroid is (3,3).

AB has slope $-3/5$ , while AC has slope $1/3$ at the recovered t.
The distinct slopes show that A, B, and C are not collinear.

Answer

$\begin{gathered}t=10\\G=(3,3).\end{gathered}$
The recovered triangle is nondegenerate.

Key idea

A condition on the centroid links both coordinate averages to the triangle geometry.
Problem 6 Two measurements of one outline

A scale drawing uses one coordinate unit for one meter. A courtyard has simple boundary vertices A(0,0), B(11,0), C(11,5), D(7,8), and E(0,5), in that order. Find its exact fencing length and paving area.

Two measurements of one outline. Named vertices and segments show only the givens.

Text description of this figure

Two measurements of one outline. Named vertices and segments show only the givens. The drawing uses equal coordinate scales and labels only the named points.
1. Hint 1
  Trace all five outside sides for fencing.
2. Hint 2
  A horizontal cut through C and E separates two familiar area pieces.
Answer

Fencing is $26+\sqrt{58}$ meters. Paving area is $143/2$ square meters.

Full solution

The base is 11 meters and the vertical sides are each 5 meters.
The slanted side CD has gaps 4 and 3, so it is 5 meters.
Side DE has gaps 7 and 3, so it is $\sqrt{58}$ meters.

Add only the outside edges.
$P=11+5+5+\sqrt{58}+5.$
Thus the fencing length is $26+\sqrt{58}$ meters.

Cut along EC.
The lower rectangle has area 55 square meters, and the upper triangle has base 11 and perpendicular height 3.
$A=55+\tfrac12(11)(3).$
The total paving area is $143/2$ square meters.

EC splits the courtyard into a lower rectangle and an upper triangle; DF is perpendicular to EC.

Answer

Fencing is $26+\sqrt{58}$ meters. Paving area is $143/2$ square meters.

Key idea

Coordinate lengths measure the boundary while perpendicular heights measure the region.
Problem 7 Test a diagonal claim

Disprove the claim that every convex quadrilateral with perpendicular diagonals is a rhombus. Give four exact coordinate vertices in boundary order. Verify perpendicularity and show that two named side lengths differ.
1. Hint 1
  A counterexample must meet the diagonal condition and fail the side condition.
2. Hint 2
  You may place one diagonal horizontally and the other vertically, with an unequal division on one of them.
Answer

One example is A(-4,0), B(0,2), C(1,0), D(0,-2). Its diagonals are perpendicular, but AB and BC have unequal lengths.

Full solution

Choose A(-4,0), B(0,2), C(1,0), D(0,-2) in boundary order.
These are on four alternating rays from the origin, giving a convex quadrilateral.

AC is horizontal and BD is vertical.
They are perpendicular without using a slope-product test on an undefined slope.

The squared adjacent lengths are
$\begin{gathered}AB^2=20\\BC^2=5.\end{gathered}$
Thus AB and BC differ, so the quadrilateral is not a rhombus.
The example satisfies the hypothesis and violates the conclusion.

AC is horizontal and BD vertical; the side lengths need not be equal.

Answer

One example is A(-4,0), B(0,2), C(1,0), D(0,-2). Its diagonals are perpendicular, but AB and BC have unequal lengths.

Key idea

A counterexample must verify the hypothesis as carefully as it rejects the conclusion.
Problem 8 Two area descriptions

A simple hexagon has boundary vertices (0,0), (12,0), (12,h), (5,h), (5,3), and (0,3), where h is greater than 3. Its area is 57 square units. Find h and the exact perimeter. Explain why your value preserves the stated boundary.
1. Hint 1
  The short left part and tall right part can be measured separately.
2. Hint 2
  For perimeter, account for both the outer vertical edge and the interior step.
Answer

$\begin{gathered}h=6\\P=36\text{ units}.\end{gathered}$
The value satisfies h greater than 3.

Full solution

The left rectangle has width 5 and height 3, and the right rectangle has width 7 and height h.
Their interiors do not overlap.
$\begin{aligned}15+7h&=57\\h&=6.\end{aligned}$

The boundary lengths are 12, 6, 7, 3, 5, and 3.
$P=12+6+7+3+5+3.$
This gives 36 units.

The recovered height is greater than 3, as required.
All six edges have positive lengths, and the step remains in the stated boundary order.

At the recovered height, the six positive edges form the stated concave boundary.

Answer

$\begin{gathered}h=6\\P=36\text{ units}.\end{gathered}$
The value satisfies h greater than 3.

Key idea

An area equation must be checked against the geometric inequalities that defined the figure.
Problem 9 Opposite midpoints

Let a,b be positive. In rectangle ABCD, choose coordinates A(0,0), B(2a,0), C(2a,2b), D(0,2b). Let E, F, G, H be the midpoints of AB, BC, CD, DA respectively. Prove that EFGH is a rhombus, and determine exactly when it is a square.
1. Hint 1
  Begin with a numerical rectangle to see each midpoint calculation.
2. Hint 2
  Find the four midpoint coordinates and compare the squared side lengths.
3. Hint 3
  Its horizontal and vertical diagonals can distinguish a square from the other rhombi.
Answer

EFGH is always a rhombus. It is a square exactly when $a=b$ .

Full solution

First take a rectangle of width 8 and height 6, so a is 4 and b is 3.
The midpoints are E(4,0), F(8,3), G(4,6), H(0,3).
Each adjacent side has square 25.
The diagonals EG and FH have lengths 6 and 8, so the rhombus is not a square.

For the general rectangle the midpoints are
$\begin{gathered}E=(a,0)\\F=(2a,b)\\G=(a,2b)\\H=(0,b).\end{gathered}$
Each side has squared length $a^2+b^2$ .
Positivity keeps the quadrilateral convex and its sides nonzero, so four equal sides establish a rhombus.

The diagonals EG and FH have lengths 2b and 2a.
If the rhombus is a square, it is a rectangle and its diagonals are congruent, forcing a to equal b.

Conversely, if a equals b, the rhombus has congruent diagonals.
A rhombus is a parallelogram, so this gives a rectangle as well and hence a square.
No step required the sample dimensions.

In the numerical instance, the four side midpoints form the inner quadrilateral; EG and FH follow perpendicular symmetry directions.

Answer

EFGH is always a rhombus. It is a square exactly when $a=b$ .

Key idea

A general calculation may establish one family while an extra equality selects a stronger family.
Problem 10 A median and an equal-side constraint

Let a,h be positive and t real. A triangle has A(0,0), B(2a,0), and C(t,h). Its centroid is G. Determine every value of t for which CG is perpendicular to AB, and prove that these are exactly the values for which CA and CB are congruent. Your argument must cover all allowed a,h.
1. Hint 1
  Try a numerical base and height before keeping them arbitrary.
2. Hint 2
  Compare the first coordinates of C and the opposite side midpoint, or of C and G.
3. Hint 3
  For the equal-side condition, equate squared distances and preserve both directions.
Answer

$t=a$ is the unique value. For every positive a,h, this is exactly the condition $CA=CB$ .

Full solution

Test a base of length 10 and height 7.
The midpoint is (5,0).
A perpendicular to the horizontal base is vertical, so the median from C is perpendicular precisely when t is 5.
The side-square equality is
$t^2+49=(t-10)^2+49.$
Expanding gives t equal to 5 as well.

In general the centroid coordinates are
$\begin{gathered}x_G=(2a+t)/3\\y_G=h/3.\end{gathered}$
Since h is positive, G and C are distinct.
The line CG is perpendicular to horizontal AB precisely when it is vertical.

Its first coordinates agree precisely when
$\begin{aligned}t&=(2a+t)/3\\3t&=2a+t\\t&=a.\end{aligned}$

The equal-side condition is equivalent to equality of positive side squares.
$\begin{aligned}t^2+h^2&=(t-2a)^2+h^2\\4at&=4a^2\\t&=a.\end{aligned}$
Division is valid because a is positive.

Conversely, t equal to a gives both a vertical CG and matching side squares.
All transformations were reversible, so there are no other values.
The positive height keeps the triangle nondegenerate.

In the numerical solution, C, G, and the base midpoint M lie on one vertical median.

Answer

$t=a$ is the unique value. For every positive a,h, this is exactly the condition $CA=CB$ .

Key idea

A shared coordinate constraint can prove both directions of a geometric equivalence.

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