Classifying figures from coordinates
Four points are plotted on a grid. How can you be sure which shape they make, without trusting your eyes?
Where you are 9 in Coordinate geometry
When you reach for it
A drawing can only suggest a shape. Two sides that look parallel may differ in slope by a hair, and a corner that looks square may be 87^\circ. Once the corners are given as coordinates, though, there is no need to guess: every property in the quadrilateral family tree can be tested with arithmetic.
- Parallel sides are a question about slopes.
- Right angles are a question about slopes too: perpendicular lines have slopes that multiply to -1.
- Equal sides and equal diagonals are a question about lengths.
- Diagonals that cut each other in half are a question about midpoints.
Reach for this whenever points are plotted and a question asks what figure they make: is this triangle right, is it isosceles, is this quadrilateral a parallelogram, a rectangle, a rhombus, a square, a trapezoid or a kite? The answer is then not “it looks like one” but a short argument that it meets the definition, or that it fails one property every such figure must have.
The procedure
Three measurements
Everything rests on three formulas you already have.
The slope of the segment from (x_1, y_1) to (x_2, y_2) is
and it is undefined when x_1 = x_2, for a vertical segment. Two segments are parallel, \parallel, when their slopes are equal and they do not lie on one line, or when both are vertical and not on one line. Two segments are perpendicular, \perp, when their slopes multiply to -1, or when one is horizontal and the other vertical (Slopes of perpendicular lines). That last case has to be checked separately, since a vertical slope cannot be multiplied.
The The distance formula gives the length of a segment, and its square is simpler still:
To decide whether two segments are equal there is no need to take square roots. Lengths are never negative, and two numbers that are not negative are equal exactly when their squares are equal, so AB = CD exactly when AB^2 = CD^2. Working with squared lengths keeps every number an integer when the coordinates are integers, and it is also the form the converse of the Pythagorean theorem wants.
The midpoint of a segment is the average of its endpoints,
and two segments bisect each other exactly when they have the same midpoint.
Triangles
A triangle is right when two of its sides are perpendicular, and the right angle sits at the vertex those two sides share. So compute the three slopes and test the three pairs. A second test needs no slopes at all: by the converse of the Pythagorean theorem, the triangle is right exactly when the largest squared length equals the sum of the other two. The same comparison also sorts the other triangles: if the largest squared length is bigger than the sum of the other two, the triangle is obtuse, and if it is smaller, the triangle is acute (Classifying triangles).
A triangle is isosceles when two of its squared side lengths are equal, and equilateral when all three are.
Before any of this, the three points must not lie on one line, or there is no triangle. If m_{AB} = m_{AC}, or \overline{AB} and \overline{AC} are both vertical, the points A, B and C are on one line, since the two segments share the point A.
Quadrilaterals
A quadrilateral ABCD is named by going round it, so its sides are \overline{AB}, \overline{BC}, \overline{CD} and \overline{DA}. The opposite pairs are \overline{AB} with \overline{CD} and \overline{BC} with \overline{DA}, and the diagonals are \overline{AC} and \overline{BD}. The order of the letters matters: the same four points taken in another order can make a different figure, or no figure at all (see Boundaries).
Each name in the family tree has a definition, and some have a shorter test that is a theorem. Either is enough to show the name fits. These are the ones used below.
- Trapezoid: one pair of opposite sides parallel. That is the definition, which asks for at least one pair.
- Parallelogram: both pairs of opposite sides parallel, which is the definition. Two theorems give other routes: diagonals that share a midpoint, or both pairs of opposite sides congruent (Proving a quadrilateral is a parallelogram).
- Rectangle: a parallelogram with one right angle. Neighboring angles of a parallelogram add to 180^\circ and opposite angles are equal, so one right angle makes all four right. Or a parallelogram with congruent diagonals (Proving a parallelogram is a rectangle or a rhombus).
- Rhombus: four congruent sides. Opposite sides congruent already make it a parallelogram, and then it meets the definition. Or a parallelogram with perpendicular diagonals (the rhombus test).
- Square: a rhombus with one right angle. A rhombus is a parallelogram, so the right angle makes it a rectangle as well.
- Isosceles trapezoid: one pair of opposite sides parallel, and the other pair congruent and not parallel. That is the definition.
- Kite: AB = AD and CB = CD, or BA = BC and DA = DC. That is the definition, two pairs of congruent neighbors.
Notice which tests begin with “a parallelogram with”. Congruent diagonals on their own do not make a rectangle, since an isosceles trapezoid has them too, and perpendicular diagonals on their own do not make a rhombus, since a kite has them too. Those tests only work once the parallelogram has been shown.
To show a figure is not something, one property that every such figure has, and this one lacks, is enough. Every parallelogram has both pairs of opposite sides parallel, so one non-parallel pair rules it out. Every rectangle has a right angle at each corner, so one neighboring pair of sides that is not perpendicular rules it out. Every rhombus has four equal sides, so two neighbors of different lengths rule it out.
Because the definitions are inclusive, a square is also a rhombus, a rectangle, a parallelogram, a kite and a trapezoid. The useful answer is the most specific name the figure earns, together with the names it misses. That is why it pays to work down the tree: first look for parallel pairs, then right angles, then equal sides.
Worked examples
Is the triangle right?
Decide whether the triangle with vertices A(-2, 1), B(2, 3) and C(3, 1) is a right triangle. If it is, name the vertex of the right angle.
SolutionFind the three slopes.
Test the pairs. \displaystyle m_{AB} \cdot m_{BC} = \frac{1}{2} \cdot (-2) = -1, so \overline{AB} \perp \overline{BC}, and the right angle is at B, the vertex the two sides share (Figure 1).
The converse of the Pythagorean theorem gives the same verdict without slopes. The squared lengths are
and 20 + 5 = 25, so the angle opposite the longest side \overline{AC}, which is the angle at B, is a right angle. The three squared lengths are all different, so the triangle is a right triangle that is not isosceles.
An isosceles triangle that is not right
Classify the triangle with vertices A(1, 1), B(5, 4) and C(-3, 4) by its sides and by its angles.
SolutionStart with the squared lengths, since they settle the sides and the angles together.
Two of them are equal, so AB = AC = 5 and the triangle is isosceles, with BC = 8 as its base.
For the angles, compare the largest squared length with the sum of the other two: 64 > 25 + 25 = 50. By the converse of the Pythagorean theorem and its companion, the angle opposite \overline{BC}, the angle at A, is obtuse.
The slopes agree. \overline{BC} is horizontal, and neither of the other sides is vertical, so there is no right angle at B or at C. At A, \displaystyle m_{AB} = \frac{3}{4} and \displaystyle m_{AC} = \frac{3}{-4} = -\frac{3}{4}, whose product is \displaystyle -\frac{9}{16}, not -1. The triangle is an isosceles obtuse triangle.
A trapezoid with two right angles
Classify the quadrilateral with vertices A(-2, 0), B(4, 2), C(3, 5) and D(0, 4).
SolutionWork down the tree, beginning with the slopes of the four sides.
\overline{AB} and \overline{CD} have equal slopes. They are not on one line, because the line through A with slope \displaystyle \frac{1}{3} meets x = 0 at height \displaystyle \frac{2}{3}, not at D’s height 4. So \overline{AB} \parallel \overline{CD} and ABCD is a trapezoid. The other pair has slopes -3 and 2, so it is not parallel, and ABCD is not a parallelogram. That also rules out rectangle, rhombus and square, which are all parallelograms.
Could it be an isosceles trapezoid or a kite? The squared side lengths are
The legs, \overline{BC} and \overline{DA}, differ, so it is not an isosceles trapezoid. A kite needs two pairs of equal neighbors. \overline{BC} and \overline{CD} are an equal pair, but the other two sides, \overline{AB} and \overline{DA}, are not equal, and the other way of pairing, AB = BC with CD = DA, fails at once. So it is not a kite.
The products of neighboring slopes find something else: \displaystyle m_{AB} \cdot m_{BC} = \frac{1}{3} \cdot (-3) = -1 and m_{BC} \cdot m_{CD} = -1. The angles at B and C are right angles (Figure 2). ABCD is a trapezoid, and no more specific name in the tree fits it, though it has two right angles; such a figure is often called a right trapezoid.
A rhombus that is not a square
Show that A(0, 0), B(5, 1), C(6, 6), D(1, 5) is a rhombus, and decide whether it is a square.
SolutionThe squared side lengths are
All four sides are congruent. Opposite sides congruent already makes ABCD a parallelogram, and a parallelogram with four congruent sides is a rhombus by definition.
A square would also need a right angle. The neighboring sides \overline{AB} and \overline{BC} have slopes \displaystyle \frac{1}{5} and 5, whose product is 1, not -1. So the angle at B is not a right angle, ABCD is not a rectangle, and so it is not a square.
The diagonals confirm the picture (Figure 3). \displaystyle m_{AC} = \frac{6}{6} = 1 and \displaystyle m_{BD} = \frac{5 - 1}{1 - 5} = -1, so the diagonals are perpendicular, as every rhombus’s are. Their squared lengths are AC^2 = 72 and BD^2 = 32, which differ, as a rectangle’s never would.
A rectangle shown by its diagonals
Use the diagonals to show that A(-1, 0), B(3, 2), C(0, 8), D(-4, 6) is a rectangle, then decide whether it is a square.
SolutionThe congruent-diagonals test needs a parallelogram first, and the diagonals can show that too. The midpoint of \overline{AC} is
and the midpoint of \overline{BD} is
The diagonals share a midpoint, so they bisect each other and ABCD is a parallelogram.
Their squared lengths are
so the diagonals are congruent, and a parallelogram with congruent diagonals is a rectangle (Figure 4).
For a square, two neighboring sides would have to be equal. AB^2 = 4^2 + 2^2 = 20 and BC^2 = (-3)^2 + 6^2 = 45 differ, so ABCD is a rectangle but not a square. As a check, the slopes \displaystyle m_{AB} = \frac{1}{2} and \displaystyle m_{BC} = \frac{6}{-3} = -2 multiply to -1, the right angle at B that every rectangle has.
Finishing a square
Three vertices of a square PQRS are P(1, -2), Q(5, 0) and R(3, 4). Find S, then prove that PQRS is a square.
SolutionA square is a parallelogram, so its diagonals \overline{PR} and \overline{QS} share a midpoint. The midpoint of \overline{PR} is \displaystyle \left(\frac{1 + 3}{2}, \frac{-2 + 4}{2}\right) = (2, 1). For (2, 1) to be the midpoint of \overline{QS}, S must be as far beyond (2, 1) as Q is short of it:
Finding S this way only guarantees a parallelogram. Whether it is a square depends on the three given points, so the proof is still needed. The squared side lengths are
so PQRS has four congruent sides and is a rhombus. The neighboring sides at Q have slopes \displaystyle m_{PQ} = \frac{2}{4} = \frac{1}{2} and \displaystyle m_{QR} = \frac{4}{-2} = -2, which multiply to -1, so the angle at Q is right. A rhombus with a right angle is a square.
Had \angle Q not been a right angle, or PQ and QR not been equal, no point S at all would have finished a square, and the parallelogram found this way would have been the end of it.
A vertex that slides along a line
The triangle has vertices A(0, 0), B(6, 2) and C(k, 8), where k can be any number. Find every value of k that makes the triangle a right triangle.
SolutionThe right angle could be at any of the three vertices, so test each one. Here \displaystyle m_{AB} = \frac{2}{6} = \frac{1}{3} does not depend on k.
Right angle at A. For k \ne 0, \displaystyle m_{AC} = \frac{8}{k}, and perpendicularity needs
When k = 0, \overline{AC} is vertical, but \overline{AB} is not horizontal, so there is no right angle then.
Right angle at B. For k \ne 6, \displaystyle m_{BC} = \frac{8 - 2}{k - 6} = \frac{6}{k - 6}, and
When k = 6, \overline{BC} is vertical and \overline{AB} again is not horizontal.
Right angle at C. For k \ne 0 and k \ne 6 the condition is
whose discriminant is 36 - 192 = -156 < 0, so no real k works. The two vertical cases fail too: at k = 0, \displaystyle m_{CB} = \frac{6}{-6} = -1, and at k = 6, \displaystyle m_{CA} = \frac{8}{6}, and neither side is horizontal.
So the triangle is right for exactly two values: k = 4, with the right angle at B, and \displaystyle k = -\frac{8}{3}, with the right angle at A. For k = 4, C is (4, 8) and the squared lengths 40, 40 and 80 satisfy 40 + 40 = 80, so that triangle is also isosceles.
Boundaries
The points must make the figure. Three points on one line make no triangle, and the slope tests above would then report “parallel” sides that are really one line. Four points make a quadrilateral ABCD only if no three of them lie on one line and the sides, taken in the order named, do not cross. The points A(0, 0), B(4, 0), C(0, 3), D(4, 3) are the corners of a rectangle, yet ABCD is not one: its sides \overline{BC} and \overline{DA} cross (Figure 5). The rectangle is ABDC.
Equal slopes mean parallel or the same line. Two sides with the same slope are parallel only if they do not lie on one line; checking one point, as in Example 3, settles it.
Vertical lines have no slope. The test “slopes multiply to -1” covers only sides that are neither vertical nor horizontal. A vertical side is perpendicular exactly to the horizontal ones, and parallel exactly to the other vertical ones.
Some tests need a parallelogram first. Congruent diagonals belong to isosceles trapezoids as well as rectangles, and perpendicular diagonals belong to kites as well as rhombuses. On their own they decide nothing.
The arithmetic must be exact. Coordinates read off a picture, or lengths rounded to decimals, can make unequal sides look equal. The argument holds only for exact coordinates and exact squared lengths. The formulas themselves assume the two axes are perpendicular and marked in the same unit; on a stretched grid the picture lies, though the numbers do not.
Where this goes next
Every test here was run on numbers, so each answer is about one figure. Put letters in place of the numbers and the same slopes, lengths and midpoints prove a statement about every figure of a kind at once: that the diagonals of any parallelogram bisect each other, or that the midpoints of any quadrilateral’s sides form a parallelogram. That is coordinate proof with variables, where the work is choosing coordinates that keep the figure general and the algebra short.
Once a figure is classified, its measurements are already half done. The squared lengths found here give its perimeter, and knowing that a figure is a rectangle or a rhombus says which area formula applies (Perimeter and area from coordinates, Area of a polygon from coordinates).
Exercises
Triangles
For each triangle, decide whether it is right, and if so where the right angle is, and whether it is isosceles.
- A(1, 2), B(4, 6), C(8, 3)
- D(-3, -1), E(3, 1), F(1, 5). If it is not right, say whether it is acute or obtuse.
- G(-2, 3), H(4, 3), I(1, -1). If it is not right, say whether it is acute or obtuse.
- J(0, 0), K(2, 5), L(12, 1)
- M(-2, -1), N(-2, 4), P(5, -1)
Name the quadrilateral
Give the most specific name for ABCD (trapezoid, isosceles trapezoid, parallelogram, rectangle, rhombus, square or kite), with your evidence, or show that none of them fits.
- A(0, 0), B(6, 0), C(8, 4), D(2, 4)
- A(-2, -1), B(2, -3), C(4, 1), D(0, 3)
- A(0, 0), B(3, 3), C(1, 5), D(-2, 2)
- A(-4, 0), B(0, -2), C(4, 0), D(0, 2)
- A(0, 0), B(4, 1), C(6, 6), D(1, 4)
- A(-3, 0), B(5, 0), C(3, 4), D(-1, 4)
- A(0, 0), B(5, 0), C(6, 3), D(1, 4)
Is it, or is it not
- Use the diagonals to show that A(-1, 0), B(3, -2), C(4, 0), D(0, 2) is a rectangle. Is it a square?
- Kai finds that the diagonals of P(0, 0), Q(6, 0), R(4, 4), S(2, 4) are congruent and concludes that PQRS is a rectangle. Is he right? Name the figure.
- The diagonals of W(-2, 0), X(0, -3), Y(2, 0), Z(0, 5) are perpendicular. Is WXYZ a rhombus? Name the figure.
Finding a missing point or value
- Three vertices of parallelogram ABCD are A(-1, 2), B(3, 1) and C(6, 4). Find D.
- A(1, 1) and B(5, 3) are neighboring vertices of square ABCD, which lies above the line AB. Find C and D.
- Find k so that the triangle with vertices A(1, 2), B(5, 4) and C(k, 0) has a right angle at B.
- The points P(0, 0), Q(5, 0), R(8, y), S(3, y) with y \ne 0 always form a parallelogram PQRS. For which values of y is it a rhombus? Can it ever be a square?
Reasoning
- Explain why showing that a quadrilateral has four congruent sides is enough to show it is a rhombus, while showing that its diagonals are perpendicular is not.
- Explain why comparing AB^2 with CD^2 decides whether AB = CD, so that no square root is ever needed for that question.
- Lee shows that a quadrilateral has one right angle and both pairs of opposite sides congruent. Has Lee shown that it is a rectangle? Explain.
- The points A(0, 0), B(4, 0), C(0, 3), D(4, 3) are the four corners of a rectangle. Explain why the quadrilateral ABCD is nevertheless not a rectangle, and give an order of the letters that does name the rectangle.
- Explain why no equilateral triangle has all three of its vertices at points whose coordinates are both integers. (Use two facts: the square of each side length is then an integer, and the area of a triangle with integer vertices can be found by boxing it in a rectangle and subtracting right triangles and rectangles.)
Answers
- Right angle at B, and isosceles. \displaystyle m_{AB} = \frac{4}{3} and \displaystyle m_{BC} = -\frac{3}{4} multiply to -1; AB^2 = 25 = BC^2, and AC^2 = 50 = 25 + 25.
- Not right; scalene and acute. Slopes \displaystyle m_{DE} = \frac{1}{3}, m_{EF} = -2, \displaystyle m_{DF} = \frac{3}{2}: no pair multiplies to -1 and no side is vertical. DE^2 = 40, EF^2 = 20, DF^2 = 52, and 52 < 40 + 20 = 60.
- Isosceles, not right, acute. GI^2 = 25 = HI^2 and GH^2 = 36. \overline{GH} is horizontal and no side is vertical; \displaystyle m_{GI} \cdot m_{HI} = -\frac{4}{3} \cdot \frac{4}{3} = -\frac{16}{9}. 36 < 25 + 25.
- Right angle at K, not isosceles. \displaystyle m_{JK} = \frac{5}{2}, \displaystyle m_{KL} = -\frac{2}{5}, product -1; JK^2 = 29, KL^2 = 116, JL^2 = 145 = 29 + 116.
- Right angle at M, not isosceles. \overline{MN} is vertical and \overline{MP} is horizontal, so the slope product cannot be used, but the sides are perpendicular. MN = 5, MP = 7, NP^2 = 74.
- Parallelogram, not a rectangle or rhombus. \overline{AB} and \overline{CD} are horizontal; m_{BC} = m_{DA} = 2. AB^2 = 36 \ne BC^2 = 20, and a horizontal side with a side of slope 2 makes no right angle.
- Square. Every side has squared length 20, so it is a rhombus, and \displaystyle m_{AB} = -\frac{1}{2}, m_{BC} = 2 multiply to -1, a right angle at B.
- Rectangle, not a square. Slopes 1, -1, 1, -1 make both opposite pairs parallel and the neighbors perpendicular; AB^2 = 18 \ne BC^2 = 8.
- Rhombus, not a square. Every side has squared length 20; \displaystyle m_{AB} = -\frac{1}{2} and \displaystyle m_{BC} = \frac{1}{2} multiply to \displaystyle -\frac{1}{4}, so the angle at B is not right.
- Kite, not a rhombus. AB^2 = AD^2 = 17 and CB^2 = CD^2 = 29. Slopes \displaystyle \frac{1}{4}, \displaystyle \frac{5}{2}, \displaystyle \frac{2}{5}, 4: no opposite pair is parallel, so it is not a trapezoid.
- Isosceles trapezoid. \overline{AB} and \overline{DC} are horizontal; m_{BC} = -2 and m_{DA} = 2 are not equal, and BC^2 = DA^2 = 20.
- None of them. Slopes 0, 3, \displaystyle -\frac{1}{5}, 4 give no parallel pair, so no trapezoid or parallelogram; squared sides 25, 10, 26, 17 give no equal neighbors, so no kite.
- Rectangle, not a square. Both diagonals have midpoint \displaystyle \left(\frac{3}{2}, 0\right), so it is a parallelogram, and AC^2 = BD^2 = 25. AB^2 = 20 \ne BC^2 = 5.
- No; it is an isosceles trapezoid. PR^2 = QS^2 = 32, but m_{QR} = -2 and m_{PS} = 2, so PQRS is not a parallelogram. \overline{PQ} \parallel \overline{SR} (both horizontal) and QR^2 = PS^2 = 20.
- No; it is a kite. WX^2 = XY^2 = 13 and YZ^2 = ZW^2 = 29, so the four sides are not all equal. The diagonals lie on the axes, which is why they are perpendicular.
- D(2, 5). The diagonals share a midpoint, \displaystyle \left(\frac{5}{2}, 3\right), so D = (-1 + 6 - 3,\ 2 + 4 - 1). Check: \displaystyle m_{AB} = m_{DC} = -\frac{1}{4}, m_{BC} = m_{AD} = 1.
- C(3, 7) and D(-1, 5). From A to B is 4 across and 2 up; turning that a quarter turn gives 2 back and 4 up, so C = B + (-2, 4) and D = A + (-2, 4). Every side has squared length 20 and \displaystyle m_{AB} \cdot m_{BC} = \frac{1}{2} \cdot (-2) = -1.
- k = 7. \displaystyle m_{BA} = \frac{1}{2} and \displaystyle m_{BC} = \frac{-4}{k - 5}; their product \displaystyle \frac{-2}{k - 5} equals -1 when k - 5 = 2. Check: AB^2 = BC^2 = 20, AC^2 = 40.
- y = 4 or y = -4; never a square. A rhombus needs PS^2 = PQ^2, so 9 + y^2 = 25. A square needs \overline{PS} \perp \overline{PQ}, and \overline{PQ} is horizontal while \overline{PS} runs 3 across, so it is never vertical.
- Four congruent sides means both pairs of opposite sides are congruent, which makes the quadrilateral a parallelogram; a parallelogram with four congruent sides is a rhombus by definition. Perpendicular diagonals also occur in kites that are not rhombuses, such as Exercise 15, so that test needs a parallelogram first.
- Lengths are never negative, and for numbers that are not negative, a = b exactly when a^2 = b^2 (squaring keeps their order). So equal squared lengths mean equal lengths, and unequal squared lengths mean unequal lengths.
- Yes. Both pairs of opposite sides congruent makes it a parallelogram. In a parallelogram, neighboring angles add to 180^\circ and opposite angles are equal, so one right angle forces all four to be right, and a parallelogram with four right angles is a rectangle.
- Taken in order, ABCD has sides \overline{AB}, \overline{BC}, \overline{CD}, \overline{DA}, and \overline{BC} (from (4, 0) to (0, 3)) crosses \overline{DA} (from (4, 3) to (0, 0)) at \displaystyle \left(2, \frac{3}{2}\right), so the figure is a crossed bow tie, not a quadrilateral. The order ABDC goes round the rectangle.
- If the side is s, then s^2 is a sum of two squares of integers, so a whole number. An equilateral triangle has area \displaystyle \frac{\sqrt{3}}{4}s^2, which is irrational because \displaystyle \sqrt{3} is irrational and \displaystyle \frac{s^2}{4} is a non-zero fraction. But a triangle with integer vertices has area equal to its bounding rectangle (a whole number) minus right triangles with integer legs (each a multiple of \displaystyle \frac{1}{2}) and rectangles with integer sides, so its area is a multiple of \displaystyle \frac{1}{2}, a rational number. The two cannot agree, so no such triangle exists.