Coordinate Geometry Basics: Analyzing Triangles on the Coordinate Plane

Capítulo 8

Estimated reading time: 5 minutes

+ Exercise

1) Plot vertices from coordinates and label sides

When analyzing a triangle on the coordinate plane, start by naming the vertices and connecting them in order. Use consistent labels so your calculations match your diagram.

Example A: Build triangle ABC

Let A(-1, 1), B(5, 1), and C(2, 5).

  • Plot points A, B, C.
  • Draw segments AB, BC, and CA.
  • Label the sides: AB is opposite C, BC is opposite A, CA is opposite B.

To keep work organized, create a quick reference table of the three sides you will analyze.

SideEndpoints
ABA(-1,1) to B(5,1)
BCB(5,1) to C(2,5)
CAC(2,5) to A(-1,1)

2) Compute side lengths to classify the triangle

To classify a triangle by side lengths, compute the three side lengths and compare them.

  • Scalene: all three side lengths are different.
  • Isosceles: two side lengths are equal.
  • Equilateral: all three side lengths are equal (this can happen on the coordinate plane, though it may involve non-integer lengths).

Tip: compare squared lengths to avoid radicals

Because distance calculations often produce square roots, it is efficient to compare squared side lengths. If AB^2 = AC^2, then AB = AC without needing to simplify square roots.

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Example A continued: classify triangle ABC by side lengths

Compute squared lengths:

AB^2 = (5 - (-1))^2 + (1 - 1)^2 = 6^2 + 0^2 = 36
BC^2 = (2 - 5)^2 + (5 - 1)^2 = (-3)^2 + 4^2 = 9 + 16 = 25
CA^2 = (-1 - 2)^2 + (1 - 5)^2 = (-3)^2 + (-4)^2 = 9 + 16 = 25

Since BC^2 = CA^2 = 25, the triangle has two equal sides (BC = CA), so it is isosceles. Because AB^2 is different, it is not equilateral.

Example B: a scalene triangle

Let P(0,0), Q(4,1), R(1,5). Compute squared lengths:

PQ^2 = (4-0)^2 + (1-0)^2 = 16 + 1 = 17
QR^2 = (1-4)^2 + (5-1)^2 = 9 + 16 = 25
RP^2 = (0-1)^2 + (0-5)^2 = 1 + 25 = 26

All three squared lengths are different, so triangle PQR is scalene.

Example C: an equilateral triangle (when applicable)

Equilateral triangles can be formed using coordinates, though the coordinates may not all be integers. For instance, take A(0,0), B(2,0), and C(1, √3).

AB^2 = (2-0)^2 + (0-0)^2 = 4
AC^2 = (1-0)^2 + (√3-0)^2 = 1 + 3 = 4
BC^2 = (1-2)^2 + (√3-0)^2 = 1 + 3 = 4

All three squared lengths match, so the triangle is equilateral.

3) Use slopes to identify right angles and parallel sides

Slopes help you verify angle relationships without measuring angles directly.

Right angles via perpendicular slopes

Two non-vertical lines are perpendicular if their slopes are negative reciprocals: if m1 · m2 = -1. Special cases:

  • A horizontal line (slope 0) is perpendicular to a vertical line (undefined slope).
  • If one segment is vertical, check whether the other is horizontal to confirm perpendicularity.

Example D: verify a right triangle using slopes

Let A(1,1), B(1,6), C(5,1).

  • AB is vertical (same x-coordinate), so its slope is undefined.
  • AC is horizontal (same y-coordinate), so its slope is 0.

A vertical segment is perpendicular to a horizontal segment, so angle A is a right angle and triangle ABC is a right triangle.

Parallel sides in related constructions

In triangles, you may create or analyze additional segments (like midsegments) where parallelism matters. Two non-vertical segments are parallel if their slopes are equal. Two vertical segments are parallel to each other as well.

Example E: midsegment parallel to a side (uses midpoint + slope)

Let triangle ABC have A(0,0), B(6,0), C(2,4). Find the midpoints of AB and AC, then compare slopes.

M = midpoint of AB = ((0+6)/2, (0+0)/2) = (3,0)
N = midpoint of AC = ((0+2)/2, (0+4)/2) = (1,2)

Compute slopes:

slope(MN) = (2-0)/(1-3) = 2/(-2) = -1
slope(BC) = (4-0)/(2-6) = 4/(-4) = -1

Since the slopes match, MN is parallel to BC. This is a coordinate-based verification of a common triangle construction: the segment joining midpoints of two sides is parallel to the third side.

4) Validation tasks: calculate, then confirm by accurate graphing

Validation means you do the algebraic checks first (distance and slope), then you graph carefully to see whether the picture matches your results. The graph is a confirmation, not a substitute for calculation.

Task 1: Determine whether a triangle is right

Given A(-2,1), B(3,4), C(1,-1).

Step 1: compute slopes of two sides that meet at a vertex (try vertex B using BA and BC).

m_BA = (1-4)/(-2-3) = (-3)/(-5) = 3/5
m_BC = (-1-4)/(1-3) = (-5)/(-2) = 5/2

Step 2: check perpendicular condition

m_BA · m_BC = (3/5) · (5/2) = 3/2 ≠ -1

So the angle at B is not right. Check another vertex (for example, slopes AB and AC at A).

m_AB = (4-1)/(3-(-2)) = 3/5
m_AC = (-1-1)/(1-(-2)) = (-2)/3 = -2/3
m_AB · m_AC = (3/5) · (-2/3) = -2/5 ≠ -1

Not right at A. Check at C:

m_CA = (1-(-1))/(-2-1) = 2/(-3) = -2/3
m_CB = (4-(-1))/(3-1) = 5/2
m_CA · m_CB = (-2/3) · (5/2) = -5/3 ≠ -1

Result: triangle ABC is not a right triangle.

Step 3: confirm by graphing

  • Plot A, B, C accurately.
  • Draw the sides and visually inspect angles; none should appear to be exactly 90°.
  • If your drawing seems to show a right angle, re-check plotting accuracy and slope arithmetic.

Task 2: Determine whether a triangle is isosceles

Given P(-1,2), Q(3,2), R(1,6).

Step 1: compute squared side lengths

PQ^2 = (3-(-1))^2 + (2-2)^2 = 4^2 + 0^2 = 16
PR^2 = (1-(-1))^2 + (6-2)^2 = 2^2 + 4^2 = 4 + 16 = 20
QR^2 = (1-3)^2 + (6-2)^2 = (-2)^2 + 4^2 = 4 + 16 = 20

Step 2: compare

PR^2 = QR^2, so PR = QR. Triangle PQR is isosceles with equal sides meeting at R (so R is the vertex of the equal sides, and PQ is the base).

Step 3: confirm by graphing

  • Plot P and Q on the horizontal line y=2, then plot R above them.
  • Draw segments PR and QR; they should look symmetric around the vertical line through the midpoint of PQ.

Task 3: Combine midpoint and slope to validate a construction

Given A(0,1), B(8,3), C(2,7). Let M be the midpoint of AB and N the midpoint of AC. Verify that MN is parallel to BC.

Step 1: midpoints

M = ((0+8)/2, (1+3)/2) = (4,2)
N = ((0+2)/2, (1+7)/2) = (1,4)

Step 2: slopes

slope(MN) = (4-2)/(1-4) = 2/(-3) = -2/3
slope(BC) = (7-3)/(2-8) = 4/(-6) = -2/3

Result: slopes match, so MN is parallel to BC.

Step 3: confirm by graphing

  • Plot A, B, C, then plot M and N.
  • Draw BC and MN; they should have the same tilt and never meet.

Now answer the exercise about the content:

Given triangle PQR with P(-1,2), Q(3,2), and R(1,6), which conclusion is supported by comparing squared side lengths?

You are right! Congratulations, now go to the next page

You missed! Try again.

Compute squared lengths: PQ^2=16, PR^2=20, QR^2=20. Since PR^2=QR^2, two sides are equal, so the triangle is isosceles and PQ is the base.

Next chapter

Coordinate Geometry Basics: Rectangles and Parallelograms Using Slopes and Distances

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