Determinants · Lesson 2 of 6
Determinants and the Area of a Triangle
“Use one coordinate determinant to calculate area, recognise collinearity, and describe a line.”
• Connect the coordinate formula for a triangle’s area to an order-three determinant. • Distinguish a signed determinant from the nonnegative geometric area. • Use zero area to test collinearity and form a line equation. • Find unknown coordinates from a specified area and retain both possible signs. • Check determinant answers against the geometry of the given points.
Coordinates can carry geometric information
Three points give us six coordinates. A determinant arranges them so that one calculation answers a geometric question: what is the area of the triangle they form? The calculation also tells us when the three points lie on a straight line, which will let us express a line by a determinant equation.
Let the vertices be P(x₁,y₁), Q(x₂,y₂), and R(x₃,y₃). An earlier coordinate-area expression combines x₁(y₂−y₃), x₂(y₃−y₁), and x₃(y₁−y₂). Place the x-coordinates in column 1, the corresponding y-coordinates in column 2, and a 1 in each row of column 3. Each row now belongs to one complete point. Mixing a point’s x-coordinate with another point’s y-coordinate would change the triangle.
Expand D along its first row. It becomes x₁(y₂−y₃) − y₁(x₂−x₃) + (x₂y₃−y₂x₃). Regroup the middle terms to obtain x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂). Thus the determinant packages the familiar coordinate expression; it is not a different area rule. The final absolute value ensures that the area does not become negative because of the order in which the vertices were listed.
We take the absolute value after completing the determinant calculation. Replacing negative coordinates by positive ones would move the vertices and generally change the triangle. Similarly, taking absolute values of separate terms destroys cancellations. Work with the original signed coordinates, obtain D, and only then use T = |D|/2.
Problem
Find the area of the triangle with vertices (3,8), (−4,2), and (5,1).
- 1.Form D = det [3,8,1;−4,2,1;5,1,1], keeping the coordinates of each vertex together.
- 2.Expand the first row: D = 3(2·1−1·1) − 8((−4)·1−1·5) + 1((−4)·1−2·5).
- 3.The three bracket values are 1, −9, and −14. Hence D = 3 − 8(−9) − 14 = 61.
- 4.Take half the magnitude: T = 61/2 square units. If the vertices were listed in another order that gave D = −61, the same area would result.
A geometric check on the formula
The determinant method works for a sloping triangle as well as one with a horizontal base. When a convenient base and height are visible, they provide an independent check. Such checks are useful because a correct-looking array can still contain a copied coordinate or a sign error.
Problem
Find the area of A(0,0), B(5,0), and C(3,3) by determinants and check it geometrically.
- 1.Write D = det [0,0,1;5,0,1;3,3,1]. In the first row only the last entry contributes, with a positive position sign.
- 2.The remaining determinant is det [5,0;3,3] = 15. Thus T = 15/2 square units.
- 3.The horizontal segment AB has length 5, and C is 3 units above its line. Therefore one-half base times height gives (1/2)·5·3 = 15/2, confirming the result.
- 4.If C were (3,−3), the determinant would be −15 but the perpendicular distance and area would still be 3 and 15/2. The sign is not a negative geometric area.
The determinant D equals twice the signed area. If the required area is 4, the determinant can be 8 or −8; it is not generally 4. Write the relationship T = |D|/2 before solving an unknown-coordinate problem.
Zero area detects a straight line
If three vertices lie on one straight line, there is no two-dimensional region between them and their triangle has zero area. Conversely, a zero-area triangle has collinear vertices. The determinant therefore gives a condition for collinearity without calculating slopes or dividing by coordinate differences.
To describe the line through two distinct points A and B, let its third point P have variable coordinates (x,y). Put A, B, and P into the three determinant rows and set the value equal to zero. Expanding produces an equation relating x and y. Every point satisfying that equation is collinear with A and B. This is why an area calculation can generate a line equation.
Problem
Find the line through A(1,3) and B(0,0). Then find k if D(k,0) makes triangle ABD have area 3.
- 1.Let P(x,y) lie on AB. Set det [0,0,1;1,3,1;x,y,1] = 0 because the three points are collinear.
- 2.Expanding along the first row gives y−3x = 0. The line equation is therefore y = 3x. Both given points satisfy it.
- 3.For the area condition, use det [1,3,1;0,0,1;k,0,1] = −3k. Then (1/2)|−3k| = 3.
- 4.Multiplying by 2 gives 3|k| = 6, so |k| = 2. Thus k = 2 or k = −2.
- 5.The two possible points lie on opposite sides of B along the horizontal axis. Their triangles have equal area, so discarding the negative value would lose a valid geometric possibility.
Problem
Show that A(a,b+c), B(b,c+a), and C(c,a+b) are collinear.
- 1.Use the expanded determinant expression: D = a[(c+a)−(a+b)] + b[(a+b)−(b+c)] + c[(b+c)−(c+a)].
- 2.Simplify the differences inside the brackets: D = a(c−b) + b(a−c) + c(b−a).
- 3.Distribute: D = ac−ab + ab−bc + bc−ac = 0. Thus the area is zero, proving collinearity.
- 4.There is also a coordinate check: each point satisfies x+y = a+b+c. This supports the same conclusion even when some of a, b, and c are equal.
Unknown coordinates require both signs
When the area is given, the determinant is constrained by a magnitude. We can remove that magnitude by treating two equations, D = 2T and D = −2T. These branches usually describe points on opposite sides of a fixed line, so both belong to the original geometric question.
Problem
The area of the triangle with vertices (k,0), (4,0), and (0,2) is 4 square units. Find k.
- 1.Form D = det [k,0,1;4,0,1;0,2,1]. Expansion along the first row gives D = −2k + 8.
- 2.Since T = |D|/2, the condition is |8−2k| = 8.
- 3.For 8−2k = 8, obtain k = 0. For 8−2k = −8, obtain k = 8.
- 4.Check with base and height: the horizontal base has length |4−k| and the height is 2, so its area is |4−k|. Both 0 and 8 give area 4; k = 4 would instead collapse the base.
Before finishing an area question, check the units, the factor one-half, and whether all sign branches were retained. For a collinearity question, check that the final signed determinant is zero rather than merely negative. For a line equation, substitute both fixed points. These checks use the meaning of the calculation, so they can expose mistakes that repeating the same arithmetic may miss.
Quiz
A coordinate determinant D has value −18. What is the triangle’s area?
Which determinant condition expresses collinearity?
The area is 5 square units. Which equation must the determinant D satisfy?
For A(0,0), B(4,0), C(1,3), what is the area?
To find the line through two distinct given points by this method, which third row is used?
What should you do with a negative coordinate in the area determinant?
Practice Problems
- Find the triangle areas for these three vertex sets: (1,0),(6,0),(4,3); (2,7),(1,1),(10,8); and (−2,−3),(3,2),(−1,−8). Check the first using a horizontal base.
- Prove by expanding a determinant that (a,b+c),(b,c+a),(c,a+b) are collinear. Identify the terms that cancel.
- Find every real k when the area is 4 square units for each set: (k,0),(4,0),(0,2); and (−2,0),(0,4),(0,k). Explain the two sign branches.
- Use determinants to obtain the line through (1,2) and (3,6), then the line through (3,1) and (9,3). Verify both given points in each resulting equation.
- The vertices (2,−6),(5,4),(k,4) form a triangle of area 35 square units. Find all possible k, then check the result using its horizontal base.
- Use the determinant method to describe the line through (2,−1) and (2,5). Explain why the method remains valid although its slope is not a finite real number.
- Calculate D for (0,0),(4,0),(1,3), and then for (0,0),(1,3),(4,0). Compare the signed values and the geometric areas.
Key Takeaways
• Arrange one complete point per row, with a final column of ones. • Triangle area is half the magnitude of the completed coordinate determinant. • A zero coordinate determinant means the three points are collinear. • A variable third point gives a determinant equation for the line through two distinct fixed points. • A prescribed nonzero area usually requires both D = 2T and D = −2T. • Geometric checks help confirm signs, units, and the factor one-half.