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Chapter 4 · Determinants

Exercise 4.2 · Q1: Areas of three triangles

Exercise 4.2Area of a triangle · Section 4.3
The question

Find area of the triangle with vertices at the point given in each of the following:

(i) (1,0)(1, 0)⁠, (6,0)(6, 0)⁠, (4,3)(4, 3)
(ii) (2,7)(2, 7)⁠, (1,1)(1, 1)⁠, (10,8)(10, 8)
(iii) (−2,−3)(-2, -3)⁠, (3,2)(3, 2)⁠, (−1,−8)(-1, -8)

Exam answer

(i) Area =∣Δ∣= |\Delta|⁠, where

Δ=12∣101601431∣\Delta = \frac12\begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}

Expanding along C2C_2⁠:

Δ=12[−0+0−3∣1161∣]=12[−3(1−6)]=152\begin{aligned} \Delta &= \frac12\big[-0 + 0 - 3\begin{vmatrix} 1 & 1 \\ 6 & 1 \end{vmatrix}\big] \\ &= \frac12\big[-3(1 - 6)\big] \\ &= \frac{15}{2} \end{aligned}

∴\therefore Area =152= \frac{15}{2} sq. units.

(ii) Area =∣Δ∣= |\Delta|⁠, where

Δ=12∣2711111081∣\Delta = \frac12\begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix}

Expanding along R1R_1⁠:

Δ=12[2(1−8)−7(1−10)+1(8−10)]=12(−14+63−2)=472\begin{aligned} \Delta &= \frac12\big[2(1 - 8) - 7(1 - 10) \\ &\qquad + 1(8 - 10)\big] \\ &= \frac12(-14 + 63 - 2) \\ &= \frac{47}{2} \end{aligned}

∴\therefore Area =472= \frac{47}{2} sq. units.

(iii) Area =∣Δ∣= |\Delta|⁠, where

Δ=12∣−2−31321−1−81∣\Delta = \frac12\begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix}

Expanding along R1R_1⁠:

Δ=12[−2(2+8)+3(3+1)+1(−24+2)]=12(−20+12−22)=−15\begin{aligned} \Delta &= \frac12\big[-2(2 + 8) + 3(3 + 1) \\ &\qquad + 1(-24 + 2)\big] \\ &= \frac12(-20 + 12 - 22) \\ &= -15 \end{aligned}

∴\therefore Area =∣−15∣=15= |-15| = 15 sq. units.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The area of a triangle is half a 3×33\times3 determinant. Write each vertex as a row x, y, 1x,\ y,\ 1⁠, evaluate the determinant, halve it, and take the absolute value, because the determinant can be negative (its sign only records the order in which the vertices are listed). Each part is then one careful 3×33\times3 expansion.

Given, to find, and the plan
Given
Three triangles, each by the coordinates of its vertices.
To find
The area of each triangle, in square units.
Plan
For each part: (1) put the vertices into the determinant of Section 4.3; (2) choose the row or column to expand along (zeros first); (3) evaluate every 2×22\times2 minor and add the signed terms; (4) halve and take the absolute value; (5) check on a grid with base and height or the box method.

Basics you need

1Area as a determinant

For vertices (x1,y1)(x_1,y_1)⁠, (x2,y2)(x_2,y_2)⁠, (x3,y3)(x_3,y_3)⁠, Section 4.3, equation (1), gives

Δ=12∣x1y11x2y21x3y31∣,\Delta = \frac12\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix},

and the area is ∣Δ∣|\Delta| (Remark (i): area is positive, so take the absolute value). Basics: determinant as area

2Expanding a 3×33\times3 determinant

Pick any row or column. For each entry aija_{ij} in it, multiply the entry, the sign (−1)i+j(-1)^{i+j}⁠, and the 2×22\times2 determinant left when row ii and column jj are deleted (the minor MijM_{ij}⁠); then add. Every row and column gives the same value, so choose one with zeros (Section 4.2.3, Remark (i)). Lesson: expanding a 3 × 3 determinant

3The sign pattern

(−1)i+j(-1)^{i+j} is ++ when i+ji+j is even and −- when it is odd, which gives the chessboard

+−+−+−+−+\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}

Basics: the + − + sign pattern

4The 2×22\times2 rule
∣pqrs∣=ps−qr:\begin{vmatrix} p & q \\ r & s \end{vmatrix} = ps - qr:

main-diagonal product minus other-diagonal product (Section 4.2.2). Basics: where the determinant comes from

Part (i): vertices (1, 0), (6, 0), (4, 3)

  1. (i) Put the vertices into the determinant

    Δ=12∣101601431∣\Delta = \frac12\begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}
    Row 1 is the vertex (1,0)(1, 0)⁠, row 2 is (6,0)(6, 0)⁠, row 3 is (4,3)(4, 3)⁠; each row ends in 1. Listing the vertices in another order can only change the sign of the determinant, and the absolute value at the end removes that.
    Section 4.3, equation (1). Basics: determinant as area

  2. (i) Choose the second column: two zeros

    Column 2 holds 0,0,30, 0, 3⁠. Its signs, read down the middle column of the chessboard, are −,+,−-, +, -⁠. So

    ∣101601431∣=−0⋅M12+0⋅M22−3⋅M32=−3 M32.\begin{aligned} \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix} &= -0\cdot M_{12} + 0\cdot M_{22}\\ &\quad - 3\cdot M_{32}\\ &= -3\,M_{32}. \end{aligned}
    An entry 0 makes its whole term 0, whatever its minor is. Expanding along C2C_2 therefore needs only one 2×22\times2 determinant instead of three. The entry 3 sits in row 3, column 2, and 3+2=53+2 = 5 is odd, so its sign is −-⁠.
    Expand along the line with the most zeros (Section 4.2.3, Remark (i)). Basics: the + − + sign pattern

  3. (i) The one minor we need

    Delete row 3 and column 2:

    M32=∣1161∣=1⋅1−1⋅6=−5.M_{32} = \begin{vmatrix} 1 & 1 \\ 6 & 1 \end{vmatrix} = 1\cdot1 - 1\cdot6 = -5.

    Hence the determinant is −3×(−5)=15-3\times(-5) = 15⁠.

    After deleting row 3 and column 2, what is left is rows 1 and 2 with columns 1 and 3, in their original order: 1,11, 1 over 6,16, 1⁠. The pattern sign −- and the negative minor combine to a positive term.
    2×22\times2 rule (Basic 4). Basics: where the determinant comes from

  4. (i) Halve and take the absolute value

    Δ=12×15=152,Area=∣152∣=152.\begin{aligned} \Delta &= \tfrac12\times15 = \tfrac{15}{2},\\ \text{Area} &= \left|\tfrac{15}{2}\right| = \tfrac{15}{2}. \end{aligned}

    The area is 152=7.5\tfrac{15}{2} = 7.5 square units.

    y x −1 0 1 2 3 4 5 6 7 −1 0 1 2 3 4 height 3 base 5 (1, 0) (6, 0) (4, 3) Δ = ½ × 15 = 15/2 Area = |Δ| = 7.5 Check: ½ × base × height = ½ × 5 × 3 = 7.5
    Part (i) on a square grid (equal scales): base 5 along the xx-axis, height 3, area 12⋅5⋅3=7.5\tfrac12\cdot5\cdot3 = 7.5⁠.
    Two vertices lie on the xx-axis, so the triangle has a base of length 6−1=56 - 1 = 5 along the axis, and the third vertex is 3 units above it. Half of base times height is 12⋅5⋅3=7.5\tfrac12\cdot5\cdot3 = 7.5⁠, the same answer. This is also why column 2 had two zeros: the zeros are the yy-coordinates of the two vertices on the axis.
    Area =∣Δ∣= |\Delta| (Section 4.3, Remark (i)).

Part (ii): vertices (2, 7), (1, 1), (10, 8)

  1. (ii) Put the vertices into the determinant

    Δ=12∣2711111081∣\Delta = \frac12\begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix}
    Same recipe: one row per vertex, in the order given, with a 1 at the end of each row.

  2. (ii) No zeros: expand along the first row

    No row or column contains a 0, so every choice needs three minors. We use R1R_1⁠, with entries 2,7,12, 7, 1 and signs +,−,++, -, +⁠:

    ∣2711111081∣=2M11−7M12+1⋅M13.\begin{aligned} \begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix} &= 2M_{11} - 7M_{12}\\ &\quad + 1\cdot M_{13}. \end{aligned}
    When there are no zeros, no line saves work, and the first row is the expansion the book writes out in Section 4.2.3.
    Signs along R1R_1⁠: (−1)1+1,(−1)1+2,(−1)1+3(-1)^{1+1}, (-1)^{1+2}, (-1)^{1+3}⁠, that is +,−,++, -, +⁠. Basics: the + − + sign pattern

  3. (ii) The three minors

    Delete row 1 and, in turn, column 1, 2 and 3:

    M11=∣1181∣=1⋅1−1⋅8=−7,M12=∣11101∣=1⋅1−1⋅10=−9,M13=∣11108∣=1⋅8−1⋅10=−2.\begin{aligned} M_{11} &= \begin{vmatrix} 1 & 1 \\ 8 & 1 \end{vmatrix} = 1\cdot1 - 1\cdot8 = -7,\\ M_{12} &= \begin{vmatrix} 1 & 1 \\ 10 & 1 \end{vmatrix} = 1\cdot1 - 1\cdot10 = -9,\\ M_{13} &= \begin{vmatrix} 1 & 1 \\ 10 & 8 \end{vmatrix} = 1\cdot8 - 1\cdot10 = -2. \end{aligned}
    Each minor uses rows 2 and 3, (1,1,1)(1, 1, 1) and (10,8,1)(10, 8, 1)⁠, with one column removed. Keep the remaining entries in their original positions; swapping two of them would change the sign of the minor.
    2×22\times2 rule (Basic 4).

  4. (ii) Combine, halve, take the absolute value

    det=2(−7)−7(−9)+1(−2)=−14+63−2=47,\begin{aligned} \text{det} &= 2(-7) - 7(-9) + 1(-2)\\ &= -14 + 63 - 2 = 47, \end{aligned} Δ=12×47=472,Area=472.\begin{aligned} \Delta &= \tfrac12\times47 = \tfrac{47}{2},\\ \text{Area} &= \tfrac{47}{2}. \end{aligned}

    The area is 472=23.5\tfrac{47}{2} = 23.5 square units.

    y x 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 (2, 7) (1, 1) (10, 8) Δ = ½ × 47 = 47/2 Area = |Δ| = 23.5 Check (dashed box): 63 − 3 − 4 − 31.5 − 1 = 23.5
    Part (ii) on a square grid (equal scales). The dashed box has area 63; the pieces outside the triangle have total area 39.5, leaving 23.5.
    The middle term is where signs go wrong: the pattern sign is −-⁠, the entry is 7 and the minor is −9-9⁠, so the term is −7×(−9)=+63-7\times(-9) = +63⁠. The determinant is positive, so halving it already gives the area.
    Minus times minus is plus. Basics: algebra the chapter leans on

Part (iii): vertices (−2, −3), (3, 2), (−1, −8)

  1. (iii) Put the vertices into the determinant

    Δ=12∣−2−31321−1−81∣\Delta = \frac12\begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix}
    Negative coordinates go in with their signs; the third column is still all 1s.

  2. (iii) No zeros: expand along the first row

    Entries −2,−3,1-2, -3, 1⁠, signs +,−,++, -, +⁠:

    ∣−2−31321−1−81∣=(−2)M11−(−3)M12+1⋅M13.\begin{aligned} \begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix} &= (-2)M_{11} - (-3)M_{12}\\ &\quad + 1\cdot M_{13}. \end{aligned}
    Write the pattern sign and the entry's own sign separately, as here, and combine them only at the end. That avoids the most common slip, losing one of the two minus signs in −(−3)-(-3)⁠.
    Basics: the + − + sign pattern

  3. (iii) The three minors

    M11=∣21−81∣=2⋅1−1⋅(−8)=10,M12=∣31−11∣=3⋅1−1⋅(−1)=4,M13=∣32−1−8∣=3⋅(−8)−2⋅(−1)=−22.\begin{aligned} M_{11} &= \begin{vmatrix} 2 & 1 \\ -8 & 1 \end{vmatrix}\\ &= 2\cdot1 - 1\cdot(-8) = 10,\\ M_{12} &= \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix}\\ &= 3\cdot1 - 1\cdot(-1) = 4,\\ M_{13} &= \begin{vmatrix} 3 & 2 \\ -1 & -8 \end{vmatrix}\\ &= 3\cdot(-8) - 2\cdot(-1) = -22. \end{aligned}
    In each minor the second product contains a negative number, so subtracting it adds: 2−(−8)=102 - (-8) = 10⁠, 3−(−1)=43 - (-1) = 4⁠, −24−(−2)=−22-24 - (-2) = -22⁠.
    Subtracting a negative number adds its size. Basics: algebra the chapter leans on

  4. (iii) Combine, halve, take the absolute value

    det=(−2)(10)−(−3)(4)+1(−22)=−20+12−22=−30,\begin{aligned} \text{det} &= (-2)(10) - (-3)(4) + 1(-22)\\ &= -20 + 12 - 22 = -30, \end{aligned} Δ=12×(−30)=−15,Area=∣−15∣=15.\begin{aligned} \Delta &= \tfrac12\times(-30) = -15,\\ \text{Area} &= |-15| = 15. \end{aligned}

    The area is 15 square units.

    y x −5 −4 −3 −2 −1 0 1 2 3 4 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 (−2, −3) (3, 2) (−1, −8) Δ = ½ × (−30) = −15 Area = |Δ| = 15 Check (dashed box): 50 − 12.5 − 20 − 2.5 = 15
    Part (iii) on a square grid (equal scales). The dashed box has area 50; the three corner triangles have total area 35, leaving 15.
    Here Δ\Delta is negative. That does not mean anything has gone wrong: the vertices, in the order given, go round clockwise (from (−2,−3)(-2,-3) up to (3,2)(3,2)⁠, then down to (−1,−8)(-1,-8)⁠), and the determinant is negative for a clockwise order. Remark (i) of Section 4.3 tells us to take the absolute value, so the area is 15, not −15-15⁠.
    Area =∣Δ∣= |\Delta|⁠. Basics: determinant as area
Result

(i) 152\tfrac{15}{2} square units; (ii) 472\tfrac{47}{2} square units; (iii) 15 square units.

(Textbook answer: (i) 152\tfrac{15}{2}⁠, (ii) 472\tfrac{47}{2}⁠, (iii) 15.)

Check it with numbers

(i) Base and height. Base from (1,0)(1,0) to (6,0)(6,0) is 5, height of (4,3)(4,3) above the axis is 3: 12⋅5⋅3=7.5=152\tfrac12\cdot5\cdot3 = 7.5 = \tfrac{15}{2}⁠. ✓

(ii) Box method. The dashed box runs from x=1x = 1 to 10 and y=1y = 1 to 8, area 9×7=639\times7 = 63⁠. Outside the triangle but inside the box are three right triangles and one small square:

PieceSizeArea
between (1,1)(1,1) and (2,7)(2,7)legs 1 and 63
between (2,7)(2,7) and (10,8)(10,8)legs 8 and 14
between (1,1)(1,1) and (10,8)(10,8)legs 9 and 731.5
square at corner (1,8)(1, 8)1×11\times11

63−3−4−31.5−1=23.5=47263 - 3 - 4 - 31.5 - 1 = 23.5 = \tfrac{47}{2}⁠. ✓

(iii) Box method. Box from x=−2x = -2 to 3 and y=−8y = -8 to 2, area 5×10=505\times10 = 50⁠. The three corner triangles have legs 5 and 5 (area 12.5), 4 and 10 (area 20), 1 and 5 (area 2.5): 50−12.5−20−2.5=1550 - 12.5 - 20 - 2.5 = 15⁠. ✓

Bracket formula, part (iii). The formula from earlier classes, quoted at the start of Section 4.3, gives the same Δ\Delta⁠:

12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=12[(−2)(10)+3(−5)+(−1)(−5)]=12(−20−15+5)=−15.\begin{aligned} &\tfrac12\big[x_1(y_2-y_3) + x_2(y_3-y_1)\\ &\qquad + x_3(y_1-y_2)\big]\\ &= \tfrac12\big[(-2)(10) + 3(-5)\\ &\qquad + (-1)(-5)\big]\\ &= \tfrac12(-20 - 15 + 5) = -15. \end{aligned}

This is the value found in Part (iii), Step 4. ✓

Common mistakes
  1. Forgetting the 12\tfrac12⁠. The determinant in (ii) is 47; the area is half of it.
  2. Reporting a negative area. In (iii) Δ=−15\Delta = -15⁠; the area is ∣Δ∣=15|\Delta| = 15⁠.
  3. Losing a minus sign. In (iii) the middle term is −(−3)(4)=+12-(-3)(4) = +12⁠, and M11=2−(−8)=10M_{11} = 2 - (-8) = 10⁠, not −6-6⁠.
  4. Using the wrong sign for a single term. In (i) the entry 3 is at position (3, 2), where the pattern sign is −-⁠, not ++⁠.
What to take away
  1. Area =∣Δ∣= |\Delta| with Δ=12\Delta = \tfrac12 of the determinant whose rows are xi, yi, 1x_i,\ y_i,\ 1⁠.
  2. Look for zeros before expanding: a vertex on an axis gives a 0 in the determinant and saves work.
  3. A sketch on a grid, with base and height or the box method, is a quick independent check.

Every fact used, and where it comes from

FactUsed inSource
Δ=12\Delta = \tfrac12 of the determinant with rows xi,yi,1x_i, y_i, 1Step 1 of each partSection 4.3, equation (1) (Basics: determinant as area)
Area is the absolute valueLast step of each partSection 4.3, Remark (i)
Expansion along any row or column gives the same value; choose zerosPart (i), Step 2; Parts (ii), (iii), Step 2Section 4.2.3 and its Remark (i) (Lesson: expanding a 3 × 3 determinant)
Sign (−1)i+j(-1)^{i+j}Step 2 of each partSection 4.2.3, Remark (ii)
2×22\times2 rulePart (i), Step 3; Parts (ii), (iii), Step 3Section 4.2.2
Area of a triangle =12×= \tfrac12\times base ×\times heightPart (i), Step 4; CheckMensuration (earlier classes)
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