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Chapter 4 · Determinants

Exercise 4.2 · Q3: Find kk for a triangle of area 4

Exercise 4.2Area of a triangle · Section 4.3
The question

Find values of kk if area of triangle is 4 sq. units and vertices are

(i) (k,0)(k, 0)⁠, (4,0)(4, 0)⁠, (0,2)(0, 2)
(ii) (−2,0)(-2, 0)⁠, (0,4)(0, 4)⁠, (0,k)(0, k)

Exam answer

(i) Area =4= 4 sq. units, so

12∣k01401021∣=±4\frac12\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4

Expanding along C2C_2⁠:

12[−0+0−2(k−4)]=±4⇒ 4−k=±4\begin{aligned} &\frac12\big[-0 + 0 - 2(k - 4)\big] = \pm 4 \\ &\Rightarrow\ 4 - k = \pm 4 \end{aligned}

∴ k=0\therefore\ k = 0 or k=8k = 8⁠.

(ii) Area =4= 4 sq. units, so

12∣−2010410k1∣=±4\frac12\begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} = \pm 4

Expanding along C1C_1⁠:

12[−2(4−k)−0+0]=±4⇒ k−4=±4\begin{aligned} &\frac12\big[-2(4 - k) - 0 + 0\big] = \pm 4 \\ &\Rightarrow\ k - 4 = \pm 4 \end{aligned}

∴ k=0\therefore\ k = 0 or k=8k = 8⁠.

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

The one idea this problem is built on

The area is the absolute value of Δ\Delta⁠, half the area determinant. So "area =4= 4" does not say Δ=4\Delta = 4⁠; it says Δ=4\Delta = 4 or Δ=−4\Delta = -4⁠. Each sign gives one equation in kk⁠, and both values of kk are answers (Section 4.3, Remark (ii)).

Given, to find, and the plan
Given
Two triangles, each with one coordinate unknown (kk⁠), and area 4 square units.
To find
All values of kk in each part.
Plan
For each part: (1) write Δ\Delta as half a determinant; (2) expand along the line with two zeros, so Δ\Delta becomes a simple expression in kk⁠; (3) solve Δ=4\Delta = 4 and Δ=−4\Delta = -4⁠; (4) check each kk by drawing the triangle and using base and height.

Basics you need

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

area =∣Δ∣= |\Delta| (Section 4.3, equation (1) and Remark (i)). Basics: determinant as area

2When the area is given, use both signs

∣Δ∣=4|\Delta| = 4 means Δ=4\Delta = 4 or Δ=−4\Delta = -4⁠, because the numbers whose absolute value is 4 are exactly 4 and −4-4 (Section 4.3, Remark (ii)). Basics: algebra the chapter leans on

3Expansion and the sign pattern

Expand along the row or column with the most zeros; position (i,j)(i, j) has sign (−1)i+j(-1)^{i+j}⁠:

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

Basics: the + − + sign pattern

Part (i): vertices (k, 0), (4, 0), (0, 2)

  1. (i) Write the area determinant

    Δ=12∣k01401021∣\Delta = \frac12\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix}
    One row per vertex, in the order given: (k,0)(k, 0)⁠, (4,0)(4, 0)⁠, (0,2)(0, 2)⁠, each followed by 1. The unknown kk is just an entry like any other.
    Section 4.3, equation (1). Basics: determinant as area

  2. (i) Expand along the second column

    Column 2 is 0,0,20, 0, 2⁠, with signs −,+,−-, +, -⁠. Only the entry 2, at position (3, 2), survives, and its sign is (−1)3+2=−1(-1)^{3+2} = -1⁠:

    M32=∣k141∣=k⋅1−1⋅4=k−4,M_{32} = \begin{vmatrix} k & 1 \\ 4 & 1 \end{vmatrix} = k\cdot1 - 1\cdot4 = k - 4, det=−2 M32=−2(k−4)=8−2k.\begin{aligned} \text{det} &= -2\,M_{32} = -2(k - 4)\\ &= 8 - 2k. \end{aligned}

    So Δ=12(8−2k)=4−k\Delta = \tfrac12(8 - 2k) = 4 - k⁠.

    Two zeros in column 2 kill two of the three terms, leaving one 2×22\times2 determinant. M32M_{32} is what remains after deleting row 3 and column 2: rows 1 and 2, columns 1 and 3.
    Expand along the line with the most zeros (Section 4.2.3, Remark (i)). Lesson: expanding a 3 × 3 determinant

  3. (i) Use both signs and solve

    Area 4 means ∣4−k∣=4|4 - k| = 4⁠, that is,

    4−k=4⇒ k=0,4−k=−4⇒ k=8.\begin{aligned} 4 - k &= 4 &&\Rightarrow\ k = 0,\\ 4 - k &= -4 &&\Rightarrow\ k = 8. \end{aligned}
    If we solved only Δ=4\Delta = 4⁠, we would find k=0k = 0 and miss k=8k = 8⁠. The sign of Δ\Delta depends on whether the vertices, in the listed order, go round anticlockwise or clockwise, and moving (k,0)(k, 0) from one side of (4,0)(4, 0) to the other reverses the direction. Both triangles are genuine answers.
    ∣u∣=4  ⟺  u=4|u| = 4 \iff u = 4 or u=−4u = -4 (Section 4.3, Remark (ii)). Basics: algebra the chapter leans on

  4. (i) Check each value on the grid

    k=0k = 0⁠: vertices (0,0)(0, 0)⁠, (4,0)(4, 0)⁠, (0,2)(0, 2)⁠, a right triangle with legs 4 and 2, area 12⋅4⋅2=4\tfrac12\cdot4\cdot2 = 4⁠. ✓ Also Δ=4−0=4\Delta = 4 - 0 = 4⁠. ✓

    k=8k = 8⁠: vertices (8,0)(8, 0)⁠, (4,0)(4, 0)⁠, (0,2)(0, 2)⁠. The base from (4,0)(4, 0) to (8,0)(8, 0) has length 4 and the height of (0,2)(0, 2) above the xx-axis is 2, so the area is 12⋅4⋅2=4\tfrac12\cdot4\cdot2 = 4⁠. ✓ Also Δ=4−8=−4\Delta = 4 - 8 = -4⁠. ✓

    y x −2 −1 0 1 2 3 4 5 6 7 8 9 −1 0 1 2 3 (0, 0) (4, 0) (8, 0) (0, 2) k = 0: blue, area ½ × 4 × 2 = 4 k = 8: gold, area ½ × 4 × 2 = 4
    Part (i) on a square grid (equal scales). For both values, the side on the xx-axis has length ∣k−4∣=4|k - 4| = 4 and the height is 2.
    The figure shows why there are exactly two answers: the vertex (k,0)(k, 0) slides along the xx-axis, the height stays 2, and the area is 4 exactly when the base ∣k−4∣|k - 4| is 4, which happens 4 units to the left or right of (4,0)(4, 0)⁠.

Part (ii): vertices (−2, 0), (0, 4), (0, k)

  1. (ii) Write the area determinant

    Δ=12∣−2010410k1∣\Delta = \frac12\begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix}
    Same recipe. This time two vertices have x=0x = 0⁠, so the zeros fall in column 1.

  2. (ii) Expand along the first column

    Column 1 is −2,0,0-2, 0, 0⁠, with signs +,−,++, -, +⁠. Only −2-2⁠, at position (1, 1), survives, with sign (−1)1+1=+1(-1)^{1+1} = +1⁠:

    M11=∣41k1∣=4⋅1−1⋅k=4−k,M_{11} = \begin{vmatrix} 4 & 1 \\ k & 1 \end{vmatrix} = 4\cdot1 - 1\cdot k = 4 - k, det=(−2) M11=−2(4−k)=2k−8.\begin{aligned} \text{det} &= (-2)\,M_{11} = -2(4 - k)\\ &= 2k - 8. \end{aligned}

    So Δ=12(2k−8)=k−4\Delta = \tfrac12(2k - 8) = k - 4⁠.

    M11M_{11} is what is left after deleting row 1 and column 1. The pattern sign is ++⁠; the minus in front of the 2 is the entry's own sign.
    Basics: the + − + sign pattern

  3. (ii) Use both signs and solve

    Area 4 means ∣k−4∣=4|k - 4| = 4⁠:

    k−4=4⇒ k=8,k−4=−4⇒ k=0.\begin{aligned} k - 4 &= 4 &&\Rightarrow\ k = 8,\\ k - 4 &= -4 &&\Rightarrow\ k = 0. \end{aligned}
    As in part (i), each sign of Δ\Delta gives one value, and both are needed.
    Section 4.3, Remark (ii).

  4. (ii) Check each value on the grid

    k=0k = 0⁠: vertices (−2,0)(-2, 0)⁠, (0,4)(0, 4)⁠, (0,0)(0, 0)⁠. The side from (0,0)(0, 0) to (0,4)(0, 4) lies on the yy-axis and has length 4; the vertex (−2,0)(-2, 0) is 2 units from that axis, so the area is 12⋅4⋅2=4\tfrac12\cdot4\cdot2 = 4⁠. ✓

    k=8k = 8⁠: vertices (−2,0)(-2, 0)⁠, (0,4)(0, 4)⁠, (0,8)(0, 8)⁠. The side from (0,4)(0, 4) to (0,8)(0, 8) has length 4, and the height is again 2, so the area is 4. ✓

    y x −4 −3 −2 −1 0 1 2 −1 0 1 2 3 4 5 6 7 8 9 (−2, 0) (0, 4) (0, 0) (0, 8) k = 0 (blue): base 4 on the y-axis, height 2, area 4 k = 8 (gold): base 4 (from 4 to 8), height 2, area 4
    Part (ii) on a square grid (equal scales). The side on the yy-axis has length ∣k−4∣=4|k - 4| = 4⁠; the distance of (−2,0)(-2, 0) from the yy-axis is 2.
    Here the base is vertical: the vertex (0,k)(0, k) slides along the yy-axis, and the area is 4 exactly when it is 4 units above or below (0,4)(0, 4)⁠.
Result

(i) k=0k = 0 or k=8k = 8⁠; (ii) k=0k = 0 or k=8k = 8⁠.

(Textbook answer: (i) 0, 8; (ii) 0, 8.)

Check it with numbers

Substitute back into the full determinants (expanded along R1R_1⁠, not the column used above):

Part (i), k=8k = 8⁠: 8(0−2)−0+1(8−0)=−16+8=−88(0 - 2) - 0 + 1(8 - 0) = -16 + 8 = -8⁠, so Δ=−4\Delta = -4 and the area is 4. ✓

Part (ii), k=8k = 8⁠: −2(4−8)−0+1(0−0)=8-2(4 - 8) - 0 + 1(0 - 0) = 8⁠, so Δ=4\Delta = 4 and the area is 4. ✓

The values k=0k = 0 were checked in Step 4 of each part, and the verification script repeats all four with exact arithmetic.

Common mistakes
  1. Solving only Δ=4\Delta = 4⁠. That finds one value of kk in each part and misses the other.
  2. Forgetting the 12\tfrac12⁠. Setting the determinant itself equal to ±4\pm4 gives k=2,6k = 2, 6 in part (i), which are wrong (those triangles have area 2).
  3. Wrong sign on the single surviving term. In part (i) the entry 2 is at position (3, 2), where the sign is −-⁠.
What to take away
  1. Given an area SS⁠, solve Δ=S\Delta = S and Δ=−S\Delta = -S⁠.
  2. A vertex on an axis puts a zero in the determinant; two such zeros reduce the work to one minor.
  3. When the unknown vertex slides along a line that contains one side (here an axis), the height to that side stays fixed and only the base changes, which is why there are two answers, one on each side.

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, 1⁠; area =∣Δ∣= |\Delta|Step 1 of each partSection 4.3, equation (1), Remark (i) (Basics: determinant as area)
If the area is given, use both signsStep 3 of each partSection 4.3, Remark (ii); Example 7
Expansion along a column; signs (−1)i+j(-1)^{i+j}Step 2 of each partSection 4.2.3, Remarks (i), (ii) (Lesson: expanding a 3 × 3 determinant)
2×22\times2 ruleStep 2 of each partSection 4.2.2
Area =12×= \tfrac12\times base ×\times heightStep 4 of each partMensuration (earlier classes)
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