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Chapter 4 · Determinants

Exercise 4.2 · Q5 (MCQ): Find kk for a triangle of area 35

Exercise 4.2Area of a triangle · Section 4.3
The question

If area of triangle is 35 sq units with vertices (2,−6)(2, -6)⁠, (5,4)(5, 4) and (k,4)(k, 4)⁠. Then kk is

(A) 12
(B) −2-2
(C) −12,−2-12, -2
(D) 12,−212, -2

Exam answer

Area =35= 35 sq. units, so

12∣2−61541k41∣=±35\frac12\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 35

Expanding along R1R_1⁠:

12[2(4−4)+6(5−k)+1(20−4k)]=±35⇒ 25−5k=±35\begin{aligned} &\frac12\big[2(4 - 4) + 6(5 - k) \\ &\qquad + 1(20 - 4k)\big] = \pm 35 \\ &\Rightarrow\ 25 - 5k = \pm 35 \end{aligned}

So k=−2k = -2 or k=12k = 12⁠.

Answer: (D) 12,−212, -2

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

The one idea this problem is built on

Area is the absolute value of Δ\Delta⁠, so area 35 means Δ=35\Delta = 35 or Δ=−35\Delta = -35⁠. Each sign gives one value of kk⁠, and the correct option must list both (Section 4.3, Remark (ii)).

Given, to find, and the plan
Given
Vertices (2,−6)(2, -6)⁠, (5,4)(5, 4)⁠, (k,4)(k, 4)⁠; area 35 square units.
To find
All values of kk⁠, and the option that lists exactly those.
Plan
(1) Write Δ\Delta as half a determinant. (2) Expand it to an expression in kk⁠. (3) Solve Δ=±35\Delta = \pm35⁠. (4) Check with base and height. (5) Examine every option.

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

2Both signs when the area is given

∣Δ∣=35|\Delta| = 35 means Δ=35\Delta = 35 or Δ=−35\Delta = -35 (Section 4.3, Remark (ii)). Basics: algebra the chapter leans on

3Expansion along a row

Entry ×\times sign (−1)i+j(-1)^{i+j} ×\times minor, added; signs along R1R_1 are +,−,++, -, +⁠. Lesson: expanding a 3 × 3 determinant

Solution

  1. Write the area determinant

    Δ=12∣2−61541k41∣\Delta = \frac12\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix}
    One row per vertex, xx⁠, yy⁠, 1, in the order given. The unknown kk is the xx-coordinate of the third vertex.
    Section 4.3, equation (1). Basics: determinant as area

  2. Expand along the first row

    There are no zeros, so we follow the book's usual choice, R1R_1⁠, with entries 2,−6,12, -6, 1 and signs +,−,++, -, +⁠. The minors:

    M11=∣4141∣=4⋅1−1⋅4=0,M12=∣51k1∣=5⋅1−1⋅k=5−k,M13=∣54k4∣=5⋅4−4k=20−4k.\begin{aligned} M_{11} &= \begin{vmatrix} 4 & 1 \\ 4 & 1 \end{vmatrix} = 4\cdot1 - 1\cdot4 = 0,\\ M_{12} &= \begin{vmatrix} 5 & 1 \\ k & 1 \end{vmatrix} = 5\cdot1 - 1\cdot k = 5 - k,\\ M_{13} &= \begin{vmatrix} 5 & 4 \\ k & 4 \end{vmatrix} = 5\cdot4 - 4k = 20 - 4k. \end{aligned}

    So the determinant is

    2⋅0−(−6)(5−k)+1⋅(20−4k)=30−6k+20−4k=50−10k,\begin{aligned} &2\cdot0 - (-6)(5 - k) + 1\cdot(20 - 4k)\\ &= 30 - 6k + 20 - 4k\\ &= 50 - 10k, \end{aligned}

    and Δ=12(50−10k)=25−5k\Delta = \tfrac12(50 - 10k) = 25 - 5k⁠.

    M11=0M_{11} = 0 because rows 2 and 3 both have y=4y = 4⁠: the two vertices (5,4)(5, 4) and (k,4)(k, 4) lie on the same horizontal line. The middle term has pattern sign −- and entry −6-6⁠, so it becomes +6(5−k)+6(5 - k)⁠.
    Signs (−1)1+j(-1)^{1+j}⁠: +,−,++, -, +⁠. Basics: the + − + sign pattern

  3. Use both signs and solve

    Area 35 means ∣25−5k∣=35|25 - 5k| = 35⁠:

    25−5k=35⇒5k=−10⇒k=−2,25−5k=−35⇒5k=60⇒k=12.\begin{aligned} 25 - 5k = 35 &\Rightarrow 5k = -10\\ &\Rightarrow k = -2,\\ 25 - 5k = -35 &\Rightarrow 5k = 60\\ &\Rightarrow k = 12. \end{aligned}

    So k=12k = 12 or k=−2k = -2⁠.

    Each equation is linear, so each gives exactly one value; together they give every kk with ∣Δ∣=35|\Delta| = 35⁠. Leaving out either sign would lose a correct triangle.
    Section 4.3, Remark (ii).

  4. Check with base and height

    The vertices (5,4)(5, 4) and (k,4)(k, 4) lie on the line y=4y = 4⁠, so take that side as the base: its length is ∣k−5∣|k - 5|⁠. The third vertex (2,−6)(2, -6) is 4−(−6)=104 - (-6) = 10 units below that line, so the height is 10. Then

    Area=12⋅∣k−5∣⋅10=5 ∣k−5∣.\text{Area} = \tfrac12\cdot|k - 5|\cdot10 = 5\,|k - 5|.

    Area 35 gives ∣k−5∣=7|k - 5| = 7⁠, so k=12k = 12 or k=−2k = -2⁠, the same two values. ✓

    y x −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 height 10 (2, −6) (5, 4) (12, 4) (−2, 4) k = 12 (blue): base 7, height 10, area 35 k = −2 (gold): base 7, height 10, area 35
    Both triangles on a square grid (equal scales). The third vertex lies on y=4y = 4⁠, 7 units to the right or left of (5,4)(5, 4)⁠; the height from (2,−6)(2, -6) is 10 in both cases, so each area is 12⋅7⋅10=35\tfrac12\cdot7\cdot10 = 35⁠.
    This independent check also explains the answer: the area depends only on how far (k,4)(k, 4) is from (5,4)(5, 4)⁠, and there are two points on y=4y = 4 at distance 7.

  5. Examine every option

    OptionClaimVerdict
    (A)k=12k = 12Incomplete. k=12k = 12 does give area 35 (Δ=25−60=−35\Delta = 25 - 60 = -35⁠), but k=−2k = -2 works too, so "kk is 12" is not the full answer.
    (B)k=−2k = -2Incomplete for the same reason: Δ=25+10=35\Delta = 25 + 10 = 35 works, but it misses k=12k = 12⁠.
    (C)k=−12,−2k = -12, -2Wrong. k=−2k = -2 is right, but k=−12k = -12 gives Δ=25+60=85\Delta = 25 + 60 = 85⁠, an area of 85, not 35.
    (D)k=12,−2k = 12, -2Correct. Exactly the two solutions of ∣25−5k∣=35|25 - 5k| = 35 found in Step 3.
    The question asks for the values of kk⁠, so the right option lists every solution and nothing else. (A) and (B) are what you get by using only one sign; (C) comes from a sign slip when solving 25−5k=−3525 - 5k = -35⁠.
Result

The correct answer is (D): k=12k = 12 or k=−2k = -2⁠. (Textbook answer: D.)

Check it with numbers

Substitute each value into the full determinant and expand along R1R_1 again, with the numbers in place:

k=12:  2(4−4)+6(5−12)  +(20−48)  =0−42−28=−70,k=−2:  2(4−4)+6(5+2)  +(20+8)  =0+42+28=70.\begin{aligned} &k = 12:\\ &\ \ 2(4 - 4) + 6(5 - 12)\\ &\ \ \quad + (20 - 48)\\ &\ \ = 0 - 42 - 28 = -70,\\ &k = -2:\\ &\ \ 2(4 - 4) + 6(5 + 2)\\ &\ \ \quad + (20 + 8)\\ &\ \ = 0 + 42 + 28 = 70. \end{aligned}

So Δ=−35\Delta = -35 and Δ=35\Delta = 35⁠, and both areas are 35. ✓

Common mistakes
  1. Solving only Δ=35\Delta = 35 and choosing (B).
  2. Losing the sign of −(−6)-(-6): the middle term is +6(5−k)+6(5 - k)⁠, not −6(5−k)-6(5 - k)⁠.
  3. Forgetting the 12\tfrac12: setting 50−10k=±3550 - 10k = \pm35 gives k=1.5k = 1.5 or 8.58.5⁠, which are not offered and are wrong.
What to take away
  1. Given an area, set Δ\Delta equal to plus and minus that area.
  2. Two vertices with the same yy give a horizontal base, which makes base ×\times height a quick check.
  3. In an MCQ, test each listed value: it rules out options like (C) at once.

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|Steps 1, 2Section 4.3, equation (1), Remark (i) (Basics: determinant as area)
Given area: use both signsSteps 3, 5Section 4.3, Remark (ii); Example 7
Expansion along R1R_1⁠; signs (−1)i+j(-1)^{i+j}Step 2Section 4.2.3 (Lesson: expanding a 3 × 3 determinant)
2×22\times2 ruleStep 2Section 4.2.2
Area =12×= \tfrac12\times base ×\times heightStep 4Mensuration (earlier classes)
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