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Chapter 4 · Determinants

Exercise 4.2 · Q4: Equations of lines using determinants

Exercise 4.2Area of a triangle and the equation of a line · Section 4.3
The question

(i) Find equation of line joining (1,2)(1, 2) and (3,6)(3, 6) using determinants.

(ii) Find equation of line joining (3,1)(3, 1) and (9,3)(9, 3) using determinants.

Exam answer

(i) Let P(x,y)P(x, y) be any point on the line through (1,2)(1, 2) and (3,6)(3, 6)⁠. The three points are collinear, so the area of the triangle they form is 0:

12∣121361xy1∣=0\frac12\begin{vmatrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{vmatrix} = 0

Expanding along R3R_3⁠:

12[x(2−6)−y(1−3)+1(6−6)]=0⇒ −4x+2y=0\begin{aligned} &\frac12\big[x(2 - 6) - y(1 - 3) \\ &\qquad + 1(6 - 6)\big] = 0 \\ &\Rightarrow\ -4x + 2y = 0 \end{aligned}

∴\therefore The equation of the line is y=2xy = 2x⁠.

(ii) Let P(x,y)P(x, y) be any point on the line through (3,1)(3, 1) and (9,3)(9, 3)⁠. As in (i),

12∣311931xy1∣=0\frac12\begin{vmatrix} 3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1 \end{vmatrix} = 0

Expanding along R3R_3⁠:

12[x(1−3)−y(3−9)+1(9−9)]=0⇒ −2x+6y=0\begin{aligned} &\frac12\big[x(1 - 3) - y(3 - 9) \\ &\qquad + 1(9 - 9)\big] = 0 \\ &\Rightarrow\ -2x + 6y = 0 \end{aligned}

∴\therefore The equation of the line is x−3y=0x - 3y = 0⁠.

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

The one idea this problem is built on

A point P(x,y)P(x, y) lies on the line through two given points exactly when the three points are collinear, that is, when the "triangle" they form has zero area. Writing that area as half a determinant and setting it equal to 0 gives an equation in xx and yy⁠: the equation of the line. This is the method of Example 7.

Given, to find, and the plan
Given
Two distinct points in each part: (1,2)(1, 2) and (3,6)(3, 6)⁠; (3,1)(3, 1) and (9,3)(9, 3)⁠.
To find
The equation of the line through each pair, using determinants.
Plan
For each part: (1) let P(x,y)P(x, y) be any point and write Δ=0\Delta = 0 for the triangle formed by the two given points and PP⁠; (2) expand along the row that contains xx and yy⁠; (3) simplify to the equation of the line; (4) check that both given points satisfy it.

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}

(Section 4.3, equation (1)); the area is ∣Δ∣|\Delta|⁠. Basics: determinant as area

2Zero area means collinear

Three collinear points give area 0 (Section 4.3, Remark (iii)). Conversely, three points that are not collinear form a genuine triangle with a positive base and a positive height, so a positive area, and by Section 4.3 that area is ∣Δ∣|\Delta|⁠; so Δ≠0\Delta \ne 0⁠. So for two distinct points QQ⁠, RR⁠: PP is on the line QRQR exactly when the area of QRPQRP is 0.

3Expansion along a row

Entry ×\times sign (−1)i+j(-1)^{i+j} ×\times minor, added along the row. Along R3R_3 the signs are +,−,++, -, +⁠. Lesson: expanding a 3 × 3 determinant

Part (i): the line through (1, 2) and (3, 6)

  1. (i) A point P(x, y) on the line gives zero area

    Let P(x,y)P(x, y) be any point. PP lies on the line through (1,2)(1, 2) and (3,6)(3, 6) exactly when

    12∣121361xy1∣=0.\frac12\begin{vmatrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{vmatrix} = 0.
    Basic 2 turns "on the line" into "zero area", and Basic 1 turns "zero area" into a determinant. The two given points are different, so they do fix one line. We do not need the absolute value here: ∣Δ∣=0|\Delta| = 0 is the same as Δ=0\Delta = 0⁠.
    Section 4.3, Remark (iii); Example 7 uses the same idea. Basics: determinant as area

  2. (i) Expand along the third row

    Row 3 is x,y,1x, y, 1⁠, with signs +,−,++, -, +⁠. Its minors:

    M31=∣2161∣=2⋅1−1⋅6=−4,M32=∣1131∣=1⋅1−1⋅3=−2,M33=∣1236∣=1⋅6−2⋅3=0.\begin{aligned} M_{31} &= \begin{vmatrix} 2 & 1 \\ 6 & 1 \end{vmatrix} = 2\cdot1 - 1\cdot6 = -4,\\ M_{32} &= \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} = 1\cdot1 - 1\cdot3 = -2,\\ M_{33} &= \begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix} = 1\cdot6 - 2\cdot3 = 0. \end{aligned}

    So the determinant is

    x M31−y M32+1⋅M33=x(−4)−y(−2)+0=−4x+2y.\begin{aligned} &x\,M_{31} - y\,M_{32} + 1\cdot M_{33}\\ &= x(-4) - y(-2) + 0\\ &= -4x + 2y. \end{aligned}
    The unknowns xx and yy sit in row 3, and the minors of row 3 contain only numbers. Expanding along this row gives the coefficients of xx and yy straight away. (In Example 7 the first row is 0,0,10, 0, 1⁠, so expanding along R1R_1 is quickest there; any row gives the same value.)
    Signs along R3R_3⁠: (−1)3+1=+(-1)^{3+1} = +⁠, (−1)3+2=−(-1)^{3+2} = -⁠, (−1)3+3=+(-1)^{3+3} = +⁠. Basics: the + − + sign pattern

  3. (i) Simplify to the equation

    12(2y−4x)=0  ⟺  2y=4x  ⟺  y=2x.\begin{aligned} \tfrac12(2y - 4x) = 0 &\iff 2y = 4x\\ &\iff y = 2x. \end{aligned}

    The required line is y=2xy = 2x.

    y x −1 0 1 2 3 4 5 −1 0 1 2 3 4 5 6 7 8 (1, 2) (3, 6) P(x, y) Any P(x, y) on the line: the "triangle" is flat, so its area is 0. ½ (2y − 4x) = 0 y = 2x
    Part (i) on a square grid (equal scales). The line y=2xy = 2x passes through (1,2)(1, 2)⁠, (3,6)(3, 6) and the origin. Any point PP on it makes a flat triangle with the two given points.
    Multiplying by 2 and then dividing by 2 are reversible, so no solutions are lost or gained. The constant term was M33=0M_{33} = 0⁠, which says the line passes through the origin: (0,0)(0, 0) satisfies y=2xy = 2x⁠.

  4. (i) Check both points

    (1,2)(1, 2)⁠: 2=2⋅12 = 2\cdot1⁠. ✓ (3,6)(3, 6)⁠: 6=2⋅36 = 2\cdot3⁠. ✓

    A line through two points must contain both of them; if either failed, there would be an arithmetic slip in the expansion.

Part (ii): the line through (3, 1) and (9, 3)

  1. (ii) A point P(x, y) on the line gives zero area

    12∣311931xy1∣=0.\frac12\begin{vmatrix} 3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1 \end{vmatrix} = 0.
    Same reasoning as Part (i), Step 1: P(x,y)P(x, y) is on the line through the distinct points (3,1)(3, 1) and (9,3)(9, 3) exactly when the area of the triangle they form is 0.

  2. (ii) Expand along the third row

    M31=∣1131∣=1⋅1−1⋅3=−2,M32=∣3191∣=3⋅1−1⋅9=−6,M33=∣3193∣=3⋅3−1⋅9=0.\begin{aligned} M_{31} &= \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} = 1\cdot1 - 1\cdot3 = -2,\\ M_{32} &= \begin{vmatrix} 3 & 1 \\ 9 & 1 \end{vmatrix} = 3\cdot1 - 1\cdot9 = -6,\\ M_{33} &= \begin{vmatrix} 3 & 1 \\ 9 & 3 \end{vmatrix} = 3\cdot3 - 1\cdot9 = 0. \end{aligned} x M31−y M32+1⋅M33=x(−2)−y(−6)+0=−2x+6y.\begin{aligned} &x\,M_{31} - y\,M_{32} + 1\cdot M_{33}\\ &= x(-2) - y(-6) + 0\\ &= -2x + 6y. \end{aligned}
    Each minor is found by deleting row 3 and one column, keeping the remaining entries of rows 1 and 2 in place. The middle term carries the pattern sign −-⁠, and with M32=−6M_{32} = -6 it becomes +6y+6y⁠.
    Basics: the + − + sign pattern

  3. (ii) Simplify to the equation

    12(6y−2x)=0  ⟺  6y=2x  ⟺  x−3y=0.\begin{aligned} \tfrac12(6y - 2x) = 0 &\iff 6y = 2x\\ &\iff x - 3y = 0. \end{aligned}

    The required line is x−3y=0x - 3y = 0 (equivalently y=13xy = \tfrac13 x⁠).

    y x −1 0 1 2 3 4 5 6 7 8 9 10 11 −1 0 1 2 3 4 5 (3, 1) (9, 3) P(x, y) ½ (6y − 2x) = 0, that is, x − 3y = 0
    Part (ii) on a square grid (equal scales). The line x−3y=0x - 3y = 0 passes through (3,1)(3, 1)⁠, (9,3)(9, 3) and the origin; P(6,2)P(6, 2) is one more point on it.
    From 6y=2x6y = 2x⁠, divide by 2 to get 3y=x3y = x⁠, then move 3y3y across. Again M33=0M_{33} = 0⁠, so this line also passes through the origin.

  4. (ii) Check both points

    (3,1)(3, 1)⁠: 3−3⋅1=03 - 3\cdot1 = 0⁠. ✓ (9,3)(9, 3)⁠: 9−3⋅3=09 - 3\cdot3 = 0⁠. ✓

Result

(i) y=2xy = 2x⁠; (ii) x−3y=0x - 3y = 0⁠.

(Textbook answer: (i) y=2xy = 2x⁠, (ii) x−3y=0x - 3y = 0⁠.)

Check it with numbers

Slopes. (i) The slope through (1,2)(1, 2) and (3,6)(3, 6) is 6−23−1=2\frac{6-2}{3-1} = 2⁠, and y=2xy = 2x has slope 2. ✓ (ii) The slope through (3,1)(3, 1) and (9,3)(9, 3) is 3−19−3=13\frac{3-1}{9-3} = \frac13⁠, and x−3y=0x - 3y = 0 has slope 13\frac13⁠. ✓

A point off the line. (1,1)(1, 1) is not on y=2xy = 2x⁠; with x=1x = 1⁠, y=1y = 1 the determinant in Part (i) is −4+2=−2≠0-4 + 2 = -2 \ne 0⁠, so the triangle (1,2)(1, 2)⁠, (3,6)(3, 6)⁠, (1,1)(1, 1) has area 1, not 0. ✓

Common mistakes
  1. Dropping the sign on the middle term. −y M32-y\,M_{32} with M32=−2M_{32} = -2 is +2y+2y⁠, not −2y-2y⁠.
  2. Using "=±= \pm" here. Zero area means Δ=0\Delta = 0⁠; the ±\pm of Remark (ii) is for a given nonzero area.
  3. Putting P(x,y)P(x, y) in a column instead of a row. Each point, including PP⁠, is one row x, y, 1x,\ y,\ 1⁠.
What to take away
  1. Line through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2)⁠: set the determinant with rows (x1,y1,1)(x_1, y_1, 1)⁠, (x2,y2,1)(x_2, y_2, 1)⁠, (x,y,1)(x, y, 1) equal to 0.
  2. Expanding along the row of x,y,1x, y, 1 gives the coefficients of xx⁠, of yy and the constant term directly.
  3. Always check that both given points satisfy the equation you found.

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)
Collinear exactly when the area is zeroStep 1 of each partSection 4.3, Remark (iii); Example 7
Expansion along R3R_3⁠; signs (−1)i+j(-1)^{i+j}Step 2 of each partSection 4.2.3, Remark (ii) (Lesson: expanding a 3 × 3 determinant)
2×22\times2 ruleStep 2 of each partSection 4.2.2
Slope of a line through two pointsCheckCoordinate geometry (Class XI)
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