Chapter 4 · Determinants

Example 7: Line through $A(1,3)$ ⁠, $B(0,0)$ and $k$ for area 3

ExamplesArea of a triangle · Section 4.3

The question

Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants and find $k$ if $D(k, 0)$ is a point such that area of triangle $ABD$ is 3 sq units.

Exam answer

(a) Let $P(x, y)$ be any point on $AB$ ⁠. Then the area of the triangle with vertices $B(0, 0)$ ⁠, $A(1, 3)$ ⁠, $P(x, y)$ is 0:

\frac12\begin{vmatrix} 0 & 0 & 1 \\ 1 & 3 & 1 \\ x & y & 1 \end{vmatrix} = 0

Expanding along $R_1$ ⁠:

\begin{aligned} &\frac12\big[0 - 0 + 1(y - 3x)\big] = 0 \\ &\Rightarrow\ y - 3x = 0 \end{aligned}

So the equation of $AB$ is $y = 3x$ ⁠.

(b) Area of triangle $ABD$ is 3 sq. units, so

\frac12\begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 3

Expanding along $R_2$ ⁠:

\begin{aligned} &\frac12\big[-0 + 0 - 1(0 - 3k)\big] = \pm 3 \\ &\Rightarrow\ \frac{3k}{2} = \pm 3 \end{aligned}

$\therefore\ k = \pm 2$ ⁠.

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

The one idea this problem is built on

The area determinant answers two questions. Zero area means the three points lie on one line, so a point $P(x, y)$ is on line $AB$ exactly when the area of "triangle" $ABP$ is 0: that condition is the equation of the line. A given area $a$ means the determinant is $+2a$ or $-2a$ (its sign depends only on the order of the vertices), so both signs must be solved, giving two values of $k$ ⁠.

Given, to find, and the plan

Given: Points $A(1, 3)$ ⁠, $B(0, 0)$ ⁠, and $D(k, 0)$ on the $x$ -axis with area of triangle $ABD = 3$ ⁠.
To find: (a) The equation of line $AB$ ⁠. (b) Every value of $k$ ⁠.
Plan: (a) Set the area determinant of $B, A, P$ equal to 0 and expand along the row with two zeros. (b) Set the area determinant of $A, B, D$ equal to $\pm3$ ⁠, expand along the row with two zeros, solve both equations, and check with base and height.

Basics you need

1Area as a determinant

Area of the triangle with vertices $(x_i, y_i)$ is $|\Delta|$ ⁠, where

\Delta = \frac12\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}.

Basics: determinant as area

2Collinear points give zero area

Three points on one line form a "flat" triangle with area 0 (Remark (iii), Section 4.3). Conversely, if the area is 0 the three points cannot form a genuine triangle, so they lie on one line.

3Given area: use both signs

If the area is $a$ ⁠, then $\Delta = a$ or $\Delta = -a$ (Remark (ii), Section 4.3), because $|\Delta| = a$ has both solutions. Basics: determinant as area

4Expansion along a row with zeros

Only entries that are not 0 contribute; each carries the sign $(-1)^{i+j}$ ⁠. Basics: the + − + sign pattern

Part (a): the equation of line AB

Turn "P is on AB" into "area is zero"

Let $P(x, y)$ be any point. This answers the book's "(Why?)":

$P$ lies on line $AB$ $\iff$ $B$ ⁠, $A$ ⁠, $P$ are collinear $\iff$ area of triangle $ABP$ is 0.

So, with the rows in the order $B, A, P$ used by the book,
$\frac12\begin{vmatrix} 0 & 0 & 1 \\ 1 & 3 & 1 \\ x & y & 1 \end{vmatrix} = 0.$
If $P$ is on the line through $A$ and $B$ ⁠, the three points are in a straight line and enclose no area. If $P$ is off the line, $ABP$ is a genuine triangle with positive area. Because $A \ne B$ ⁠, there is exactly one line through them, so "area 0" describes exactly the points of that line, no more and no less. The order of the rows does not matter here, since a sign change does not affect " $= 0$ ".

Section 4.3, Remark (iii): three collinear points give area 0. Basics: determinant as area
Expand along the first row

Row 1 is $0, 0, 1$ ⁠: only $a_{13} = 1$ contributes, with sign $(-1)^{1+3} = +1$ ⁠. Delete $R_1$ and $C_3$ ⁠:
$\begin{vmatrix} 1 & 3 \\ x & y \end{vmatrix} = 1\cdot y - x\cdot 3 = y - 3x.$
So
$\frac12\big[\,0 - 0 + 1\,(y - 3x)\,\big] = \frac12(y - 3x).$
Row 1 holds the point $B(0, 0)$ ⁠, which gives two zeros, so it is the cheapest row: one $2\times2$ determinant instead of three. This is why it is convenient to put the origin first.

Expand along the line with the most zeros, Section 4.2.3, Remark (i). Lesson: expanding a 3 × 3 determinant
Solve for the equation of the line
$\begin{aligned} \tfrac12(y - 3x) = 0 &\iff y - 3x = 0\\ &\iff y = 3x. \end{aligned}$
Check: $A(1, 3)$ gives $3 = 3\cdot 1$ ⁠, true; $B(0, 0)$ gives $0 = 3\cdot0$ ⁠, true. So line $AB$ is $y = 3x$ .

Multiplying by 2 does not change which $x, y$ satisfy the equation. A line through the origin has the form $y = mx$ ⁠; here the slope is $m = \tfrac{3-0}{1-0} = 3$ ⁠, which agrees.

Part (b): the values of k

Write the area condition with both signs

With the vertices in the order $A(1, 3)$ ⁠, $B(0, 0)$ ⁠, $D(k, 0)$ ⁠:
$\frac12\begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 3.$
The area is 3, and the area is the absolute value of $\Delta$ ⁠. So $\Delta$ itself may be $3$ or $-3$ ⁠, depending on whether $A, B, D$ go round anticlockwise or clockwise. We do not know which in advance, because we do not know $k$ ⁠, so both must be solved.

Section 4.3, Remark (ii): if area is given, use both positive and negative values of the determinant. Basics: determinant as area
Expand along the second row

Row 2 is $0, 0, 1$ ⁠: only $a_{23} = 1$ contributes, with sign $(-1)^{2+3} = -1$ ⁠. Delete $R_2$ and $C_3$ ⁠:
$\begin{vmatrix} 1 & 3 \\ k & 0 \end{vmatrix} = 1\cdot 0 - k\cdot 3 = -3k.$
The signs along $R_2$ are $-, +, -$ ⁠, and the entries $a_{21} = a_{22} = 0$ contribute nothing, so the determinant is
$-0 + 0 - 1\cdot(-3k) = 3k,$
and the condition becomes $\tfrac12(3k) = \pm 3$ ⁠.

A note on the book's line. The textbook's working shows $\tfrac{-3k}{2}$ at this point. That is the value for the rows in the order $B, A, D$ ⁠: expanding along $R_1 = (0, 0, 1)$ ⁠, only $a_{13} = 1$ (sign $+$ ⁠) contributes, and
$\begin{vmatrix} 0 & 0 & 1 \\ 1 & 3 & 1 \\ k & 0 & 1 \end{vmatrix} = +1\begin{vmatrix} 1 & 3 \\ k & 0 \end{vmatrix} = -3k.$
With the rows in the order $A, B, D$ ⁠, as printed in the book's determinant, the value is $+\tfrac{3k}{2}$ ⁠.

Row 2 holds $B(0, 0)$ ⁠, so it has two zeros. The position $(2, 3)$ has an odd index sum, so its sign is $-$ ⁠, and minus times $-3k$ gives $+3k$ ⁠. The forms $\tfrac{3k}{2}$ and $\tfrac{-3k}{2}$ differ only in sign, and since both signs $\pm3$ are solved, either form leads to exactly the same pair of values of $k$ ⁠.

Sign of position $(2,3)$ ⁠: $(-1)^5 = -1$ ⁠. Basics: the + − + sign pattern
Solve both equations
$\begin{aligned} \tfrac{3k}{2} = 3 &\iff 3k = 6 \iff k = 2,\\ \tfrac{3k}{2} = -3 &\iff 3k = -6 \iff k = -2. \end{aligned}$
So $k = 2$ or $k = -2$ ⁠, that is $k = \pm 2$ (the book writes $k = \mp 2$ ⁠, the same pair).

Each equation is linear in $k$ and has exactly one solution. Together they give all values of $k$ for which $|\Delta| = 3$ ⁠.
Check both values with base and height

$B(0, 0)$ and $D(k, 0)$ both lie on the $x$ -axis, so take $BD$ as the base: its length is $|k|$ ⁠. The height is the distance of $A(1, 3)$ from the $x$ -axis, which is 3.
$\text{Area} = \tfrac12\cdot|k|\cdot 3 = \tfrac{3|k|}{2}.$
For $k = 2$ and for $k = -2$ ⁠, $|k| = 2$ ⁠, so the area is $\tfrac{3\cdot 2}{2} = 3$ ⁠. Both values are correct.

This check uses plain geometry and no determinant, so it is genuinely independent. It also shows why there are two answers: $D$ can sit 2 units to the right or 2 units to the left of $B$ ⁠, and the height stays 3 either way.

Equal scales on both axes. Both positions of

D

give a base

BD

of length 2 on the

x

-axis and height 3 (the dashed line), so both triangles have area 3.

Result

The line $AB$ is $y = 3x$ ⁠. The area of triangle $ABD$ is 3 square units when $k = 2$ or $k = -2$ ⁠, that is $k = \pm 2$ ⁠.

This agrees with the textbook's solution.

Check it with numbers

Another point of $y = 3x$ ⁠, say $P(2, 6)$ ⁠, should give zero area with $A$ and $B$ ⁠:

\begin{vmatrix} 0 & 0 & 1 \\ 1 & 3 & 1 \\ 2 & 6 & 1 \end{vmatrix} = +1\,(1\cdot 6 - 2\cdot 3) = 0.

A point off the line, say $P(2, 5)$ ⁠, gives $1\cdot 5 - 2\cdot 3 = -1 \ne 0$ ⁠, area $\tfrac12$ ⁠.

For $k = 2$ the full determinant with rows $A, B, D$ is $3k = 6$ ⁠, so $\Delta = 3$ ⁠; for $k = -2$ it is $-6$ ⁠, so $\Delta = -3$ ⁠. Both have $|\Delta| = 3$ ⁠.

Common mistakes

Solving only $\Delta = 3$ ⁠. That gives $k = 2$ alone and misses $k = -2$ ⁠.
Forgetting the sign of position $(2,3)$ ⁠. Expanding along $R_2$ ⁠, the entry $a_{23}$ carries $(-1)^{2+3} = -1$ ⁠.
Stopping at " $\tfrac12(y-3x) = 0$ ". The equation of the line should be simplified to $y = 3x$ and checked at $A$ and $B$ ⁠.

What to take away

Equation of the line through two points: set the area determinant with a general point $(x, y)$ equal to 0.
Given area $a$ ⁠: solve $\Delta = a$ and $\Delta = -a$ separately. The sign of $\Delta$ only records the order of the vertices.
Put a vertex with zero coordinates in a row of its own: its zeros make the expansion short.

Every fact used, and where it comes from

Fact	Used in	Source
Area $= \|\Delta\|$ ⁠, $\Delta = \tfrac12\det$ with rows $(x_i, y_i, 1)$	Part (a), Step 1; Part (b), Step 1	Section 4.3, equation (1) and Remark (i) (Basics: determinant as area)
Collinear points give area 0	Part (a), Step 1	Section 4.3, Remark (iii)
Given area: use both signs	Part (b), Steps 1 and 3	Section 4.3, Remark (ii)
Expansion along a row; sign $(-1)^{i+j}$	Part (a), Step 2; Part (b), Step 2	Section 4.2.3 and Remark (ii) (Lesson: expanding a 3 × 3 determinant)
Area $= \tfrac12\times$ base $\times$ height	Part (b), Step 4	Mensuration (earlier classes)

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Example 7: Line through A(1,3)A(1,3)A(1,3)⁠, B(0,0)B(0,0)B(0,0) and kkk for area 3

Basics you need

Part (a): the equation of line AB

Turn "P is on AB" into "area is zero"

Expand along the first row

Solve for the equation of the line

Part (b): the values of k

Write the area condition with both signs

Expand along the second row

Solve both equations

Check both values with base and height

Every fact used, and where it comes from

Example 7: Line through $A(1,3)$ ⁠, $B(0,0)$ and $k$ for area 3