Example 6: Area of the triangle $(3,8)$⁠, $(-4,2)$⁠, $(5,1)$

Basics you need

Solution

1. Where the area formula comes from

   From earlier classes, the area of the triangle is

   $$\begin{aligned} \tfrac12\big[\,&x_1(y_2 - y_3) + x_2(y_3 - y_1)\\ &+ x_3(y_1 - y_2)\big]. \end{aligned}$$

   Expanding the determinant of Basic 1 along its third column (all 1s, signs $+, -, +$⁠) gives

   $$\begin{aligned} &(x_2y_3 - x_3y_2) - (x_1y_3 - x_3y_1)\\ &\quad + (x_1y_2 - x_2y_1), \end{aligned}$$

   and regrouping by $x_1, x_2, x_3$ gives exactly the bracket above. So the determinant is just a tidy way of writing the familiar formula.

   Knowing why the formula holds tells us how to use it: one row per vertex, the third column all 1s, and a factor $\tfrac12$ outside. It also explains the sign: the bracket is positive when the vertices go round anticlockwise and negative when they go clockwise (a fact from coordinate geometry, not proved in this chapter), which is why Remark (i) takes the absolute value.
   Section 4.3, equation (1). Basics: determinant as area

2. Put the vertices into the determinant

   $$\Delta = \frac12\begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix}.$$
   Row 1 is the first vertex $(3, 8)$⁠, row 2 is $(-4, 2)$⁠, row 3 is $(5, 1)$⁠, each followed by a 1. Any order of the vertices is acceptable: in the bracket of Step 1, swapping two vertices changes the sign of every term (for example, swapping vertices 1 and 2 turns $x_1(y_2 - y_3)$ into $x_2(y_1 - y_3)$⁠, the negative of the old $x_2(y_3 - y_1)$⁠), so a different order can change the sign of the determinant but never its size.

3. Choose a row and write the signs

   No row or column contains a zero, so every expansion costs three $2\times2$ determinants. We follow the book and expand along $R_1$⁠, with entries $3, 8, 1$ and signs $+, -, +$⁠.

   With no zeros there is nothing to gain by choosing another line, and matching the book makes the working easy to compare. Every row and column gives the same value (Section 4.2.3).
   Signs $(-1)^{1+j}$ along $R_1$⁠: $+, -, +$⁠. Basics: the + − + sign pattern

4. The three $2\times2$ determinants

   Delete $R_1$ and $C_1$⁠, then $C_2$⁠, then $C_3$⁠:

   $$\begin{aligned} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} &= 2(1) - 1(1) = 1,\\ \begin{vmatrix} -4 & 1 \\ 5 & 1 \end{vmatrix} &= -4(1) - 5(1) = -9,\\ \begin{vmatrix} -4 & 2 \\ 5 & 1 \end{vmatrix} &= -4(1) - 5(2) = -14. \end{aligned}$$
   Each comes from the rows $R_2$⁠, $R_3$ with one column removed, the remaining entries kept in order. In the second and third, the main-diagonal product is $-4$ and the other-diagonal product is positive, so both results are negative.
   $2\times2$ rule, Section 4.2.2. Basics: where the determinant comes from

5. Combine with the signs

   $$\begin{aligned} \Delta &= \tfrac12\big[\,3(1) - 8(-9) + 1(-14)\,\big]\\ &= \tfrac12\,(3 + 72 - 14)\\ &= \tfrac12\,(61) = \tfrac{61}{2}. \end{aligned}$$
   The middle term has pattern sign $-$ and its $2\times2$ value is $-9$⁠, so $-8(-9) = +72$⁠. The last term is $+1\times(-14) = -14$⁠. This is the book's line $\tfrac12[3(2-1) - 8(-4-5) + 1(-4-10)]$⁠.
   Minus times minus is plus. Basics: algebra the chapter leans on

6. Take the absolute value

   $\Delta = \tfrac{61}{2}$ is already positive, so

   $$\text{Area} = \left|\tfrac{61}{2}\right| = \tfrac{61}{2} = 30.5 \text{ square units}.$$
   Area is never negative (Remark (i), Section 4.3). Here the value is positive because the vertices, in the order given, go round anticlockwise (see the figure). Listing them clockwise, for instance $(3,8)$⁠, $(5,1)$⁠, $(-4,2)$⁠, would give $-\tfrac{61}{2}$⁠, and the absolute value would still give $\tfrac{61}{2}$⁠.
   Area $= |\Delta|$⁠. Basics: determinant as area

Every fact used, and where it comes from