Area of a triangle from its vertices, and why three collinear points give zero

What to assume they know

• Determinants of order three, and expanding one along a chosen line
• Plotting a point from a pair of numbers, and the axes it is measured against
• The area of a triangle from three points, as met in an earlier class
• What it means for three points to lie on one straight line
• The absolute value of a number, and why it is never negative
• Reading an equation of a straight line, and checking whether a point satisfies it
• Solving a linear equation for one unknown, including one that yields two signs

What they should be able to do

• Write the earlier-class area expression and rearrange it into a three-row determinant with a column of ones
• Say which row holds which point, and why every row ends in a one
• Explain why the chapter attaches an absolute value to the determinant before calling the result an area
• Explain why the reverse problem — area given, a coordinate wanted — must keep both signs, and produce both answers
• Test three given points for collinearity by evaluating one determinant
• Derive the equation of the line through two given points by setting the area of the triangle they make with a general point to zero
• Solve for a missing coordinate when the area of the triangle is prescribed
• Recognise the same determinant in the chapter's Summary and read it back into the geometry

Where it usually goes wrong

• "The determinant is the area." Half of its absolute value is the area. Two operations sit between them and students routinely drop one; the chapter prints the half inside the display and the absolute value in a Remark below it, which is exactly the arrangement that makes the second easy to miss.
• "A negative answer means I made a mistake." A negative determinant is normal and depends only on the order the three points were written in. It is the area that cannot be negative.
• "So I should always just make it positive." Not when the area is given and something else is wanted. Remark (ii) exists for that case, and the chapter's own Example 7 and Exercise 4.2 Q3 and Q5 all return two answers because of it. Discarding the sign too early loses one of them every time.
• "The column of ones is padding to make the array square." It is doing arithmetic work. Replace the ones with anything else and the expansion no longer reproduces the area expression.
• "Zero area means the three points are the same point." It means they are in a straight line, which includes but does not require coincidence. Three distinct points strung along one line give zero.
• "Collinearity is a separate topic from this formula." It is the same formula read in the other direction, and the chapter's own Example 7 uses it that way within one page of stating it.
• "The line derived in Example 7 came from a line formula." It came from setting an area to zero. Nothing about lines was assumed; the equation fell out of the determinant. That is the whole trick and it is worth saying twice.
• "The vertices have to be listed anticlockwise." The chapter says nothing about order and the absolute value makes the question moot for areas. Order only decides the sign, which is discarded.

Questions to check understanding

• Compute the area of a triangle from three given points, reporting a positive number and showing where the absolute value was applied
• Decide whether three given points are collinear by evaluating one determinant
• Given the area and two of the three points, find the missing coordinate, and report both values — the form of Exercise 4.2 Q3
• Derive the equation of the line through two given points by the area-zero argument — the form of Exercise 4.2 Q4
• Prove a collinearity that involves letters rather than numbers — the form of Exercise 4.2 Q2
• Choose the correct option when an area is prescribed and one coordinate is unknown — the form of Exercise 4.2 Q5
• State what the column of ones contributes, and show what breaks if it is replaced

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• The earlier-class expression (§4.3, Part I p. 82). The chapter recalls the area of a triangle on three points as a half of a bracketed sum of three terms, each a first coordinate multiplied by a difference of two second coordinates. It is hard to hold in the head and easy to mis-copy, and saying so out loud is the honest motivation for what follows.
• The rewrite (§4.3, Part I p. 82). The same expression becomes a half times a three-by-three determinant whose rows are the three points, each padded with a one in the third position. Read off the printed page: the array sits inside vertical bars, not brackets — this is a determinant from the first line, never a matrix. The chapter tags the display with an equation number and refers back to it in the Remarks immediately below. Verified: expanding along the last column reproduces the earlier-class expression exactly, term for term and sign for sign. That check is a good ninety seconds of video and the chapter does not perform it.
• Why the column of ones (not in the book). The chapter prints the ones and never says what they are for. Supply the reason: the earlier-class expression is a sum of products of one coordinate with a difference of two others, and a column of ones is exactly what makes the three-by-three expansion produce differences of that shape. Flag it as the explanation's account; the chapter offers none.
• Remark (i) (§4.3, Part I p. 82). Because an area is a positive quantity, what the determinant returns is always stripped of its sign first. The wording matters: it is the determinant that may come out negative, and the area that may not.
• Remark (ii) (§4.3, Part I p. 82). When the area is handed to you and something inside the triangle is wanted, both the positive and the negative value are in play. This is the counterpart of Remark (i) and it is why the chapter's own Example 7 produces two answers rather than one.
• Remark (iii) (§4.3, Part I p. 82). Three points on one line enclose no area, so the determinant is zero. This one line is the whole reason the section belongs in a determinants chapter rather than in a coordinate-geometry one.
• Example 6 (§4.3, Part I pp. 82–83). Three points, one of them with a negative first coordinate. Verified: expanding along the top row, the three terms are three times one, minus eight times minus nine, and one times minus fourteen — that is three plus seventy-two minus fourteen, or sixty-one — and half of that is sixty-one halves. The chapter leaves the answer as a fraction and does not attach a unit to it, where the exercises that follow quote units. Spoken form: "sixty-one by two".
• Example 7, first half (Part I p. 83). Two named points and a general point are made the three rows of a determinant, the area is declared zero because the general point is to lie on the line through the other two, and the resulting equation is the equation of that line. Verified: with the origin in one row and the point one, three in another, expanding gives a half of the second coordinate minus three times the first, so the line is the second coordinate equalling three times the first. The chapter marks the declaration with a parenthetical "Why?" and does not answer it — that answer is Remark (iii) two paragraphs above.
• Example 7, second half (Part I p. 83). A third point on the horizontal axis is chosen so that the triangle has area three, and the missing coordinate is solved for; the answer is two values of opposite sign. Verified independently: the determinant with rows one and three, then the origin, then the unknown point comes to three times the unknown, so a half of it equals plus or minus three gives the unknown as plus or minus two. The printed intermediate line carries a leading minus sign that an added expansion does not produce — see Notes. The answer pair is unaffected, because the plus-or-minus on the other side absorbs the sign, but the intermediate line should not be shown as printed.
• Exercise 4.2 (Part I p. 83). Five items in four shapes. Q1 is three straight area computations; verified: the first comes to fifteen halves, the second to forty-seven halves, and the third to fifteen. Q2 asks for a collinearity proof on three points built from three letters; verified: expanding the determinant directly gives zero identically, with no need for any machinery beyond §4.2.3 — see Notes on what this item was designed for. Q3 is the reverse problem twice, area given and a coordinate wanted; verified: both parts land on the same pair, zero or eight, which is worth saying out loud because two differently arranged triangles giving one answer set looks like a copying slip and is not. Q4 is two line-through-two-points constructions in the style of Example 7; verified: the first gives the second coordinate as twice the first, and the second gives the first coordinate as three times the second. Q5 is a multiple choice with a prescribed area and an unknown first coordinate; verified: the determinant comes to fifty minus ten times the unknown, so the area condition gives the unknown as twelve or minus two, and the intended option is the one carrying both.
• The Summary bullet (Part I p. 101). The area determinant is reproduced in the Summary exactly as §4.3 prints it, bars and all. What the Summary does not carry is any of the three Remarks — no absolute value, no both-signs instruction, no collinearity. A student revising only from the Summary has the formula and none of the three things that make it usable. Make this a beat.

Figures to have open

• Three plotted points with the triangle drawn between them, for sections 1 and 7. The chapter draws no axes and no triangle anywhere in §4.3, so this is added here and it is the picture the section most needs.
• The same three points shown step by step onto one straight line, with the enclosed region thinning to nothing, for section 6. An added device; the chapter states the fact in one Remark and draws nothing.
• Two triangles sharing a base, one on each side of it, for section 9's two roots. Not in the book; the chapter reports the two values and does not draw them.
• A five-row table of exercise items against question shape for section 10. Build it with the repo's DataTable component.
• No figure in this chapter is numbered, captioned or labelled Fig. Verified on the page image of every one of the twenty-eight pages. Every picture in this topic is added here.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 4 "Determinants", Part I pp. 76–103
• §4.3 Area of a Triangle, the earlier-class expression and its determinant form, Part I p. 82
• The three Remarks following that display, Part I p. 82
• Example 6, Part I pp. 82–83; Example 7, Part I p. 83
• Exercise 4.2, questions 1 to 5, Part I p. 83
• Summary, the area bullet, Part I p. 101