One point and a slope, or two points: the same condition written twice

What to assume they know

• Steepness as the tangent of an angle, and the one line that has none — slope as the tangent of the inclination, and the quotient of coordinate differences
• Lines parallel to an axis, where one coordinate never changes — §9.3's requirement that an equation be true on the line and false off it, and the vertical family that has no slope
• The collinearity of three points expressed as equal slopes, from Equal slopes mean parallel; slopes multiplying to minus one mean perpendicular
• That multiplying both sides of an equation by an expression can change its solution set when the expression can vanish
• Rearranging a linear equation into a form with all terms on one side

What they should be able to do

• Derive the point-slope equation from the slope of a segment joining a fixed point to a variable one
• Explain why the quotient form excludes the fixed point and why the cleared form does not
• State what §9.3.2 has to check about points off the line, and why that half matters
• Derive the two-point equation from a collinearity statement
• Show that the two-point equation is the point-slope equation with the slope supplied by §9.2.1
• Identify the hypothesis both forms need, and name the family of lines they cannot produce
• Write the equation of a line from a point and a slope, or from a point and an inclination
• Write the equation of a line through two given points, and rearrange it to a form with integer coefficients
• Combine these forms with the perpendicularity criterion to write medians, altitudes and perpendiculars
• Fit a linear relation to two measured data pairs and use it to predict a third

Where it usually goes wrong

• "Point-slope and two-point are two formulas to memorise." They are one condition. If you can compute a slope from two points you can turn the second into the first in a line of work, and remembering only the first is enough.
• "The two forms differ in which point you subtract." They differ in whether the slope is given or computed. The choice of which given point to call the fixed one is free in both, and changes the written equation but not the line.
• "Multiplying out is just tidying." It changes the solution set. The quotient form is silent at the fixed point; the cleared form is true there. That is a repair, not a rearrangement.
• "An equation of a line is a formula for y." It is a condition on a pair. Some lines cannot be written as a formula for y at all, which is why the vertical family needed its own section.
• "3x − y − 4 = 0 and −3x + y + 4 = 0 are different lines." They have identical solution sets. Multiplying an equation by a non-zero constant leaves the line alone, and the chapter prints Example 6's answer in the sign opposite to the one the natural working produces.
• "You need the two given points in order, left to right." You do not. Swapping them negates both differences in the slope, and the slope survives.
• "The two-point form works for any two distinct points." Not for two points with the same first coordinate. That is the vertical case and it has no slope; §9.3.1 handles it.
• "Word problems about rods and milk are a different topic." They are the two-point form with units attached. Two measurements determine the line; the third value is read off it.

Questions to check understanding

• Write the equation of a line from a point and a slope, or from a point and an inclination
• Write the equation of a line through two given points and reduce it to integer coefficients
• Write the equation of a median or an altitude of a triangle given its three vertices
• Write the equation of a line through a given point perpendicular or parallel to a line specified by two other points
• Given a point on a line and the foot of the perpendicular from the origin, find the line
• Fit a linear relation to two data pairs and predict a third value
• Prove three points collinear by producing the equation of the line through two of them
• Explain why the point-slope form cannot produce a vertical line

Examples worth working on the board

Inputs only. Values marked verified are worked out here on data printed inside pp. 159–175.

• Fig 9.10 (§9.3.2, p. 160). One line rising to the right, with two points marked on it: the fixed point lower left, the arbitrary point upper right, and the words "Slope m" printed beside the line. Read off the printed page: only the fixed point carries a subscript, which is the figure's way of saying which of the two is given and which is being tested.
• The quotient and its blind spot (§9.3.2, p. 160). Verified, and the chapter does not remark on it: writing the slope of the segment from the fixed point to the arbitrary point as a quotient requires the two first coordinates to differ, so the quotient form says nothing about the fixed point itself. After multiplying through, the fixed point satisfies the equation trivially — both sides become zero. So the cleared form covers one point the quotient form cannot, and it is the one point we already know lies on the line.
• The other half (§9.3.2, pp. 160–161). Verified: the chapter is explicit that no point outside the line satisfies the cleared equation, which is what earns the phrase "if and only if" in the statement that follows. Suppose a point satisfies it and differs from the fixed point in its first coordinate; then the slope from the fixed point to it is m, so it lies on the line through the fixed point with slope m. Suppose instead it agrees in the first coordinate; then the equation forces it to agree in the second too, so it is the fixed point.
• Example 5 (p. 161). The line through (−2, 3) with slope −4. Verified: the point-slope form gives y − 3 = −4(x + 2), which rearranges to 4x + y + 5 = 0.
• The mislabel in Example 5. Read off p. 161: the sentence introducing the solution names the formula it is using "slope-intercept form", but the formula it cites is the point-slope equation of §9.3.2, and slope-intercept form is not reached until §9.3.4 on the following page. Cite the heading, not the sentence. This is worth showing students because it is a live instance of the rule that a form is identified by its section heading and never by a passing reference.
• Fig 9.11 (§9.3.3, p. 161). One rising line carrying three marked points: the first given point low on the left, the variable point in the middle, the second given point high on the right. Read off the printed page: all three sit on the same drawn line, so the figure asserts the collinearity the derivation starts from rather than deriving it.
• The two-point derivation (§9.3.3, p. 161). Verified: setting the slope from the first given point to the variable point equal to the slope from the first given point to the second, and clearing the denominator, gives the printed equation. Substituting the §9.2.1 quotient for m in the point-slope equation gives the same thing character for character, so §9.3.3 introduces no new idea — it supplies a missing input.
• Example 6 (p. 161). The line through (1, −1) and (3, 5). Verified: the slope is 6/2 = 3, and the two-point form gives y + 1 = 3(x − 1), i.e. 3x − y − 4 = 0. The chapter prints the same line as −3x + y + 4 = 0, which is this equation multiplied by −1. Both name the same set of points; nothing distinguishes them mathematically. Point this out rather than letting students think their sign is wrong.
• The shared hypothesis. Verified: both forms assume the line is not vertical — the point-slope form because it needs a slope, the two-point form because its quotient needs the two first coordinates to differ. Neither can produce x = b. That family was settled in §9.3.1 and this is why it had to be.
• Exercise 9.2 items belonging here, with their data intact:
  • q2: through (−4, 3) with slope 1/2. Verified: x − 2y + 10 = 0.
  • q3: through the origin with slope m. Verified: y = mx — the point-slope form with both fixed coordinates zero, which is the cleanest possible instance.
  • q4: through (2, 2√3), inclined to the x-axis at 75°. Verified: tan 75° = 2 + √3, so the line is y − 2√3 = (2 + √3)(x − 2).
  • q5: meeting the x-axis 3 units to the left of the origin, with slope −2. Verified: the point is (−3, 0) and the line is 2x + y + 6 = 0.
  • q7: through (−1, 1) and (2, −4). Verified: slope −5/3, line 5x + 3y + 2 = 0.
  • q8: the median through R in the triangle with P(2, 1), Q(−2, 3), R(4, 5). Verified: the midpoint of PQ is (0, 2), the slope from R to it is 3/4, and the median is 3x − 4y + 8 = 0.
  • q9: through (−3, 5), perpendicular to the line through (2, 5) and (−3, 6). Verified: 5x − y + 20 = 0.
  • q10: perpendicular to the segment joining (1, 0) and (2, 3) and cutting it in the ratio 1 : n. Verified: the cutting point is ((2 + n)/(1 + n), 3/(1 + n)), the segment has slope 3, so the required slope is −1/3 and the line follows by point-slope. The answer depends on n, which is the point of the question.
  • q13: through (0, 2) at an angle of 2π/3 measured from the positive x-direction, and then the parallel line crossing the y-axis 2 units below the origin. Verified: the slope is tan 120° = −√3, so the lines are √3·x + y − 2 = 0 and √3·x + y + 2 = 0.
  • q14: a line is met by the perpendicular dropped from the origin at the point (−2, 9). Verified: that perpendicular has slope −9/2, so the line's slope is 2/9 and the line is 2x − 9y + 85 = 0.
  • q15: a copper rod's length L in centimetres is linear in its Celsius temperature C, with L = 124.942 at C = 20 and L = 125.134 at C = 110. Verified: the slope is 0.192/90 = 0.0021333… centimetres per degree, so L = 124.942 + (0.192/90)(C − 20).
  • q16: a shop sells 980 litres a week at Rs 14 per litre and 1220 litres at Rs 16, with a linear relation assumed; how many at Rs 17? Verified: the slope is 240/2 = 120 litres per rupee, so the prediction is 1220 + 120 = 1340 litres. Note the relation as posed has demand rising with price, which is worth remarking on rather than smoothing over.
  • q19: prove (3, 0), (−2, −2) and (8, 2) collinear using the concept of the equation of a line. Verified: the line through the first two is 2x − 5y − 6 = 0, and the third satisfies it since 16 − 10 − 6 = 0.
• The Summary entries (p. 174). Both forms are restated there, the point-slope one with its "if and only if" intact. Read off the printed page.

Figures to have open

• Fig 9.10 redrawn with the fixed point and the arbitrary point distinguished by their labelling. The chapter's own figure.
• Fig 9.11 redrawn with three points on one line, the variable point visibly between the two given ones. The chapter's own figure.
• A movement of the arbitrary point sliding along the line while the computed slope stays fixed, then leaving the line and the slope changing. Standard schematic; it is the only way to show the "false off the line" half moving rather than asserted.
• A scatter of two measured points with the fitted line and an extrapolated third value marked, built from Exercise 9.2 q15 or q16. Standard schematic using the chapter's data.

Where this sits in the book

• NCERT Mathematics, Textbook for Class XI, Chapter 9 "Straight Lines", §9.3.2 Point-slope form, pp. 160–161 — Fig 9.10, the derivation and the biconditional statement
• §9.3.3 Two-point form, p. 161 — Fig 9.11 and the derivation from collinearity
• Example 5, p. 161, and Example 6, p. 161
• Exercise 9.2, pp. 163–164, questions 2, 3, 4, 5, 7, 8, 9, 10, 13, 14, 15, 16 and 19
• The chapter Summary, p. 174 — both forms restated