Equal slopes mean parallel; slopes multiplying to minus one mean perpendicular

What to assume they know

• Steepness as the tangent of an angle, and the one line that has none — the inclination, its range, slope as the tangent of it, and the quotient of coordinate differences proved in §9.2.1
• That the tangent of a right angle does not exist, and that the vertical line is therefore outside every slope statement in §9.2
• The tangent of an angle increased by a right angle, from Chapter 3 — the identity that turns a tangent into a negative cotangent
• Corresponding angles on a transversal cutting two parallel lines
• Reciprocals and negative reciprocals of real numbers, including the fact that zero has none
• Reading an "if and only if" as two separate claims that each need their own argument

What they should be able to do

• State the parallelism criterion of §9.2.2 in your own words, as two claims rather than one
• Prove the forward half — that parallel lines share an inclination and hence a slope
• Name the property of the tangent function that carries the converse, and say why the converse fails without it
• Explain why the two ends of the inclination range both give a tangent of zero, and why that costs the argument nothing
• Derive the perpendicularity criterion from the quarter-turn relation between the two inclinations
• Show that the criterion forces both slopes to be non-zero, and say which perpendicular pair is thereby placed outside it
• Reconstruct the converse of the perpendicularity criterion from the product being minus one
• Decide from two pairs of points whether the lines they determine are parallel, perpendicular, or neither
• Solve for an unknown coordinate that makes two lines perpendicular
• Use equal slopes at a shared point as a collinearity test, and say why it is the same test as the vanishing area of §9.1

Where it usually goes wrong

• "Equal slopes means the two lines are parallel and therefore distinct." The criterion cannot tell a line from itself: a line has the same slope as it does. What equal slopes buy is equal direction; whether the lines are distinct is settled by a point, not by a slope.
• "Perpendicular slopes are reciprocals." They are negative reciprocals. The minus sign is not decoration — it arrives from the tangent of an angle increased by a right angle, and dropping it converts perpendicular into a pair of lines symmetric about the diagonal.
• "Every perpendicular pair satisfies m₁m₂ = −1." The axes do not. Any horizontal line and any vertical line meet at a right angle with no product available. The criterion is stated for two lines that both have slopes precisely because that pair must be excluded.
• "The converse of the parallel criterion is automatic." It is the half that needs an argument. Equal tangents give equal angles only because tangent does not repeat values over the directions a line can take.
• "Tangent is one-to-one from zero to a straight angle." Not on the closed range as §9.2 prints it — both ends give zero. The two ends name the same family of lines, so no line is misidentified, but a student who has been told a false statement will not be able to say why it is safe.
• "You can see perpendicularity in a sketch." Fig 9.5 marks the right angle with a square because you cannot. On unequal axis scales a right angle does not look like one.
• "The right-angle test needs the two lines to actually cross in the picture." Slopes are properties of direction. Example 3's two lines are specified by four points and the intersection is never located.
• "Collinearity by slopes is a different test from the area formula." It is the same test. A triangle of zero area and three points with matching slopes are two descriptions of one situation.

Questions to check understanding

• Given four points, decide whether the two lines they determine are parallel, perpendicular or neither, and justify with slopes
• Given three points of a quadrilateral condition, prove a parallelogram or a right triangle using slopes only
• Solve for an unknown coordinate that makes two lines perpendicular
• Find the reflection of a point in a given line, using perpendicularity and the midpoint
• State both criteria as biconditionals and prove the direction you are asked for
• Explain why a horizontal line and a vertical line satisfy no slope product, and why that is not a counterexample to the criterion
• Test three points for collinearity by slopes and by area, and say why the answers agree

Examples worth working on the board

Inputs only. Values marked verified are worked out here on data printed inside pp. 152–174.

• Fig 9.4 (§9.2.2, p. 154). Two parallel lines that fall as you read left to right: l₁ is the upper-right of the pair, l₂ the lower-left, each drawn with arrowheads at both ends. Each meets the x-axis at its own point, and an angle arc is drawn at each crossing — α at l₁'s crossing, β at l₂'s. I inspected this rather than on the printed page, because the two arcs are the whole point: both sweep anticlockwise from the positive x-direction round to the line's upward arm, so both are drawn obtuse. A figure showing two acute angles would quietly teach the wrong convention.
• The forward argument (§9.2.2, p. 154). Verified: if the two lines are parallel they meet the x-axis as a transversal in equal corresponding angles, so α = β, so tan α = tan β, so m₁ = m₂. Nothing about tangent is needed here beyond its being a function.
• The converse and where its weight sits (§9.2.2, p. 154). Verified: m₁ = m₂ gives tan α = tan β, and this returns α = β only because tangent takes each value at most once over the directions available to a line. Without that, equal slopes would permit different inclinations and the criterion would collapse.
• The one repeat in the range. §9.2 prints the inclination range as running from zero to a straight angle inclusive. Verified: tangent gives zero at both of those ends, so on the closed range it is not literally one-to-one. The repeat is harmless because both ends describe the same set of lines — the horizontal ones — so the two inclinations that share a tangent were never two different directions. Every other value in the range is taken once, and the right angle is taken by nothing. Worth saying aloud, because "tangent is one-to-one there" is stated in the chapter without this qualification.
• Fig 9.5 (§9.2.2, p. 155). Two lines crossing above the x-axis: l₁ rises to the right, l₂ falls to the right, and a small square marks the right angle at their intersection. Read off the printed page: α is the acute arc at l₁'s crossing with the x-axis, drawn just left of the origin, and β is the obtuse arc at l₂'s crossing, drawn well to the right. So the figure supplies a concrete instance of the relation β = α + 90° rather than merely asserting it.
• The perpendicular derivation (§9.2.2, p. 155). Verified: from β = α + 90°, the tangent of β is the negative cotangent of α, which is minus the reciprocal of tan α; hence m₂ = −1/m₁ and m₁m₂ = −1. Verified consequence, and the chapter does not state it: the equation m₁m₂ = −1 cannot hold with either factor zero, so the criterion silently asserts that neither perpendicular line is horizontal.
• The pair outside the criterion. Verified: the x-axis and the y-axis are perpendicular; one has slope zero and the other has no slope; there is no product to form. §9.2.2 opens by restricting to lines that are not vertical, which is exactly what keeps this pair out. Treat the restriction as content, not as small print.
• Example 3 (p. 158). The line through (−2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (x, 24); find x. Verified: the first slope is 2/6 = 1/3; the second is 12/(x − 8); setting the product to −1 gives 4/(x − 8) = −1, so x − 8 = −4 and x = 4. Note the answer makes the second line's slope −3, which is indeed the negative reciprocal of 1/3.
• Exercise 9.1 q6 — show that (4, 4), (3, 5) and (−1, −1) form a right triangle without using Pythagoras. Verified: the slope from (4, 4) to (3, 5) is −1 and from (4, 4) to (−1, −1) is 1; their product is −1, so the right angle sits at (4, 4). This is the item that shows the criterion doing work Pythagoras would otherwise do.
• Exercise 9.1 q8 — show that (−2, −1), (4, 0), (3, 3) and (−3, 2) form a parallelogram without the distance formula. Verified: going round in that order the four side slopes are 1/6, −3, 1/6, −3, so each pair of opposite sides is parallel.
• Exercise 9.2 q9 — the line through (−3, 5) perpendicular to the line through (2, 5) and (−3, 6). Verified: the given slope is −1/5, so the required slope is 5, and the line is 5x − y + 20 = 0.
• Exercise 9.3 q9 — the line through (h, 3) and (4, 1) meets 7x − 9y − 19 = 0 at a right angle. Verified: the given line has slope 7/9, so the required slope is −9/7; setting −2/(4 − h) equal to that gives h = 22/9.
• Exercise 9.3 q12 — the right bisector of the segment joining (3, 4) and (−1, 2). Verified: the midpoint is (1, 3), the segment's slope is 1/2, the bisector's slope is −2, and the line is 2x + y − 5 = 0. Two separate facts are used, one from §9.1 and one from here.
• Exercise 9.3 q14 — the perpendicular dropped from the origin onto y = mx + c lands at (−1, 2). Verified: that perpendicular has slope −2, so m = 1/2, and substituting the point gives c = 5/2.
• Example 13 (pp. 169–170, Fig 9.17). The reflection of (1, 2) in the line x − 3y + 4 = 0, treating the line as a mirror. Verified: the mirror line has slope 1/3, so the segment joining a point to its image has slope −3, giving 3h + k = 5; requiring the midpoint to lie on the line gives h − 3k = −3; together these give the image (6/5, 7/5). Fig 9.17 draws the original above the line and the image below it. This is the best single application in the chapter, because it needs perpendicularity and the midpoint at once and neither alone suffices.
• The collinearity corollary (Summary, p. 174). Verified: three points are collinear exactly when the slope from the first to the second equals the slope from the second to the third. This is the parallelism criterion applied to two segments that already share a point — parallel plus a common point means one line. It is the same test as the vanishing triangle area of §9.1, expressed in slopes.
• Exercise 9.2 q19 — show (3, 0), (−2, −2) and (8, 2) are collinear. Verified: both slopes measured from (3, 0) come to 2/5.

Figures to have open

• Fig 9.4 redrawn, with both inclinations arced anticlockwise from the positive x-direction so that both read as obtuse. The chapter's own figure; a redraw that makes the angles acute teaches the wrong convention and this is the figure students copy.
• Fig 9.5 redrawn, keeping the right-angle square at the intersection and both inclination arcs at the x-axis. The chapter's own figure.
• A single movement of one line pivoting through a right angle about a point, with its slope displayed and the product of the two slopes tracked. Standard schematic; it is what makes the minus sign look like a consequence rather than a rule.
• A schematic of the axes labelled as a perpendicular pair with the product test struck out. Standard schematic built from the chapter's own restriction.

Where this sits in the book

• NCERT Mathematics, Textbook for Class XI, Chapter 9 "Straight Lines", §9.2.2 — the section headed Conditions for parallelism and perpendicularity, taken in terms of slopes, pp. 154–155 — Fig 9.4, Fig 9.5, both criteria and both converses
• Example 3, p. 158 — the perpendicular pair with an unknown coordinate
• Exercise 9.1, pp. 158–159, questions 6 and 8
• Exercise 9.2, p. 164, question 9; Exercise 9.3, pp. 167–168, questions 9, 12 and 14
• Example 13, pp. 169–170, and Fig 9.17 — reflection in a line
• The chapter Summary, p. 174 — both criteria restated, and the collinearity test in terms of slopes
• Cross-reference outside this chapter: the tangent of an angle increased by a right angle is a Chapter 3 identity, used here without re-derivation