The shortest route from a point to a line, read off its coefficients

What to assume they know

• Naming Ax + By + C = 0, the one form the distance formula ahead will take — the general form, its coefficients, and what each of them vanishing means
• The area of a triangle from the coordinates of its three vertices, recalled in §9.1
• The distance between two points, recalled in §9.1
• Equal slopes mean parallel; slopes multiplying to minus one mean perpendicular — perpendicularity in terms of slopes, needed for the foot-of-perpendicular problems
• That the perpendicular is the shortest segment from a point to a line
• The modulus of a real number, and that a square root is taken non-negative

What they should be able to do

• State what the chapter takes a point's distance from a line to mean
• Follow the two-way area computation and say which known result supplies each measurement
• Identify the three coefficient conditions the derivation quietly requires
• Check directly that the final formula still gives the right answer for a horizontal line, a vertical line and a line through the origin
• Apply the formula to a point and a line given in general form
• Convert a line given in another shape into the general form before applying the formula
• Explain what the modulus in the numerator discards, and what the unsigned numerator would have told you
• Use the sign split to find the locus of points equidistant from two given lines
• Find points on a named axis at a stated distance from a given line
• Find the foot of the perpendicular from a point to a line, and its length
• Distinguish the perpendicular distance from a distance measured along some other stated direction

Where it usually goes wrong

• "The distance from a point to a line is ambiguous, since a line has infinitely many points." That is why §9.4 opens with a definition. The perpendicular segment is the shortest of all of them, and taking it is a choice the chapter makes explicitly.
• "The formula only works if you first find the foot of the perpendicular." The whole point of the derivation is that it never locates the foot. Exercise 9.3 q13 asks for the foot separately, by a different method.
• "Plug the coefficients straight in." Only after the equation is in the general shape. Exercise 9.3 q3 arrives as a bracketed equality and gives wrong answers to anyone who reads coefficients off it as printed.
• "The derivation covers every line." It does not. It needs both axis crossings to exist and to be distinct, which fails for horizontal lines, vertical lines and lines through the origin. The formula survives all three, but that has to be checked, and the chapter does not check it.
• "The modulus is there to make the answer positive." It is there to discard a sign that carried information — which side of the line the point is on. Example 16 recovers that information by removing the modulus deliberately, and its two answers are exactly the two sides.
• "Distance measured along a given line is the distance to the line." It is not, and Example 12 is printed to make the difference concrete. Any direction other than the perpendicular gives a larger number.
• "A small numerator means small coordinates." It means the point is close to the line. Example 9's point has coordinates in the units and tens and sits three fifths of a unit from its line.
• "The denominator is the length of something in the picture." It is what remains of the base QR after the shared factor cancels. Treating it as a length in its own right leads students to try to draw it.

Questions to check understanding

• Find the distance from a given point to a line given in general form
• Rearrange a line given in another shape and then find a distance from it
• Find the points on a named axis at a stated distance from a given line
• Find the length of an altitude of a triangle from the coordinates of its vertices
• Find the foot of the perpendicular from a point to a line
• Prove a relation between the perpendicular from the origin and a line's intercepts
• Find the locus of points equidistant from two given lines
• Show that a stated distance condition confines a moving point to a line
• Find the distance from a point to a line measured along a stated direction, and compare it with the perpendicular distance

Examples worth working on the board

Inputs only. Values marked verified are worked out here on data printed inside pp. 164–175.

• Fig 9.14 (§9.4, p. 165). The line is drawn falling steeply from upper left to lower right and is labelled with its general equation. Read off the printed page: it crosses the y-axis at a point labelled R with first coordinate zero and second coordinate minus the constant over the y-coefficient, and the x-axis at a point labelled Q with second coordinate zero and first coordinate minus the constant over the x-coefficient. The given point P sits to the right of the line, joined to both Q and R by dashed segments so the triangle is visible, and the perpendicular from P meets the line at M with the segment marked d and a right-angle square at the foot. Every label in that figure is inside the artwork.
• The two measurements (§9.4, pp. 165–166). Verified: the coordinate area formula applied to P, Q and R produces, after simplification, twice the area equal to the modulus of the constant divided by the product of the two leading coefficients, multiplied by the modulus of the left-hand side evaluated at P. The base QR, computed by the distance formula on the two crossings, is the same awkward factor multiplied by the square root of the sum of the squares of the two leading coefficients. Dividing the first by the second cancels the awkward factor entirely, leaving the printed result.
• What the cancellation shows, and what it does not. Verified: the factor that drops out is pinned the moment the three coefficients are pinned, and the derivation used no pair of points on the line other than the two axis crossings. Its disappearance therefore says nothing about what some other pair would have given — that inference is not available here, and a teacher should not offer it. The reason the choice of base does not matter sits earlier than the algebra: a triangle has one area however its three sides are labelled, so twice that area divided by the length of whichever side is called the base returns the perpendicular height to that side every time.
• What the derivation assumed. Verified, and the chapter says none of it: Q exists only when the x-coefficient is non-zero, R only when the y-coefficient is non-zero, and Q and R are distinct only when the constant is non-zero. If the constant vanishes both crossings collapse onto the origin, the base has length zero, and the division that produced the formula is not available. So the derivation as printed proves the formula only for lines that cross both axes at two different points.
• Closing the gap (working added here; the chapter does not do this). Verified:
  • For a horizontal line with zero x-coefficient, the formula reduces to the modulus of the second coordinate of P plus the constant over the y-coefficient, which is exactly the vertical gap between P and the line.
  • For a vertical line with zero y-coefficient, the same reduction gives the horizontal gap.
  • For a line through the origin, translating the whole picture does not change any distance, and the formula's numerator and denominator both survive translation correctly, so the result follows from the case already proved. Doing these three checks turns a formula the students were handed into one they have verified.
• Example 9 (p. 167). The distance of (3, −5) from 3x − 4y − 26 = 0. Verified: the numerator is the modulus of 9 + 20 − 26, that is 3; the denominator is the square root of 9 + 16, that is 5; the distance is 3/5. Notice the numerator is small, which means the point is close to the line despite its coordinates being large — a useful thing to say.
• Exercise 9.3 q3 (p. 167) — the distance of (−1, 1) from 12(x + 6) = 5(y − 2). Verified: the line must first be put into the general shape, becoming 12x − 5y + 82 = 0; then the numerator is the modulus of −12 − 5 + 82, that is 65, the denominator is 13, and the distance is 5. This item exists to catch students who apply the formula to coefficients read off an unrearranged equation.
• Exercise 9.3 q4 (p. 167) — points on the x-axis whose distance from x/3 + y/4 = 1 is 4 units. Verified: the line becomes 4x + 3y − 12 = 0; a point (t, 0) gives the modulus of 4t − 12 over 5 equal to 4, so 4t − 12 is ±20 and t is 8 or −2. Two points, (8, 0) and (−2, 0), one each side of the line — which is the sign split doing visible work.
• Miscellaneous Exercise q3 (p. 173) — the same line, but points on the y-axis at distance 4. Verified: a point (0, t) gives the modulus of 3t − 12 over 5 equal to 4, so t is 32/3 or −8/3.
• Exercise 9.3 q13 (p. 168) — the foot of the perpendicular from (−1, 3) to 3x − 4y − 16 = 0. Verified: the perpendicular through the point is 4x + 3y − 5 = 0, and solving the pair gives the foot at (68/25, −49/25). This is the problem the §9.4 derivation deliberately avoids, which makes it worth setting immediately afterwards.
• Exercise 9.3 q16 (p. 168) — in the triangle with A(2, 3), B(4, −1), C(1, 2), the equation and the length of the altitude from A. Verified: BC has slope −1 and equation x + y − 3 = 0; the altitude has slope 1 and equation x − y + 1 = 0; its length is the distance from A to BC, namely the modulus of 2 + 3 − 3 over √2, that is √2.
• Exercise 9.3 q17 (p. 168) — if p is the perpendicular from the origin to the line with intercepts a and b, show that the reciprocal of p² is the sum of the reciprocals of a² and b². Verified: the line clears to bx + ay − ab = 0, so p is ab over the square root of a² + b², and squaring and inverting gives the result at once.
• Exercise 9.3 q15 (p. 168) — two perpendiculars are dropped from the origin. The first, of length p, lands on the line whose x-coefficient is cos θ, whose y-coefficient is −sin θ, and whose constant side is k cos 2θ; the second, of length q, lands on the line whose x-coefficient is sec θ, whose y-coefficient is cosec θ, and whose constant side is k. Show p² + 4q² = k². Verified: p is the modulus of k cos 2θ, since the first line's coefficients square-sum to one; and q is half the modulus of k sin 2θ, since the second line's coefficients square-sum to four over sin² 2θ. Substituting gives k² times the Pythagorean identity at the doubled angle.
• Miscellaneous Exercise q4 (p. 173) — how far the origin lies, measured along the perpendicular, from the line drawn through the point (cos θ, sin θ) and the point (cos φ, sin φ). Verified: both points lie one unit from the origin, so the segment is a chord of the unit circle and the perpendicular from the centre bisects it, giving the modulus of the cosine of half the difference of the two angles. Doing it that way is far shorter than forming the line's equation, and is worth showing as an alternative.
• Miscellaneous Exercise q22 (p. 174) — show that a product equals b², the product being of two perpendiculars: one dropped from (√(a² − b²), 0), and one dropped from (−√(a² − b²), 0), each onto the line (x/a)cos θ + (y/b)sin θ = 1. Verified: clearing denominators gives bx cos θ + ay sin θ − ab = 0; the two numerators multiply to the modulus of (a² − b²)b²cos²θ − a²b², whose size is b² times a² sin²θ + b² cos²θ, and that is b² times the squared denominator. Read the line off p. 174, not off extracted text: the two fraction bars in this equation do not survive extraction, and without them the stated identity is false.
• Example 16 (p. 172) — the path of a point equidistant from 3x − 2y = 5 and 3x + 2y = 5. Verified: the two lines have the same square-summed coefficients, so the condition is that the two signed numerators have equal size; removing the modulus gives two branches, one yielding y = 0 and the other x = 5/3, and each is a line. Note the two given lines cross at (5/3, 0), and the two answers are the two bisectors of the angles there — the chapter obtains them and does not say what they are.
• Miscellaneous Exercise q19 (p. 173) — a point whose distances from x + y − 5 = 0 and 3x − 2y + 7 = 0 always sum to 10 moves on a line. Verified: on any region where both signed numerators keep a fixed sign, the sum of the two moduli is a linear expression in the coordinates set equal to a constant, which is a line by Naming Ax + By + C = 0, the one form the distance formula ahead will take.
• Example 12 (p. 169, Fig 9.16) — the distance of 4x − y = 0 from P(4, 1), measured along the line making 135° with the positive x-direction. Verified: that direction has slope −1, so the measuring line is x + y − 5 = 0; it meets the given line at Q(1, 4); and the distance PQ is 3√2, about 4.243. Verified contrast, and the chapter does not draw it: the perpendicular distance from the same point to the same line is 15/√17, about 3.638. The two numbers differ, and the perpendicular one is the smaller — as it must be.
• Miscellaneous Exercise q14 (p. 173) — the distance of 4x + 7y + 5 = 0 from (1, 2) measured along 2x − y = 0. Verified: the given point does lie on the measuring line, the crossing is at (−5/18, −5/9), and the distance is 23√5/18, about 2.857.
• Miscellaneous Exercise q23 (p. 174) — a person standing where 2x − 3y + 4 = 0 meets 3x + 4y − 5 = 0 must get to the path 6x − 7y + 8 = 0 as quickly as possible. Data intact; the point of the question is that "least time" selects the perpendicular route, which is why §9.4 defines distance the way it does.

Figures to have open

• Fig 9.14 redrawn with P, Q, R and M all labelled, the triangle dashed and the perpendicular marked. The chapter's own figure, and every label in it lives inside the artwork, so it must be rebuilt rather than traced from text.
• A three-panel check of the excluded families — horizontal, vertical, through the origin — each showing the formula's value beside a directly measured gap. Standard schematic; the chapter omits this and it is the topic's main repair.
• A two-sided panel showing one line with a point either side, each labelled with its signed numerator, the two signs opposite. Standard schematic; it makes Example 16 inevitable.
• Fig 9.16 redrawn beside a perpendicular dropped from the same point, both lengths labelled. The chapter's own figure plus one added segment.

Where this sits in the book

• NCERT Mathematics, Textbook for Class XI, Chapter 9 "Straight Lines", §9.4 Distance of a Point From a Line, pp. 164–166 — the definition, Fig 9.14, the two-way area argument and the final formula
• Example 9, p. 167
• Exercise 9.3, pp. 167–168, questions 3, 4, 13, 15, 16 and 17
• Miscellaneous Example 12, p. 169, with Fig 9.16; Miscellaneous Example 16, p. 172
• Miscellaneous Exercise on Chapter 9, pp. 173–174, questions 3, 4, 14, 19, 22 and 23
• The chapter Summary, p. 175 — the distance result restated in the same coefficients
• §9.1, pp. 151–152 — the triangle-area and two-point-distance results the derivation reuses