Chapter 10 — Straight Lines

Introduction

In the Cartesian (xy) plane a straight line is the set of points (x,y) that satisfy a linear relation between x and y. A line can be uniquely determined by: a point and a slope (gradient), two distinct points, or the intercepts it makes on the axes.

Slope (gradient) of a line

The slope m of the line through points (x1,y1) and (x2,y2) is m = (y2 − y1)/(x2 − x1), provided x1 ≠ x2. The slope is the tangent of the angle θ the line makes with the positive x-axis: m = tan θ. Horizontal lines have m = 0 and vertical lines have undefined slope.

Common Cartesian forms and derivations

• Slope-intercept form: y = mx + c. Here m is the slope and c is the y-intercept (value of y when x = 0). Derivation: using slope m and a point (0,c).
• Point-slope form: y − y1 = m(x − x1). This gives the line through (x1,y1) with slope m.
• Two-point form: (y − y1)/(y2 − y1) = (x − x1)/(x2 − x1). This is the equation of the line through (x1,y1) and (x2,y2).
• Intercept form: x/a + y/b = 1, where a and b are x- and y-intercepts (a ≠ 0, b ≠ 0).
• General (standard) form: Ax + By + C = 0. Any non-vertical/non-horizontal line can be written in this linear polynomial form. Conversion: from y = mx + c, rearrange to mx − y + c = 0 (or Ax + By + C = 0).
• Vertical and horizontal lines: Vertical: x = k (undefined slope). Horizontal: y = k (slope 0).
• Normal form: x cos α + y sin α = p, where p is the perpendicular distance from origin to the line and α is the angle the perpendicular makes with x-axis. This is useful for distance and orientation.

Relations and properties

• Parallel lines: two lines are parallel if their slopes are equal (m1 = m2) or their direction ratios are proportional.
• Perpendicular lines: two lines are perpendicular if m1 · m2 = −1 (provided neither is vertical/horizontal special cases considered).
• Angle between two lines with slopes m1 and m2: tan φ = (m2 − m1)/(1 + m1 m2).
• Distance from point (x1,y1) to line Ax + By + C = 0: distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2).

Notes

Vertical lines cannot be represented by y = mx + c because slope is undefined; they are given by x = constant. The intercept form requires nonzero intercepts. The general form is convenient for algebraic manipulations (e.g., computing distances, testing collinearity).

What is slope?
The slope (or gradient) of a straight line measures its steepness and direction. For a non-vertical line through two points (x₁, y₁) and (x₂, y₂) the slope m is

m = (y₂ − y₁) / (x₂ − x₁)

This is the "rise over run": change in y per unit change in x. Positive slope means the line rises left-to-right; negative slope means it falls. A horizontal line has m = 0. A vertical line has undefined slope (division by zero).

Angle with the positive x-axis
If a line makes an angle α with the positive x-axis measured anticlockwise, its slope is the tangent of that angle:

m = tan α

Thus α = arctan(m). Angles can be given in degrees (°) or radians. For a vertical line α = 90° (π/2) and tan α is undefined, matching the undefined slope.

Angle with the y-axis
If β is the angle the line makes with the positive y-axis (measured from +y toward the line), then β = 90° − α and

m = cot β = tan(90° − β)

Relationship between two lines
For two lines with slopes m₁ and m₂, the angle θ between the lines satisfies:

tan θ = |(m₂ − m₁) / (1 + m₁m₂)|

Special cases:

• Lines are parallel if m₁ = m₂.
• Lines are perpendicular if m₁·m₂ = −1 (provided slopes finite). In that case m₂ = −1/m₁.

Sign and quadrant conventions
The value arctan(m) gives an angle between −90° and 90°. If a line points into the second or third quadrant (i.e., x decreases while y increases or both decrease), you may add 180° to get the direction angle in [0°,360°). Always measure the angle from the positive x-axis anticlockwise for the full direction.

Geometric interpretation
On a graph, take any two points on the line, draw the right triangle with horizontal leg Δx and vertical leg Δy; slope = Δy/Δx, and the acute angle the line makes with the horizontal satisfies tan α = Δy/Δx.

Definition. The angle between two straight lines is the smaller angle formed where they meet (intersection). If the lines make angles α1 and α2 with the positive x-axis, the angle between them is θ = |α2 − α1| (taken between 0 and π).

Slopes and angle with x-axis. If a line has slope m, its angle with the positive x-axis is α = arctan(m) (principal value). For two lines with slopes m1 and m2, let their angles be α1 = arctan(m1) and α2 = arctan(m2).

Key formula (derivation sketch). Using the tangent difference identity,

• tan(α2 − α1) = (tan α2 − tan α1) / (1 + tan α1 tan α2) = (m2 − m1) / (1 + m1 m2).
• Therefore the acute angle θ between the lines satisfies tan θ = |(m2 − m1) / (1 + m1 m2)|, and θ = arctan( |(m2 − m1) / (1 + m1 m2)| ).

Alternate form using general form Ax + By + C = 0. If two lines are a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0, their angle θ satisfies

tan θ = |(a1 b2 − a2 b1) / (a1 a2 + b1 b2)|.

Special cases and observations.

• Parallel lines: m1 = m2 → θ = 0.
• Perpendicular lines: m1 m2 = −1 → denominator 1 + m1 m2 = 0, so θ = π/2 (90°).
• Vertical lines: slope is infinite. Compute angles with x-axis directly (vertical line angle = π/2) and use θ = |α2 − α1| or use general-form formula to avoid infinities.
• Always take the acute angle (0 ≤ θ ≤ π/2) for the standard “angle between two lines” unless an oriented/obtuse angle is requested.

How to compute (stepwise).

1. Find slopes m1 and m2 (from y = mx + c or m = −A/B for Ax + By + C = 0).
2. Compute t = |(m2 − m1)/(1 + m1 m2)|.
3. Angle θ = arctan(t) (in degrees or radians). If 1 + m1 m2 = 0 → θ = 90°.

A straight line in the plane can be represented in several equivalent algebraic forms. Each form highlights different geometric features (slope, intercepts, normal, passing through points). Common forms are:

• Slope-intercept form: y = mx + c. Here m is the slope (rise/run) and c is the y-intercept (point where the line meets the y-axis).
• Point-slope (or point–gradient) form: y - y1 = m(x - x1). This is the equation of the line through a known point (x1, y1) with slope m.
• Two-point form: (y - y1)/(y2 - y1) = (x - x1)/(x2 - x1). The line through points (x1, y1) and (x2, y2).
• Intercept form: x/a + y/b = 1. The line cuts the x-axis at (a, 0) and the y-axis at (0, b), provided a and b are nonzero.
• General (or standard) form: ax + by + c = 0, where not both a, b are zero. This is a very flexible form used for algebraic manipulation.
• Normal form: x cos α + y sin α = p. Here p is the perpendicular distance from the origin to the line and α is the angle the perpendicular (normal) makes with the positive x-axis.
• Vertical and horizontal lines: A vertical line has equation x = k (undefined slope). A horizontal line has equation y = c (slope = 0).
• Parametric form: x = x0 + t·u, y = y0 + t·v where vector (u, v) is direction and t is a parameter.

Key geometric relations and conversions:

• Slope between two points: m = (y2 - y1)/(x2 - x1).
• Parallel lines: slopes equal (m1 = m2). Perpendicular lines: product of slopes = -1 (m1·m2 = -1), provided slopes are finite.
• Angle θ between two lines with slopes m1, m2: tan θ = |(m2 - m1)/(1 + m1·m2)|.
• Distance of point (x0,y0) from line ax + by + c = 0: |ax0 + by0 + c| / √(a² + b²).

How to convert between forms (examples):

• Two-point → slope → point-slope → slope-intercept: compute m then use a known point.
• General → slope-intercept: solve for y: y = (-a/b)x + (-c/b) if b ≠ 0.
• Normal form → general: multiply out x cos α + y sin α = p to get (cos α)x + (sin α)y - p = 0.

Notes:

• Vertical lines cannot be written in the slope-intercept form; use x = k.
• Intercept form requires intercepts ≠ 0. If an intercept is zero, the line passes through the origin or an axis.

Overview
In coordinate geometry, a straight line can be written in many equivalent forms. Certain forms and collections (families) of lines are particularly useful because they express simple geometric properties — e.g. being parallel to an axis, passing through a fixed point, or passing through the intersection of two lines. This topic studies these special lines and how to describe families of lines using one parameter.

Common (special) forms of a line

• Slope-intercept form: y = mx + c. Here m is the slope and c is the y-intercept.
• Point-slope form (line through (x₁,y₁)): y - y₁ = m(x - x₁).
• Intercept form: x/a + y/b = 1, where a and b are the x- and y-intercepts (a ≠ 0, b ≠ 0).
• Normal form: x cos α + y sin α = p, where p is the (positive) perpendicular distance from the origin and α is the angle the normal makes with the x-axis.
• General form: ax + by + c = 0 (a and b not both zero). Many formulas (distance, slope, perpendicularity) are easiest to use in this form.

Special single lines

• Horizontal lines: y = k (slope m = 0). Parallel to x-axis.
• Vertical lines: x = k (undefined slope). Parallel to y-axis.
• Lines through origin: y = mx (passes through (0,0)).
• Lines parallel & perpendicular to a given line: If line1 has slope m₁ and line2 has slope m₂, then
  • parallel: m₂ = m₁
  • perpendicular: m₂ = −1/m₁ (provided m₁ ≠ 0)

Families of lines (one-parameter families)
A family is a set of lines described by one parameter (often m or λ). Common families:

• All lines with a given slope m (parallel family): y = mx + c, parameter c varies. These are all lines parallel to each other.
• All lines through a fixed point (x₁,y₁): y - y₁ = m(x - x₁), parameter m varies (slope). Geometrically these are all lines radiating from that point.
• All lines through the intersection of two given lines L1 = 0 and L2 = 0: L1 + λ L2 = 0 (λ is parameter). Every λ gives a line passing through the common intersection point of L1 and L2. This is useful for families sharing a point.
• Perpendicular family using normal form: x cos α + y sin α = p, where varying p moves the line parallel to itself (same normal direction) — yields lines perpendicular to the direction given by α.

Useful related results

• Slope between two points (x₁,y₁) and (x₂,y₂): m = (y₂ − y₁)/(x₂ − x₁), provided x₂ ≠ x₁.
• Angle θ between two lines of slopes m₁ and m₂: tan θ = |(m₂ − m₁)/(1 + m₁ m₂)|.
• Distance of point (x₀,y₀) from line ax + by + c = 0: d = |ax₀ + by₀ + c|/√(a² + b²).

Why these are important
Special lines simplify calculations and geometric reasoning (e.g., finding intersections, distances, measuring angles, constructing families for locus problems). Families let us describe an infinite set of lines compactly and are widely used in locus, envelope, and optimization problems.

Intersection of two lines

Two straight lines in the plane are represented in their general form as a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0. The intersection of these two lines is the point (x, y) that satisfies both equations simultaneously.

There are three geometric possibilities:

• Unique intersection: if the lines have different slopes (i.e. a1 b2 - a2 b1 ≠ 0) they meet at exactly one point.
• No intersection (parallel): if slopes are equal (a1 b2 - a2 b1 = 0) but the constant terms are not proportional (a1/a2 = b1/b2 ≠ c1/c2), the lines are parallel and do not meet.
• Infinite intersections (coincident): if both the left-hand coefficients are proportional (a1/a2 = b1/b2 = c1/c2), the lines coincide (are the same line) and have infinitely many common points.

How to find the intersection point (algebraically)

Use substitution, elimination, or Cramer’s rule. For the pair
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0, set the system as a1 x + b1 y = -c1 and a2 x + b2 y = -c2. If D = a1 b2 - a2 b1 ≠ 0, the unique intersection is

x = (b1 c2 - b2 c1) / D, y = (a2 c1 - a1 c2) / D.

Handle vertical lines (x = constant) and horizontal lines (y = constant) by direct substitution as special cases.

Concurrency (common intersection of three or more lines)

Three (or more) lines are said to be concurrent if they all pass through a single common point. For three lines
a1 x + b1 y + c1 = 0,
a2 x + b2 y + c2 = 0,
a3 x + b3 y + c3 = 0, they are concurrent iff the following determinant is zero:

| a1 b1 c1 |
| a2 b2 c2 | = 0
| a3 b3 c3 |

This condition comes from substituting the intersection (obtained from the first two lines) into the third line (equivalently, the third is a linear combination of the first two at the intersection point). Classic geometric examples of concurrency include medians of a triangle (meet at the centroid), angle bisectors (incenter), and altitudes (orthocenter).

Practical note: For computation use elimination or Cramer’s rule; for drawing and visualization use plotting tools (Desmos, GeoGebra) or graph paper to mark lines and the common point.

What is meant by distance?
In coordinate geometry, distance means the length of the straight line segment joining two points or the shortest (perpendicular) distance from a point to a line. Distance formulas use the Pythagorean theorem and projections to give algebraic expressions for these lengths.

1. Distance between two points (x1, y1) and (x2, y2)
Consider the horizontal separation Δx = x2 − x1 and the vertical separation Δy = y2 − y1. These form the legs of a right triangle whose hypotenuse is the required distance. By Pythagoras,

Distance = √((x2 − x1)² + (y2 − y1)²).

Derivation (brief): Draw the line segment between the points and drop a vertical and horizontal to form a right triangle. The legs are |Δx| and |Δy|, so the hypotenuse length is √(Δx² + Δy²).

2. Perpendicular distance from point (x0, y0) to line ax + by + c = 0
The shortest distance from a point to a line is measured along the perpendicular. The algebraic formula is:

Distance = |a x0 + b y0 + c| / √(a² + b²).

Reason (idea): The numerator gives the signed value of the line equation at the point; dividing by the length √(a²+b²) normalizes by the line's normal vector length, producing the perpendicular length.

3. Distance between two parallel lines
If two lines are parallel and written as a x + b y + c1 = 0 and a x + b y + c2 = 0 (same a and b), then the distance between them is

Distance = |c2 − c1| / √(a² + b²).

Notes and related facts: midpoint of segment connecting (x1,y1) and (x2,y2) is ((x1+x2)/2, (y1+y2)/2) — often plotted when showing distance. Distances are always nonnegative. In three dimensions the point-to-point distance generalizes to √((Δx)²+(Δy)²+(Δz)²).

Definition (Foot of perpendicular): For a point P(x1,y1) and a line L: ax+by+c=0, the foot of the perpendicular from P to L is the point H on L such that PH is perpendicular to L. H is the nearest point on L to P.

Definition (Projection): Projection can mean:

• Scalar projection of vector u on vector v: the signed length of u onto v, given by (u·v)/|v|.
• Vector projection (component) of u on v: the vector along v equal to ((u·v)/(v·v)) v.
• Projection of a point P on a line L: the foot H (same as above).

Derivation (coordinate formula for foot): Let P(x1,y1) and L: ax+by+c=0. The signed distance from P to L is D = (ax1+by1+c)/√(a^2+b^2). The foot H(x0,y0) lies on L and on the line through P perpendicular to L, so moving from P toward L by the perpendicular distance scaled by the unit normal gives:

x0 = x1 - a*(ax1+by1+c)/(a^2+b^2),
y0 = y1 - b*(ax1+by1+c)/(a^2+b^2).

These satisfy ax0+by0+c=0 and PH is perpendicular to L.

Alternate methods:

• Using slopes: If L has slope m (y=mx+c), then perpendicular slope is -1/m; find equation of perpendicular through P and intersect with L.
• Parametric/vector method: If line L is r = a + t d (a is a point on L, d is direction), then parameter for foot t0 = d·(P-a) / (d·d), and foot H = a + t0 d.

Relation to projection of vectors: If u and v are vectors, the scalar projection of u on v is (u·v)/|v| and the vector projection (component) is proj_v(u) = ((u·v)/(v·v)) v. For a point projection on a line, this vector formula applied to position vectors gives the foot.

Special cases: For vertical line x=k, foot of (x1,y1) is (k,y1). For horizontal line y=k, foot is (x1,k). For the origin to ax+by+c=0, foot is (-ac/(a^2+b^2), -bc/(a^2+b^2)).

Geometry insight: The foot H minimizes distance from P to any point on L. The segment PH is perpendicular to L and has length |ax1+by1+c|/√(a^2+b^2).

Definition. If two straight lines meet at a point, the lines that divide each of the angles formed into two equal angles are called the angle bisectors of the two lines. The locus of points equidistant from the two given lines is the pair of angle bisectors.

Derivation / Key idea. Let the two lines be L1: a1 x + b1 y + c1 = 0 and L2: a2 x + b2 y + c2 = 0. The perpendicular distance of a point (x,y) from L1 is |a1 x + b1 y + c1|/sqrt(a1^2 + b1^2) and from L2 is |a2 x + b2 y + c2|/sqrt(a2^2 + b2^2). Points on the angle bisectors are equidistant from the two lines, so

|a1 x + b1 y + c1| / sqrt(a1^2 + b1^2) = |a2 x + b2 y + c2| / sqrt(a2^2 + b2^2).

Removing the absolute values gives the two bisector equations (the ± cases):

(a1 x + b1 y + c1)/sqrt(a1^2 + b1^2) = ± (a2 x + b2 y + c2)/sqrt(a2^2 + b2^2).

Which sign (acute / obtuse)? The two signs give the two bisectors. To identify the bisector of the acute angle: take any test point (for example the origin if it is not on either line) and evaluate the left-hand expressions E1 = a1 x + b1 y + c1 and E2 = a2 x + b2 y + c2. The bisector of the acute angle consists of points where E1 and E2 have the same sign (their product > 0) and corresponds to the + sign in the normalized equation; the obtuse-angle bisector corresponds to the − sign (where E1 and E2 have opposite signs). If the chosen test point lies on one bisector, use another test point.

Special cases. If the lines are parallel (a1:b1 = a2:b2) and distinct, the locus of points equidistant from both lines is a single line exactly midway between them (parallel to both). If the lines are perpendicular, the bisectors are also perpendicular (they coincide with the coordinate axes when the two lines are y = x and y = -x).

Connection with slopes and angles. If slopes of the lines are m1 and m2, the angle θ between them satisfies tan θ = |(m1 - m2)/(1 + m1 m2)|. The bisectors will have slopes that can be found by solving the bisector equation or by using angle half-angle formulae if needed.

Usage tip. Always put lines in the form a x + b y + c = 0 before applying the normalized formula. Use the sign-test to decide which of the two ± equations is the acute bisector.

Basic idea (homogeneous case)

A homogeneous second-degree equation in x and y,

ax² + 2hxy + by² = 0,

represents a pair of straight lines passing through the origin (0,0) provided it can be factorized as the product of two linear factors:

(l₁x + m₁y)(l₂x + m₂y) = 0.

To find the slopes of the two lines, substitute y = m x (x ≠ 0) into the homogeneous equation. This gives the quadratic in m:

b m² + 2h m + a = 0.

Its roots m₁, m₂ are the slopes of the two lines, so the lines are y = m₁x and y = m₂x.

Conditions on the nature of lines

• If h² > a b, the two lines are real and distinct.
• If h² = a b, the two lines are real and coincident (repeated line).
• If h² < a b, the pair is imaginary (no real lines).

Angle between the two lines

If θ is the acute angle between the two lines,

tan θ = 2 √(h² − ab) / (a + b),

provided a + b ≠ 0. (Use formula via slopes if a + b = 0.)

Factorization method

1) Solve b m² + 2h m + a = 0 for m₁, m₂ (slopes).
2) Then the pair is (y − m₁x)(y − m₂x) = 0, or equivalently rewrite as linear factors in x,y.

Combined forms (general pair of lines)

The general second-degree equation

ax² + 2hxy + by² + 2gx + 2fy + c = 0

represents a pair of straight lines (not necessarily through origin) iff the determinant

| a h g | | h b f | | g f c | = 0.

When that holds, the left-hand side can be factorized as (L₁)(L₂) = 0 where each L is linear in x and y: L = αx + βy + γ.

How to factor a general pair

Find two linear factors either by algebraic factorization or by finding two distinct lines whose product expands to the given quadratic. One systematic approach: if the pair meets the determinant condition, try to find the intersection point (solve the partial derivatives/linear system) or reduce by a translation to make it homogeneous and then factor as in the homogeneous case.

In Coordinate Geometry (Class 11 Straight Lines) the straight line is used to model many simple geometric and real-world relationships. Typical problem types use the line's forms and properties: finding equation from points or slope, checking parallelism/perpendicularity, computing angles and distances, finding foot of a perpendicular, loci that are straight lines, and families of lines. Key ideas: slope (gradient) measures steepness; point–slope, two–point and intercept forms give equations; the general form ax+by+c=0 is convenient for algebraic manipulations. Applications include measuring shortest distance (perpendicular), constructing parallel/perpendicular elements in design, and modeling linear relations in physics, engineering and economics.

Common problem strategies:

• Find slope m from two points: m = (y2−y1)/(x2−x1) and use point–slope form y−y1 = m(x−x1).
• Use general form ax+by+c=0 to compute distance from a point, angle between lines, and foot of perpendicular using direct formulas.
• Test parallelism (equal slopes or proportional coefficients) and perpendicularity (product of slopes = −1 or a1a2 + b1b2 = 0).
• Use determinant (area) test for collinearity or form family of lines through intersection: L1 + λL2 = 0.

These techniques solve geometric construction problems (e.g., draw a line through a point parallel to a given line), optimization of shortest paths (distance to a line), surveying and navigation (angles and bearings), and many applied tasks in engineering, architecture and robotics.