Straight Lines

Take any two dots on a sheet of graph paper and draw the one path between them that never bends: a straight line. It looks like the simplest object in geometry, and in a sense it is — yet this chapter shows that a single number and a single equation are enough to capture everything about it. That number is the slope, the tangent of the angle a line makes with the positive x-axis, and once you know a line's slope and one point on it, the entire infinite line is pinned down algebraically. This chapter is where coordinate geometry stops being a way to plot figures and becomes a way to compute with them. You already know how to find the distance between two points, divide a segment in a given ratio, and compute the area of a triangle from its vertices — the tools carried over from earlier classes. Here those tools are put to work on lines specifically. You will learn to read off a line's slope from two points or from its inclination, to test whether two lines are parallel or perpendicular purely from their slopes, and to measure the angle between any two intersecting lines. Then comes the heart of the chapter: five different ways to write down the equation of a line — point-slope, two-point, slope-intercept, intercept, and the all-purpose general form — each suited to a different kind of given information. Finally you learn to measure how far a point sits from a line, and how far apart two parallel lines are, both reduced to one clean formula. Everything here is pure algebra standing in for geometry: a line becomes a linear equation, a question about angles becomes a question about slopes, and a question about distance becomes a question about substituting coordinates into a formula. This is also the direct staircase into the next chapter, where circles, parabolas, ellipses and hyperbolas are built by asking the same kind of question — what equation captures this curve — for shapes that bend.

Key takeaways

The slope of a line is $m = \tan\theta$ , where $\theta$ is its inclination, the angle measured anticlockwise from the positive x-axis, $0° \le \theta \le 180°$ ; a horizontal line has $m = 0$ and a vertical line has undefined slope.
Given two points $(x_1,y_1)$ and $(x_2,y_2)$ on a line, the slope is $m = \dfrac{y_2-y_1}{x_2-x_1}$ , valid whenever $x_1 \neq x_2$ .
Two lines are parallel if and only if their slopes are equal, $m_1 = m_2$ ; two lines are perpendicular if and only if the product of their slopes is $-1$ , $m_1 m_2 = -1$ .
The acute angle between two lines of slopes $m_1$ and $m_2$ satisfies $\tan\theta = \left|\dfrac{m_2-m_1}{1+m_1m_2}\right|$ .
A line is fixed by a point and a slope (point-slope form, $y - y_0 = m(x-x_0)$ ), by two points (two-point form), by a slope and a y-intercept (slope-intercept form, $y = mx + c$ ), or by its two axis intercepts (intercept form, $\dfrac{x}{a} + \dfrac{y}{b} = 1$ ).
Every line, including vertical ones, can be written as the general form $Ax + By + C = 0$ , where $A$ and $B$ are not both zero.
The perpendicular distance of a point $(x_1,y_1)$ from the line $Ax+By+C=0$ is $d = \dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ .
The distance between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is $d = \dfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}$ .
Three points $A$ , $B$ , $C$ are collinear exactly when the slope of $AB$ equals the slope of $BC$ , or equivalently when the area of triangle $ABC$ works out to zero.
Key idea: a straight line is completely determined by one linear equation in x and y, and every geometric question about it — parallelism, perpendicularity, angle, or distance — reduces to algebra on the slope and the coefficients A, B, C of that equation.

The core insight

A line's slope $m = \tan\theta$ (the tangent of its inclination) fixes its direction, so parallelism means equal slopes, perpendicularity means $m_1m_2=-1$ , and any line can be written in point-slope, two-point, slope-intercept, intercept, or general form $Ax+By+C=0$ , from which the distance of any point to the line follows as $d = |Ax_1+By_1+C|/\sqrt{A^2+B^2}$ .

What this chapter covers

The chapter opens with a short recap of tools from earlier classes — the distance formula, the section formula, the midpoint formula, and the area of a triangle from its vertices — because every later result leans on them. It then defines the inclination and slope of a line, first from the angle a line makes with the x-axis and then, more usefully, from any two points on it. Building on slope, it states and proves the conditions for two lines to be parallel or perpendicular, and derives a formula for the angle between two intersecting lines. The centrepiece is a catalogue of five equivalent ways to write the equation of a line — horizontal and vertical lines as special cases, then point-slope, two-point, slope-intercept, intercept, and the general form $Ax+By+C=0$ that subsumes every case, including vertical lines that have no defined slope. It closes with two distance results: the perpendicular distance of a point from a line, and the distance between two parallel lines. Miscellaneous examples then apply all of this to concurrency of three lines, reflections of points in a line, and loci of points equidistant from two given lines.

The core ideas

Recap: distance, section formula and collinearity

Before building anything new, recall the four results on which this whole chapter rests: how far apart two points are, where a point dividing a segment in a given ratio sits, where the midpoint of a segment sits, and how to test three points for collinearity using area.

PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}, \qquad \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right), \qquad \text{Area} = \frac{1}{2}\big|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big|. \qquad (1)

where: $P(x_1,y_1)$ and $Q(x_2,y_2)$ are any two points, so the first expression is the ordinary distance formula. The second is the point dividing the segment joining $(x_1,y_1)$ and $(x_2,y_2)$ internally in the ratio $m:n$ ; setting $m=n$ gives the midpoint formula $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ . The third is the area of the triangle with vertices $(x_1,y_1)$ , $(x_2,y_2)$ , $(x_3,y_3)$ ; if this area is exactly zero, the three points lie on one straight line — the standard algebraic test for collinear points, used whenever this chapter must check that a point lies on a line.

Slope of a line

A line drawn across the coordinate plane makes two supplementary angles with the x-axis (one on each side); by convention we always measure the one obtained by rotating anticlockwise from the positive x-axis to the line. This angle carries the line's entire sense of "steepness and direction," and turning it into a single number is the first job of the chapter.

m = \tan\theta, \quad \theta \neq 90°, \qquad\qquad m = \frac{y_2-y_1}{x_2-x_1}, \quad x_1 \neq x_2. \qquad (2)

where: $\theta$ is the inclination of the line, the angle measured anticlockwise from the positive x-axis, restricted to $0° \le \theta \le 180°$ , and $m$ is the slope (or gradient). The first form defines slope directly from the inclination; the second gives the same slope computed from any two points $(x_1,y_1)$ and $(x_2,y_2)$ on the line, and the first derivation below proves the two forms agree. A line parallel to the x-axis has $\theta = 0°$ so $m=0$ ; a line parallel to the y-axis has $\theta = 90°$ , where $\tan\theta$ is undefined, so a vertical line has no slope at all — not zero, but genuinely undefined, since $x_1=x_2$ would force division by zero. This is the single most common source of error in the chapter and returns in the traps section below.

Figure

A line making an inclination angle theta with the positive x axis with two points P and Q marked on it and the right triangle formed by dropping a horizontal and a vertical leg between them showing slope as rise over run

Parallel lines, perpendicular lines, and the angle between two lines

Slope is what makes it possible to test the relationship between two lines without ever drawing them. Two non-vertical lines with slopes $m_1$ and $m_2$ have inclinations $\alpha$ and $\beta$ ; whether the lines are parallel, perpendicular, or meeting at some other angle is entirely a statement about how $\alpha$ and $\beta$ compare.

m_1 = m_2 \iff \text{lines parallel}, \qquad\qquad m_1 m_2 = -1 \iff \text{lines perpendicular}. \qquad (3)

where: parallel lines have equal inclination ( $\alpha = \beta$ ), hence equal slope, and the converse also holds; perpendicular lines have inclinations differing by exactly $90°$ , which algebraically forces $m_1 m_2 = -1$ (negative reciprocal slopes) — proved in full below. Three points $A$ , $B$ , $C$ are collinear precisely when the slope of $AB$ equals the slope of $BC$ , since two segments through a common point with equal slope must lie along the same line. When the lines are neither parallel nor perpendicular, the angle between them is still computable from their slopes alone:

\tan\theta = \left|\frac{m_2-m_1}{1+m_1m_2}\right|, \qquad 1+m_1m_2 \neq 0. \qquad (4)

where: $\theta$ is taken to be the acute angle between the two lines (the obtuse angle at the same intersection is simply $180° - \theta$ ), and the condition $1 + m_1m_2 \neq 0$ rules out the perpendicular case, where the angle formula would try to divide by zero — exactly matching the perpendicularity condition in equation $(3)$ . Notice that swapping $m_1$ and $m_2$ flips the sign inside the absolute value but not the final answer, which is why the formula is normally quoted with an absolute value or with the understanding that you report the acute angle.

The equation of a line: five standard forms

A line contains infinitely many points, so its "equation" is a single algebraic condition on $x$ and $y$ that every point of the line satisfies and no other point does. Which form is most convenient depends entirely on what information about the line you are given — a point and a slope, two points, a slope and an intercept, or two intercepts — and every one of them can be rewritten as the same general linear equation.

\begin{aligned} &\text{point-slope: } y - y_0 = m(x-x_0), &&\text{two-point: } y-y_1 = \frac{y_2-y_1}{x_2-x_1}(x-x_1),\\ &\text{slope-intercept: } y = mx+c, &&\text{intercept: } \frac{x}{a}+\frac{y}{b}=1,\\ &\text{general form: } Ax+By+C=0. && \end{aligned} \qquad (5)

where: in point-slope form, $(x_0,y_0)$ is one known point on the line and $m$ its slope; in two-point form, $(x_1,y_1)$ and $(x_2,y_2)$ are two known points, with the slope built into the formula itself. In slope-intercept form, $c$ is the y-intercept, the signed distance at which the line meets the y-axis; a line with slope $m$ and x-intercept $d$ instead satisfies $y = m(x-d)$ . In intercept form, $a$ is the x-intercept and $b$ the y-intercept, so the line passes through $(a,0)$ and $(0,b)$ ; this form fails for a line through the origin or parallel to either axis, since one or both intercepts do not then exist as finite non-zero numbers. Horizontal and vertical lines sit outside this family: a horizontal line at distance $a$ from the x-axis is $y = a$ , and a vertical line at distance $b$ from the y-axis is $x = b$ — a vertical line cannot be written in any slope-based form, precisely because its slope is undefined. Every one of these forms can be cleared of fractions and rearranged into the general form $Ax + By + C = 0$ , where $A$ and $B$ are not simultaneously zero; this is the only form flexible enough to represent every line, including vertical ( $B=0$ ) and horizontal ( $A=0$ ) ones, and it is the form the distance formulas below use.

Distance of a point from a line, and between two parallel lines

The last algebraic tool this chapter builds is a way to measure separation: how far a given point sits from a given line, and, as a direct consequence, how far apart two parallel lines are.

d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}, \qquad\qquad d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}. \qquad (6)

where: in the first formula, $d$ is the length of the perpendicular dropped from the point $(x_1,y_1)$ onto the line $Ax+By+C=0$ — the shortest possible distance from that point to any point of the line — and the absolute value ensures $d$ is never negative regardless of which side of the line the point sits on; this is proved by an area-of-a-triangle argument in the second derivation below. In the second formula, $d$ is the constant perpendicular distance between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ sharing the same coefficients $A$ and $B$ (a necessary condition for two lines to be parallel in general form) — it follows immediately from the first formula by picking any convenient point on one line and measuring its distance to the other.

Figure

A line L and a point P off the line with a dashed perpendicular segment dropped from P meeting the line at right angles at a foot M labelled with the distance d

Two derivations and a third worked in full

Derivation AThe slope formula from two points on a line

The chapter defines slope as $m=\tan\theta$ from the inclination $\theta$ , but the far more usable formula in practice is $m = (y_2-y_1)/(x_2-x_1)$ computed straight from two points, with no angle in sight. The goal here is to prove these two descriptions of slope are one and the same thing, by dropping a right triangle onto the line and reading off $\tan\theta$ from its legs, and to show the formula survives unchanged whether the inclination is acute or obtuse — a subtlety many students miss, assuming the formula secretly changes sign depending on the picture.

Step 1. Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ be two points on a non-vertical line $l$ of inclination $\theta$ , so $x_1 \neq x_2$ . Draw $QR$ perpendicular to the x-axis and $PM$ perpendicular to $QR$ , forming a right triangle $PMQ$ with legs $PM$ (horizontal) and $MQ$ (vertical) and hypotenuse $PQ$ along the line itself.

Step 2. Consider first the case where $\theta$ is acute. Then the angle at $P$ inside the triangle equals $\theta$ itself, so basic trigonometry gives $\tan\theta = \dfrac{MQ}{PM}$ . Reading the legs off the coordinates, $MQ = y_2-y_1$ (the vertical rise) and $PM = x_2-x_1$ (the horizontal run), so $m = \tan\theta = \dfrac{y_2-y_1}{x_2-x_1}$ .

Step 3. Now consider $\theta$ obtuse. The angle inside the right triangle at $P$ is no longer $\theta$ itself but its supplement, $\angle MPQ = 180° - \theta$ , because the line leans the other way. So $\theta = 180° - \angle MPQ$ , and using $\tan(180°-x) = -\tan x$ : $m = \tan\theta = -\tan\angle MPQ = -\dfrac{MQ}{PM}$ . Here the horizontal leg runs the opposite way, $PM = x_1-x_2$ , while $MQ = y_2-y_1$ still, so $m = -\dfrac{y_2-y_1}{x_1-x_2} = \dfrac{y_2-y_1}{x_2-x_1}$ — the identical formula.

Step 4. Both cases collapse to the same expression, so regardless of whether the line rises gently or steeply, or leans left or right, the slope through any two of its points is always $m = \dfrac{y_2-y_1}{x_2-x_1}$ . As a check, swapping the labels of $P$ and $Q$ flips the sign of both numerator and denominator, leaving $m$ unchanged — the formula does not care which of the two points is called "first."

Go deeper · Derivation B · 1 minPerpendicular distance from a point to a line

Derivation BPerpendicular distance from a point to a line

The chapter needs a formula for how far a point $(x_1,y_1)$ sits from a line $L: Ax+By+C=0$ , and the cleanest route uses no calculus and no trigonometric projection — only the area of a triangle computed two different ways. The goal is to set up a triangle whose one side is a segment of $L$ and whose opposite vertex is the given point, compute its area by the coordinate formula and by "half base times height," and equate the two to isolate the height, which is exactly the perpendicular distance wanted.

Step 1. Let $L$ meet the x-axis at $Q\left(-\frac{C}{A},0\right)$ and the y-axis at $R\left(0,-\frac{C}{B}\right)$ (set $y=0$ and $x=0$ in $Ax+By+C=0$ ). With $P(x_1,y_1)$ , these form a triangle $PQR$ , and $QR$ is a chord of the line $L$ whose distance from $P$ we want.

Step 2. Compute the area of $PQR$ using the coordinate area formula from equation $(1)$ , substituting the three vertices. After collecting terms, the area simplifies to $\text{Area}(PQR) = \dfrac{|C|}{2AB}\,\big|Ax_1+By_1+C\big|$ — every term factors neatly because $Q$ and $R$ both satisfy $Ax+By+C=0$ .

Step 3. Compute the same area a second way, as half the base $QR$ times the perpendicular height from $P$ to $QR$ (exactly the distance $d$ wanted, since $QR$ lies along $L$ ): $\text{Area}(PQR) = \frac12 \cdot QR \cdot d$ . The length $QR$ follows from the distance formula on $Q$ and $R$ : $QR = \dfrac{|C|}{|AB|}\sqrt{A^2+B^2}$ .

Step 4. Equate the two expressions for the area and solve for $d$ : $\dfrac{|C|}{2AB}|Ax_1+By_1+C| = \dfrac12 \cdot \dfrac{|C|}{|AB|}\sqrt{A^2+B^2} \cdot d$ . The factor $|C|/|AB|$ cancels from both sides, leaving $d = \dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ — the perpendicular distance formula, holding for every point and every line in general form, including lines through the origin (checked separately by a limiting argument).

Go deeper · Derivation C · 1 minThe angle between two lines from their slopes

Derivation CThe angle between two lines from their slopes

The chapter states that the acute angle between two lines of slopes $m_1$ and $m_2$ satisfies $\tan\theta = \left|\dfrac{m_2-m_1}{1+m_1m_2}\right|$ , and the goal is to derive this from nothing more than the tangent subtraction formula, tracking carefully which of the two angles formed at an intersection is acute and which is obtuse, since only the acute one is reported by convention.

Step 1. Let two non-vertical lines $L_1$ and $L_2$ have inclinations $\alpha_1$ and $\alpha_2$ , so $m_1 = \tan\alpha_1$ and $m_2 = \tan\alpha_2$ . Two crossing lines form two pairs of vertically opposite angles, and any two adjacent angles add to $180°$ ; call one $\theta$ and its adjacent partner $\phi = 180° - \theta$ .

Step 2. The angle $\theta$ , the angle you would rotate $L_1$ through to align it with $L_2$ , is $\theta = \alpha_2 - \alpha_1$ . Apply the tangent subtraction identity: $\tan\theta = \tan(\alpha_2-\alpha_1) = \dfrac{\tan\alpha_2 - \tan\alpha_1}{1+\tan\alpha_1\tan\alpha_2} = \dfrac{m_2-m_1}{1+m_1m_2}$ , valid whenever $1+m_1m_2 \neq 0$ .

Step 3. For the adjacent angle, $\tan\phi = \tan(180°-\theta) = -\tan\theta = -\dfrac{m_2-m_1}{1+m_1m_2}$ . So the two angles at the intersection have tangents that are negatives of each other — one positive, one negative, since neither can be zero unless the lines coincide.

Step 4. Whichever of $\theta,\phi$ has a positive tangent is the acute angle, and the other, with negative tangent, is the obtuse one. Reporting only the acute angle by convention, its tangent is the absolute value of either expression: $\tan\theta_{\text{acute}} = \left|\dfrac{m_2-m_1}{1+m_1m_2}\right|$ . When $1+m_1m_2 = 0$ , this blows up to $\tan\theta = \infty$ , meaning $\theta = 90°$ — exactly recovering the perpendicularity condition $m_1m_2=-1$ from equation $(3)$ as the special case where the angle itself is a right angle.

Worked examples

Board level - slope, perpendicularity and the unknown coordinate it forces (Example 1 and Example 3 pattern)

Find the slope of the line through the points $(3,-2)$ and $(-1,4)$ . Then, given that a second line through the points $(8,12)$ and $(x,24)$ is perpendicular to it, find $x$ .

Step 1. Apply the two-point slope formula from equation $(2)$ to the first pair: $m_1 = \dfrac{4-(-2)}{-1-3} = \dfrac{6}{-4} = -\dfrac{3}{2}$ .

Step 2. Apply the same formula to the second pair, leaving $x$ as an unknown: $m_2 = \dfrac{24-12}{x-8} = \dfrac{12}{x-8}$ .

Step 3. Since the two lines are perpendicular, equation $(3)$ requires $m_1 m_2 = -1$ : $\left(-\dfrac{3}{2}\right)\left(\dfrac{12}{x-8}\right) = -1$ , which simplifies to $\dfrac{-18}{x-8} = -1$ , so $x-8 = 18$ .

Step 4. Solve the linear equation: $x = 26$ . Check by recomputing $m_2 = 12/(26-8) = 12/18 = 2/3$ , and indeed $m_1 m_2 = (-3/2)(2/3) = -1$ , confirming perpendicularity. Answer: slope of the first line is -3/2, and x = 26.

Board level - equation of a line from a point and slope, and from two intercepts (Example 5 and Example 8 pattern)

Find the equation of the line through $(-2,3)$ with slope $-4$ . Separately, find the equation of the line which makes intercepts $-3$ and $2$ on the x-axis and y-axis respectively.

Step 1. For the first line, apply point-slope form from equation $(5)$ with $(x_0,y_0)=(-2,3)$ and $m=-4$ : $y - 3 = -4(x-(-2)) = -4(x+2)$ .

Step 2. Expand and collect into general form: $y - 3 = -4x - 8$ , so $4x + y + 5 = 0$ .

Step 3. For the second line, apply intercept form from equation $(5)$ with $a = -3$ and $b = 2$ : $\dfrac{x}{-3} + \dfrac{y}{2} = 1$ .

Step 4. Clear denominators by multiplying through by $-6$ (the LCM of $3$ and $2$ , with sign chosen to keep the leading coefficient positive): $2x - 3y + 6 = 0$ . Answer: 4x + y + 5 = 0, and 2x - 3y + 6 = 0.

Competitive level (JEE / NEET) - distance from a point to a line measured along a given direction (Example 12 pattern)

Find the distance of the line $4x-y=0$ from the point $P(4,1)$ , measured along the line making an angle of $135°$ with the positive x-axis — not the perpendicular distance, but the distance along this specific slanted path.

Step 1. Recognise that "distance measured along a given direction" is not the perpendicular-distance formula from equation $(6)$ ; instead it means finding where a second line, of the stated direction and passing through $P$ , meets the first line, and then finding the ordinary distance between $P$ and that intersection point. The slope of the second line is $\tan 135° = -1$ .

Step 2. Write the equation of this second line through $P(4,1)$ with slope $-1$ , using point-slope form: $y - 1 = -1(x-4)$ , which simplifies to $x + y - 5 = 0$ .

Step 3. Solve the two equations $4x - y = 0$ and $x+y-5=0$ simultaneously to find their intersection $Q$ . From the first, $y = 4x$ ; substituting into the second gives $x + 4x - 5 = 0$ , so $x = 1$ and $y = 4$ . Thus $Q(1,4)$ .

Step 4. The required distance is the ordinary distance between $P(4,1)$ and $Q(1,4)$ : $PQ = \sqrt{(1-4)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units. Answer: the distance is 3√2 units. Timing note: budget about two minutes. The trap here is reaching straight for the perpendicular-distance formula the instant you see "distance...from a line"; the phrase "measured along the line making an angle of 135°" signals a second line through P at that slope, an intersection, then the ordinary two-point distance formula — a two-step detour the perpendicular formula alone cannot shortcut. Underline the direction phrase before choosing a formula.

Common traps

The single most repeated error in this chapter is forgetting that a vertical line has no slope — not zero, but genuinely undefined — so no formula built from $m$ (point-slope, slope-intercept, the parallel/perpendicular tests, the angle formula) applies to it without first checking $x_1 \neq x_2$ ; the safe fallback is always $x = a$ in general form. A related slip is applying the intercept form $x/a+y/b=1$ to a line through the origin or parallel to an axis, where one or both intercepts do not exist — point-slope or general form must be used instead. Students often drop the absolute value bars in the perpendicular-distance formula, reporting a negative distance; distance is a length and must be non-negative. In the angle-between-two-lines formula, forgetting the absolute value leads to reporting an obtuse angle where an acute one was asked for. Another confusion is mixing up the parallel condition ( $m_1=m_2$ ) with the perpendicular condition ( $m_1m_2=-1$ ) — parallel lines "agree," perpendicular lines "invert and flip sign." Clearing fractions in intercept form can also multiply through by a negative number, flipping every sign in the resulting equation, so always re-check the final equation against the original point or intercepts. A last trap is reaching for the perpendicular-distance formula whenever a question says "distance...from a line," when a phrase like "measured along a direction making some angle" (as in the competitive worked example above) demands a different two-step construction, not a direct substitution.

Common trapA vertical line has no slope at all — it is not the same as a slope of zero

The single biggest misconception in this chapter is treating "slope is undefined" and "slope is zero" as interchangeable, or worse, silently assuming every line has some numerical slope you can compute. A horizontal line genuinely has slope zero ( $m=0$ , since $\tan 0° = 0$ ): it is flat, and every point-slope or slope-intercept formula works fine on it. A vertical line is a completely different situation: its inclination is $90°$ , where the tangent function is undefined, so there is no number $m$ to substitute into any slope-based formula at all. Before applying any formula from equation $(2)$ through $(5)$ that contains an $m$ , check whether the line in question could be vertical (in coordinate terms, whether $x_1 = x_2$ for the two given points); if so, the only correct equation for that line is $x =$ (the shared x-coordinate), written directly in general form as $x - x_1 = 0$ , bypassing slope entirely.

You should be able to…

State the definition of inclination and slope, and compute slope from two given points.
Explain why a horizontal line has slope zero while a vertical line has no defined slope.
Test lines for parallelism or perpendicularity using only their slopes, and find the acute angle between two intersecting lines.
Test three points for collinearity, either by comparing slopes or by the coordinate area formula.
Write the equation of a line from a point and a slope, from two points, from a slope and an intercept, or from both intercepts, and convert freely between these and the general form.
Recognise when intercept form cannot be used, and fall back on point-slope or general form instead.
Compute the perpendicular distance of a point from a line, and the distance between two parallel lines, from the general-form coefficients.
Solve applied problems on concurrency of three lines, the image of a point in a line, and distance measured along a specified direction.

Ask the tutor: Want to work through finding the equation of a line from a trickier combination of conditions, or practise the distance-along-a-direction question step by step together?

Why does a vertical line have no slope, while a horizontal line has slope exactly zero?

Slope is defined as $\tan\theta$ , where $\theta$ is the line's inclination. For a horizontal line, $\theta = 0°$ , and $\tan 0° = 0$ exactly — a well-defined number, so its slope genuinely is zero. For a vertical line, $\theta = 90°$ , and the tangent function is not defined there at all, so there is no finite number to call the slope. In the two-point formula $m=(y_2-y_1)/(x_2-x_1)$ , this shows up as division by zero, since a vertical line has $x_1=x_2$ for every pair of its points. So "slope zero" and "slope undefined" describe two opposite, equally important cases, and mixing them up is the single most common slip in this chapter.

How do I remember which condition goes with parallel lines and which with perpendicular lines?

Parallel lines point in exactly the same direction, so they share the same slope: $m_1=m_2$ . Perpendicular lines are rotated $90°$ from each other, and rotating a direction by $90°$ flips the slope upside down and changes its sign — exactly what "negative reciprocal" means. So if one line has slope $m$ , a line perpendicular to it has slope $-1/m$ , which rearranges to the symmetric form $m_1 m_2 = -1$ . A quick check: slope $2$ and slope $-1/2$ multiply to $2 \times (-1/2) = -1$ , confirming perpendicularity, while slope $2$ and slope $2$ are simply parallel, since $2 \times 2 = 4 \neq -1$ .

Why does the general form Ax + By + C = 0 need "A and B not both zero" as a condition?

If both $A$ and $B$ were zero, the equation collapses to just $C = 0$ , which is either always true (if $C=0$ too — every point satisfies it) or always false (if $C \neq 0$ , no point satisfies it). Neither situation describes a line; a genuine line needs at least one of $x$ or $y$ to appear with a nonzero coefficient, so the equation actually restricts which points can lie on it. Requiring " $A$ and $B$ not both zero" rules out this degenerate case and guarantees $Ax+By+C=0$ describes an honest straight line — possibly vertical ( $B=0$ ) or horizontal ( $A=0$ ), but always genuine.

Is there a shortcut for checking whether three given points are collinear, without computing the full area formula?

Yes — compute the slope between the first two points and the slope between the last two points, and check whether they are equal. If $A$ , $B$ , $C$ are the three points, they are collinear exactly when slope of $AB$ equals slope of $BC$ , since two line segments sharing the point $B$ and having the same slope must in fact be the same line extended. This is usually less arithmetic than expanding the full area-of-a-triangle determinant, especially when the coordinates are simple integers, though the area formula remains the more symmetric method when you need a numerical "how far from collinear" measure rather than a yes-or-no answer.

Why does the angle-between-two-lines formula stop working when 1 + m1 m2 equals zero?

Because that is exactly the perpendicularity condition, $m_1m_2=-1$ , rewritten as $1+m_1m_2=0$ . The formula $\tan\theta = (m_2-m_1)/(1+m_1m_2)$ has this expression sitting in its denominator, so as the two lines rotate toward being perpendicular, the denominator shrinks toward zero and $\tan\theta$ grows without bound — which is precisely the behaviour of the tangent function as the angle itself approaches $90°$ . So the formula is not broken at that point; it is correctly telling you the angle between the lines is a right angle, where "the tangent of the angle" is legitimately undefined, exactly as $\tan 90°$ is undefined for a single line's inclination.