Straight Lines

A line is a slope and a point

A car travels at a constant speed along a straight road. At time zero, it sits at one exact spot. After that, its position climbs at one exact rate.

Those two facts pin the car’s entire future position. The starting spot is a POINT. The rate of change is a SLOPE. Together, they fix everything about where the car is, at any later time.

A straight line works exactly the same way. Give a POINT the line passes through, and its SLOPE — how steeply it rises — and the line is completely fixed. No other information is needed.

Point-slope form writes this fact as an equation:

$y - y_1 = m(x - x_1) \cdot$

It is not one formula among several to memorise. It IS the definition of a line, written in algebra.

Every other equation for a line — slope-intercept, two-point, intercept — is this same equation, rearranged for whichever piece of information a problem happens to give you first.

A line is a slope and a point. Every other form just says so differently.

↑ Back to top

Slope and inclination

Every line makes some angle with the x-axis. That angle has a name: the INCLINATION, written $\theta$. Measure it anticlockwise from the positive x-axis. It always sits between 0 and 180 degrees.

The SLOPE of the line is $m = \tan(\theta)$. A small angle gives a small slope — a gentle line. An angle close to 90 degrees gives a huge slope — a steep, almost-vertical line.

Given two points on the line, $(x_1, y_1)$ and $(x_2, y_2)$ with $x_1 \neq x_2$, the slope is easier to compute directly: $m = (y_2 - y_1)/(x_2 - x_1)$. Rise over run.

Watch the slope grow as $\theta$ climbs. At $\theta = 45$ degrees, $m = \tan(45°) = 1$: rise equals run. At $\theta = 80$ degrees, $m \approx 5.7$: one unit of run now needs nearly six units of rise. Ten degrees further, at $\theta = 89$ degrees, $m \approx 57$. The growth is not gradual — it accelerates sharply the closer $\theta$ gets to 90 degrees.

One case breaks the pattern entirely. A VERTICAL line has $\theta = 90$ degrees, where the tangent function is undefined. A vertical line’s slope is UNDEFINED, not zero — the last step of that accelerating climb never actually lands on a number.

Slope is how steeply a line climbs. Inclination is the angle that produces it.

↑ Back to top

Parallel and perpendicular lines

Two lines can relate to each other in two simple ways: they can run in the same direction, or they can cross at a right angle. Slope alone tells you which, without drawing anything.

Two non-vertical lines with slopes $m_1$ and $m_2$ are PARALLEL exactly when $m_1 = m_2$. Same steepness, same direction, never meeting.

They are PERPENDICULAR exactly when $m_1 m_2 = -1$. One slope is the negative reciprocal of the other.

Try it on two numbers. A line of slope $m_1 = 2$ and a line of slope $m_2 = -1/2$: multiply them, $2 \times (-1/2) = -1$. That single multiplication is the entire perpendicularity test — no protractor, no drawn angle, only the two slopes.

One case sits outside both tests. A vertical line has no slope to multiply with anything. Yet a vertical line and a horizontal line are always perpendicular — one runs straight up, the other runs flat across. That fact comes from direct geometry, not from the product formula, since the product formula needs both slopes to exist first.

_Same slope means parallel. Slopes multiplying to $-1$ means perpendicular. A vertical line needs geometry instead, not the formula._

↑ Back to top

The angle between two lines

Two lines that are neither parallel nor perpendicular still cross at some definite angle. There is a formula for that too.

For two lines with slopes $m_1$ and $m_2$ (neither vertical, and $1 + m_1 m_2 \neq 0$), the angle between them satisfies

$\tan(\theta) = |(m_2 - m_1)/(1 + m_1 m_2)| \cdot$

Two crossing lines actually form two angles, not one — an acute angle and its supplement, an obtuse angle. The absolute value in the formula picks out the acute one, every time.

Check the swap claim on numbers. Slopes $m_1 = 2$ and $m_2 = 3$ give $\tan(\theta) = |(3-2)/(1+6)| = 1/7$. Swap the labels — $m_1 = 3$, $m_2 = 2$ — and $\tan(\theta) = |(2-3)/(1+6)| = |-1/7| = 1/7$. Same answer — the absolute value erases the sign the swap introduces.

One formula, one absolute value, and the acute angle falls out — no matter which line you call first.

↑ Back to top

Point-slope form

Point-slope form needs exactly two pieces of information: one point on the line, and its slope. Given those, the equation is immediate:

$y - y_1 = m(x - x_1) \cdot$

Two special cases sit at the edges of this form.

A HORIZONTAL line has slope $m = 0$. Substitute that in, and the equation collapses to $y = y_1$ — every point on the line shares the same $y$-value.

A VERTICAL line has no slope at all. There is no $m$ to substitute, because none exists. A vertical line through $(x_1, y_1)$ must be written directly as $x = x_1$ instead — every point shares the same $x$-value.

Point-slope form covers every non-vertical line directly. A vertical line needs its own equation, because it has no slope to plug in.

↑ Back to top

Two-point form

Sometimes a problem hands you two points on a line, and nothing else — no slope given directly.

Compute the slope first, the same way as before: $m = (y_2 - y_1)/(x_2 - x_1)$. Then substitute that slope, and either point, straight into point-slope form:

$y - y_1 = ((y_2 - y_1)/(x_2 - x_1))(x - x_1) \cdot$

This is not a separate rule to memorise. It is point-slope form, with the slope computed from the two points first.

Does it matter which of the two points you call $(x_1, y_1)$ and which $(x_2, y_2)$? No — swap the labels, and the slope $(y_2-y_1)/(x_2-x_1)$ becomes $(y_1-y_2)/(x_1-x_2)$, the same fraction with both signs flipped, so the ratio is unchanged. Either point plugged into point-slope form afterward gives the identical line.

Two points, one slope computed between them, then point-slope form takes over.

↑ Back to top

Slope-intercept form

Point-slope form applied to one specific point gives a familiar shortcut. Pick the point where the line crosses the y-axis, $(0, c)$, and substitute:

$y = m x + c \cdot$

$c$ is the Y-INTERCEPT — where the line crosses the y-axis. $m$ is still the slope.

Substitute $(0, c)$ into point-slope form directly: $y - c = m(x - 0)$, which simplifies to $y = m x + c$ in one line. Nothing new is being assumed here — the y-axis crossing is just one particular point, substituted into the very same formula as any other point would be.

When the line passes through the origin, $c = 0$, and the equation reduces further: $y = m x$.

Slope-intercept form is point-slope form, applied once, to the one point every line crosses the y-axis at.

↑ Back to top

Intercept form

Some lines are easiest to describe by where they cross each axis, not by their slope.

A line with x-intercept $a$ (crossing the x-axis at $(a, 0)$, $a \neq 0$) and y-intercept $b$ (crossing the y-axis at $(0, b)$, $b \neq 0$) has equation:

$x/a + y/b = 1.$

Where does this equation actually come from? Feed the two points, $(a, 0)$ and $(0, b)$, into two-point form: slope $= (b-0)/(0-a) = -b/a$, so $y - 0 = (-b/a)(x-a)$. Multiply out and divide through by $a b$, and the equation settles into the symmetric form above — the same two-point machinery from before, now carrying two intercepts instead of two arbitrary points.

This form has a real limit. A line through the origin crosses both axes at the same point, so $a = b = 0$, and $x/a + y/b$ is undefined. Such a line needs slope-intercept form instead: $y = m x$.

Intercept form names a line by its two crossings. A line through the origin has no second crossing to name.

↑ Back to top

Normal form

Every form so far can fail for some line. Slope-intercept needs a slope. Intercept form needs the line to cross both axes away from the origin. Is there a form that never fails?

Drop a perpendicular from the origin to the line. Its length is $p$ (always $p \geq 0$), and the angle that perpendicular makes with the positive x-axis is $\omega$. The NORMAL FORM writes the line as:

$x \cos(\omega) + y \sin(\omega) = p \cdot$

A perpendicular distance and a direction always exist, for any line not through the origin. So normal form never fails the way slope-intercept or intercept form can — every such line has exactly one $(p, \omega)$ pair.

Normal form measures a line by its closest approach to the origin, not by its slope or its crossings — and a closest approach always exists.

↑ Back to top

The general equation of a line

Every form covered so far handles most lines, but each has a gap: slope-intercept fails for vertical lines, intercept form fails for lines through the origin. One form covers every line, without exception.

$A x + B y + C = 0,$

for some constants $A$, $B$, $C$ with $A$ and $B$ not both zero. This is the GENERAL FORM. A vertical line has $B = 0$. A horizontal line has $A = 0$. Both are ordinary cases here, not exceptions.

Try the edge case directly. The vertical line $x = 3$ becomes $1 \cdot x + 0 \cdot y + (-3) = 0$ in general form — an ordinary case, with $B = 0$. Slope-intercept form has no way to write this line at all, since a vertical line has no slope to substitute. General form does not notice anything unusual; it just quietly sets $B$ to zero.

Every other quantity reads straight off the coefficients: slope $m = -A/B$ (when $B \neq 0$), x-intercept $= -C/A$ (when $A \neq 0$), y-intercept $= -C/B$ (when $B \neq 0$).

General form is the one equation shape no line can escape — every other form is a special case of it.

↑ Back to top

Distance from a point to a line

How far does a point sit from a line? General form has an answer for that too.

The perpendicular distance from a point $(x_1, y_1)$ to the line $A x + B y + C = 0$ is:

$d = \left|A x_1 + B y_1 + C\right|/\sqrt{A^2 + B^2} \cdot$

The absolute value is essential. A point can sit on either side of the line, so the raw expression $A x_1 + B y_1 + C$ can come out positive or negative. A distance is never negative, so the formula takes the absolute value to fix that.

Sanity-check the formula on a point that sits ON the line. Take the line $x + y - 2 = 0$ and the point $(1,1)$, which satisfies the equation: $1 + 1 - 2 = 0$. Feed that into the distance formula: $d = \left|1 + 1 - 2\right|/\sqrt{1^2+1^2} = 0/\sqrt{2} = 0$. A point on the line sits zero distance from it — precisely what the formula should give, and precisely what it does.

Distance is always positive. The absolute value in the formula makes sure the answer is too, whichever side the point sits on.

↑ Back to top

Distance between parallel lines

Two parallel lines never meet, but they do have a fixed distance between them, everywhere along their length.

Write both lines with matching coefficients — same $A$ and $B$, only the constant term different, since that is exactly what makes them parallel: $A x + B y + C_1 = 0$ and $A x + B y + C_2 = 0$.

The distance between them is:

$d = \left|C_1 - C_2\right|/\sqrt{A^2 + B^2} \cdot$

This is the point-to-line formula in disguise. Pick any point on the first line, and its distance to the second line always simplifies to this — because a point on the first line makes $A x_1 + B y_1 = -C_1$ automatically.

Check the formula’s edge behaviour. If $C_1 = C_2$, the two "parallel" lines are actually the same line, and the formula gives $d = \left|C_1 - C_2\right|/\sqrt{A^2+B^2} = 0$ — zero distance, exactly as it should be for one line measured against itself.

Parallel lines keep one fixed gap between them everywhere. This formula reads that gap straight off the two equations.

↑ Back to top

Every form, one question

Every idea in this chapter answers one question: which line, exactly?

Slope and inclination measure how steeply a line climbs. The parallel and perpendicular tests, and the angle formula, compare two lines using nothing but their slopes.

Five equation forms each pin a line down, from whichever piece of information you are given: point-slope needs a point and a slope; two-point needs two points; slope-intercept needs a slope and where the line crosses the y-axis; intercept form needs both crossings; normal form needs the line’s closest approach to the origin.

The general form, $A x + B y + C = 0$, holds every line at once — including the vertical ones none of the other forms can reach directly.

Two distance formulas finish the picture: how far a point sits from a line, and how far two parallel lines sit from each other.

In practice, the choice is mechanical once you name what a problem hands you. Given a point and a slope, reach for point-slope form. Given two points, compute the slope first, then use that same form. Given a slope and where the line meets the y-axis, slope-intercept form is already built for exactly that. Given both axis crossings, intercept form reads them off directly. Given only the coefficients themselves, general form covers every case, vertical lines included.

The question never changes. Only the form you reach for does, depending on what a problem actually gives you.

↑ Back to top

Two traps this chapter sets

THE TRAP. A vertical line’s slope is $0$, and the perpendicular condition $m_1 m_2 = -1$ can be used to check whether a vertical line and a horizontal line are perpendicular.

THE REALITY. A vertical line’s slope is UNDEFINED, not $0$. Zero is a HORIZONTAL line’s slope, not a vertical one. The tangent function is undefined at 90 degrees, so there is no number to call the slope.

The product test $m_1 m_2 = -1$ needs both slopes to exist first. It cannot be applied to a vertical line at all. A vertical line and a horizontal line are perpendicular by direct geometric fact — one runs straight up, the other runs flat across — never by the product formula.

Undefined is not the same as zero. A vertical line has no slope to plug into any formula.

THE TRAP. Every line has an intercept-form equation, $x/a + y/b = 1$. Just find the two intercepts and substitute.

THE REALITY. A line through the origin crosses both axes at the same point, $(0,0)$. Its x-intercept and its y-intercept are both $0$. Substitute that in, and $x/0 + y/0$ is undefined — there is no equation to write this way.

A line through the origin must be written in slope-intercept form instead, $y = m x$ (or as $x = 0$, if the line is vertical). Intercept form simply does not apply.

Two intercepts existing is not automatic. A line through the origin has none to name.

↑ Back to top

Practice set

↑ Back to top

Chapter-end problems

↑ Back to top

JEE-application problems