Equations of a line — slope-intercept, point-slope, two-point, standard (ax+by+c=0)

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WHY do we even want an equation for a line?

A line is an infinite set of points. We can't list them all. So we look for a condition that a point $(x,y)$ satisfies if and only if it lies on the line. That condition is the equation.

The key fact all forms rest on:

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1. Point–slope form (the parent of all forms)

WHAT: the equation of a line through a known point $(x_1,y_1)$ with slope $m$.

HOW (derive from scratch): Take any other point $(x,y)$ on the line. By the definition of slope, the slope computed between $(x_1,y_1)$ and $(x,y)$ must equal $m$:

$\frac{y-y_1}{x-x_1}=m$

Multiply both sides by $(x-x_1)$ (allowed, since we then also allow the point $(x_1,y_1)$ itself):

Why this step? Multiplying clears the denominator so the point $(x_1,y_1)$ (which would make the fraction $0/0$) is now legally included.

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2. Slope–intercept form

WHAT: slope $m$ and where the line crosses the $y$-axis (the $y$-intercept $c$).

HOW: The $y$-intercept is the point $(0,c)$. Plug $(x_1,y_1)=(0,c)$ into point–slope: $y-c=m(x-0)\;\Rightarrow\; y-c=mx$

Why this step? Using $(0,c)$ as the known point makes $x_1=0$ vanish, isolating $y$.

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3. Two-point form

WHAT: line through two known points $(x_1,y_1)$ and $(x_2,y_2)$.

HOW: First get the slope from the two points, $m=\dfrac{y_2-y_1}{x_2-x_1}$, then substitute into point–slope:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Why this step? Two points fully determine a line — so we just compute the missing $m$ and reuse the parent form. Nothing new to memorize.

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4. Standard (general) form

WHAT: every straight line — including vertical ones — can be written as $ax+by+c=0,\quad (a,b)\neq(0,0).$

HOW: Start from $y=mx+c'$ (rename intercept $c'$ to avoid clash). Move all to one side: $mx - y + c' = 0.$ This has the shape $ax+by+c=0$ with $a=m,\;b=-1,\;c=c'$. And when a line is vertical ($x=k$), we write $1\cdot x + 0\cdot y - k = 0$ — impossible in slope–intercept but fine here.

Why this step? Solving for $y$ turns it back into slope–intercept, revealing $m$ and $c$ directly.

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Worked examples

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Common mistakes (steel-manned)

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Convert-between-forms cheat map

Have	Want slope–intercept	Want standard
point + slope	plug into $y-y_1=m(x-x_1)$, expand	expand then move to one side
two points	get $m$ first, then as above	same
standard $ax+by+c=0$	$y=-\frac ab x-\frac cb$	already there

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Recall Feynman: explain to a 12-year-old

Imagine walking up a ramp. Every 1 step forward you go up the same amount — that "same amount" is the slope. If I tell you (a) where the ramp starts and (b) how steep it is, you can draw the whole ramp. Every "equation of a line" is just me telling you those two things in a slightly different wording. The $y=mx+c$ way says "start at height $c$, climb $m$ each step." The $x=5$ way is a wall — straight up, no forward walking allowed, so we can't talk about "up per step" — that's why it needs its own form.

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Active-recall flashcards

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Connections

• Slope of a line — the single number all forms depend on.
• Angle between two lines — uses $\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$.
• Parallel and perpendicular lines — $m_1=m_2$ and $m_1m_2=-1$.
• Distance of a point from a line — uses the standard form $ax+by+c=0$.
• Intercept form of a line — $\frac xa+\frac yb=1$.
• Straight line as a first-degree equation — why degree-1 always means a line.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek straight line basically ek rule hai: "jitna aage badho ($\Delta x$), utna hi fixed upar/neeche jaao ($\Delta y$)." Is fixed ratio ko hum slope $m$ bolte hain, matlab rise/run. Bas yahi ek number puri line ka steepness batata hai. Agar tumhe ek point aur slope pata hai, toh puri line draw ho sakti hai — isiliye saare forms sirf isi cheez ko alag-alag tareeke se likhte hain.

Sabse important point-slope form: $y-y_1=m(x-x_1)$. Ye "parent" form hai. Agar point $(0,c)$ le lo toh ye ban jaata hai slope-intercept $y=mx+c$ (yahan $c$ = jahan line $y$-axis ko kaatti hai). Do points diye ho toh pehle $m=\frac{y_2-y_1}{x_2-x_1}$ nikalo, phir wahi parent form. Aur standard form $ax+by+c=0$ sabse general hai — ye vertical line ($x=5$ type) ko bhi handle kar leta hai, jabki $y=mx+c$ nahi kar sakta.

Ek cheez yaad rakhna jismein log galti karte hain: standard form ka slope $-\frac{a}{b}$ hota hai, minus ke saath. Quick check: $3x+4y-12=0$ neeche ki taraf jaati hai, toh slope negative hona chahiye — $-3/4$. Sahi! Aur vertical line ka slope undefined hota hai (run zero, divide by zero), isliye uske liye $x=k$ likho.

80/20 funda: agar tum point-slope form aur slope ki definition rat lo, baaki teen forms tumhe khud derive karne aa jaayenge. Kuch alag se ratne ki zaroorat nahi — sab ek hi cheez ke roop hain.