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

The scenario matrix

Below is the full list of case classes a line-equation question can belong to. Every worked example below is tagged with which cell it fills, so you can see the coverage is complete.

Cell	Case class	Covered by
A	Positive slope, point–slope → all forms	Ex 1
B	Negative slope, line falling to the right	Ex 2
C	Two given points (slope computed)	Ex 3
D	Zero slope — horizontal line	Ex 4
E	Undefined slope — vertical line	Ex 4
F	Standard form → read slope & both intercepts	Ex 5
G	Sign trap: $b < 0$ in $a x + b y + c = 0$	Ex 6
H	Fractional / negative intercepts, all-quadrant sketch	Ex 7
I	Real-world word problem (units matter)	Ex 8
J	Limiting behaviour: slope → steeper → vertical	Ex 9
K	Exam twist: unknown coefficient from a condition	Ex 10

Notice cells D and E — the flat wall and the vertical wall — are exactly the two that break the friendly $y = m x + c$ machine. We give them their own example so you never get stuck.

Example 1 — Positive slope, build all four forms (Cell A)

Step 1 — Point–slope. Substitute $(x_{1}, y_{1}) = (- 1, 2)$ and $m = \frac{3}{2}$ into the parent form $y - y_{1} = m (x - x_{1})$ : $y - 2 = \frac{3}{2} (x - (- 1)) = \frac{3}{2} (x + 1) .$ Why this step? Point–slope is the one form that needs only a point and a slope — exactly what we were handed, so no computation is wasted.

Step 2 — Slope–intercept. Distribute and isolate $y$ : $y - 2 = \frac{3}{2} x + \frac{3}{2} \Rightarrow y = \frac{3}{2} x + \frac{7}{2} .$ Why this step? Expanding the bracket turns the point form into the machine form $y = m x + c$ , exposing the $y$ -intercept $c = \frac{7}{2}$ .

Step 3 — Standard form. Move everything to one side and clear the fraction (multiply by $2$ ): $2 y = 3 x + 7 \Rightarrow 3 x - 2 y + 7 = 0.$ Why this step? Standard form wants integer coefficients with $= 0$ on the right; multiplying by $2$ removes the halves.

Leaning: $m = \frac{3}{2} > 0$ , so the line climbs as you move right. Look at the amber line in the figure below.

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

Example 2 — Negative slope, line falling to the right (Cell B)

Step 1 — Point–slope, then expand. $y - 1 = - 2 (x - 4) = - 2 x + 8 \Rightarrow y = - 2 x + 9.$ Why this step? Same parent form; the only new thing is that the slope is negative, so distributing $- 2$ flips the sign of every term inside the bracket. Watch that carefully — dropping a minus here is the #1 error.

Step 2 — $x$ -intercept. The $x$ -intercept is where the line meets the $x$ -axis, i.e. where $y = 0$ : $0 = - 2 x + 9 \Rightarrow x = \frac{9}{2} = 4.5, point (4.5, 0) .$ Why this step? "On the $x$ -axis" means height zero, so we set $y = 0$ and solve. This is the same trick you'll reuse in Distance of a point from a line.

Because $m = - 2 < 0$ , the line falls to the right (cyan line, figure).

Example 3 — Two given points (Cell C)

Step 1 — Compute the slope. Using $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ with $A$ as point 1, $B$ as point 2: $m = \frac{- 5 - 5}{3 - ( - 2 )} = \frac{- 10}{5} = - 2.$ Why this step? Two points don't hand us the slope directly, so we must build it first — this is the only extra work two-point problems carry.

Step 2 — Point–slope with either point. Use $A = (- 2, 5)$ : $y - 5 = - 2 (x + 2) = - 2 x - 4 \Rightarrow y = - 2 x + 1.$ Why this step? Once $m$ is known, a two-point problem becomes a point–slope problem — no new formula needed.

Example 4 — The two walls: horizontal and vertical (Cells D and E)

Step 1 — Horizontal line (slope $0$ ). Every point on a horizontal line shares the same height. That height is $- 3$ , so: $y = - 3.$ Why this step? Slope $= \frac{rise}{run} = \frac{0}{run} = 0$ ; plug $m = 0$ into $y = m x + c$ and $y = c$ is left. No $x$ survives — height never changes.

Step 2 — Vertical line (slope undefined). Every point shares the same $x$ -coordinate, $6$ : $x = 6, or in standard form x - 6 = 0.$ Why this step? Here the "run" is $0$ , so $m = \frac{rise}{0}$ is division by zero — undefined. That's precisely why $y = m x + c$ fails and standard form $a x + b y + c = 0$ (here $a = 1, b = 0, c = - 6$ ) must be used. See Straight line as a first-degree equation for why $x = 6$ still counts as a genuine line.

Example 5 — Standard form → read off slope and both intercepts (Cell F)

Step 1 — Solve for $y$ to expose the slope. $5 y = - 2 x + 10 \Rightarrow y = - \frac{2}{5} x + 2.$ So $m = - \frac{2}{5}$ and $y$ -intercept $= 2$ , i.e. $(0, 2)$ . Why this step? Turning the equation back into $y = m x + c$ is the surest way to see the slope — no memorised sign to trust.

Step 2 — $x$ -intercept. Set $y = 0$ in the original: $2 x - 10 = 0 \Rightarrow x = 5, (5, 0) .$ Why this step? $x$ -intercept always means $y = 0$ ; using the original equation avoids fraction juggling.

Example 6 — The sign trap when $b$ is negative (Cell G)

Step 1 — Apply the formula carefully. Here $a = 4$ , $b = - 3$ : $m = - \frac{a}{b} = - \frac{4}{- 3} = + \frac{4}{3} .$ Why this step? The two minus signs (the formula's minus and the negative $b$ ) cancel. Rushing gives a wrong negative answer.

Step 2 — Cross-check by solving for $y$ . $- 3 y = - 4 x - 9 \Rightarrow y = \frac{4}{3} x + 3.$ Slope $= + \frac{4}{3}$ . Agrees. Why this step? When a sign feels risky, solving for $y$ is the referee. It never lies about the slope.

Example 7 — Fractional intercepts, drawing across quadrants (Cell H)

Step 1 — $y$ -intercept. Set $x = 0$ : $y = - \frac{3}{2}$ , point $(0, - 1.5)$ . Why this step? In $y = m x + c$ , the constant $c$ is the $y$ -intercept — read it, don't compute it.

Step 2 — $x$ -intercept. Set $y = 0$ : $0 = - \frac{1}{2} x - \frac{3}{2} \Rightarrow \frac{1}{2} x = - \frac{3}{2} \Rightarrow x = - 3, (- 3, 0) .$ Why this step? Two points $(0, - 1.5)$ and $(- 3, 0)$ are enough to draw the whole line.

Step 3 — Quadrant tour. With $m < 0$ and both intercepts negative, the line passes through quadrant II (upper-left, near $(- 3, 0)$ ), quadrant III (lower-left), and quadrant IV (lower-right). It never touches quadrant I. Why this step? Covering all four quadrants is the "every scenario" discipline — you should always be able to say where a line does and does not go.

Example 8 — Real-world word problem (units matter) (Cell I)

Step 1 — Match the story to $y = m x + c$ . The fare rises by a fixed amount per km — that "per km" rate is the slope: $m = 12$ ₹/km. The starting charge at $d = 0$ is the intercept: $c = 40$ ₹. So: $F = 12 d + 40.$ Why this step? "Fixed amount per unit" is the plain-English definition of slope; "value when the input is zero" is the plain-English definition of intercept. Word problems are just $y = m x + c$ wearing a costume.

Step 2 — Evaluate at $d = 7$ . $F = 12 (7) + 40 = 84 + 40 = 124 ₹ .$ Why this step? The question asks for a specific fare, so we feed the specific distance into the machine.

Units check: $[\frac{₹}{km}] \times [km] + [₹] = ₹$ . The dimensions balance — the answer is genuinely money. The ₹40 is the == $y$ -intercept==: the fare you pay for $0$ km (just for getting in).

Example 9 — Limiting behaviour: slope grows toward vertical (Cell J)

Step 1 — Interpret growing slope geometrically. Slope is $tan θ$ , where $θ$ is the angle with the positive $x$ -axis (from Slope of a line). Big $m$ means big $tan θ$ , which pushes $θ$ toward $9 0^{\circ}$ : $m = 1 \to θ = 4 5^{\circ}, m = 5 \to θ \approx 78. 7^{\circ}, m = 500 \to θ \approx 89. 9^{\circ} .$ Why this step? We use $tan θ$ because it converts a ratio (rise/run) into an angle, and angles make "approaching vertical" precise.

Step 2 — The limit. As $m \to \infty$ , $θ \to 9 0^{\circ}$ : the lines crowd toward the vertical line $x = 0$ (the $y$ -axis). But a truly vertical line has undefined slope, so it can never be $y = m x + c$ — it is the standard-form line $x = 0$ . Why this step? This shows why standard form is not optional: it is the only form that survives the limit. The vertical line is the "point at infinity" of the slope family.

Example 10 — Exam twist: find the unknown coefficient (Cell K)

Step 1 — Use "the point lies on the line". A point on a line satisfies its equation. Substitute $x = 4, y = - 2$ : $k (4) + 3 (- 2) - 6 = 0 \Rightarrow 4 k - 6 - 6 = 0 \Rightarrow 4 k = 12 \Rightarrow k = 3.$ Why this step? "Passes through" is the whole hint — it turns the geometric condition into an algebra equation for $k$ .

Step 2 — Slope of the now-complete line. With $k = 3$ : $3 x + 3 y - 6 = 0$ , so $a = 3, b = 3$ : $m = - \frac{a}{b} = - \frac{3}{3} = - 1.$ Why this step? Once $k$ is known the line is ordinary standard form, and $m = - a / b$ finishes it. This is the same skill you'll need for Parallel and perpendicular lines, where a condition on $m$ pins down an unknown coefficient.

One-line recap of the coverage

Recall Did we really hit every cell?

A ::: Ex 1 (positive slope, all forms) B ::: Ex 2 (negative slope, falling line) C ::: Ex 3 (two points → slope first) D ::: Ex 4a (horizontal, slope 0) E ::: Ex 4b (vertical, undefined slope) F ::: Ex 5 (standard → slope + both intercepts) G ::: Ex 6 (sign trap with negative $b$ ) H ::: Ex 7 (fractional intercepts, all quadrants) I ::: Ex 8 (word problem with units) J ::: Ex 9 (limit toward vertical) K ::: Ex 10 (unknown coefficient from a condition)

Connections

Parent topic — the four forms
Slope of a line — every example above starts by nailing $m$ .
Angle between two lines — Ex 9's $tan θ$ view feeds directly into this.
Parallel and perpendicular lines — Ex 10's "find the unknown" trick generalises here.
Distance of a point from a line — uses the standard form we build in Ex 1, 4, 5, 6, 10.
Intercept form of a line — Ex 5 and Ex 7's intercepts lead into $\frac{x}{a} + \frac{y}{b} = 1$ .
Straight line as a first-degree equation — why even $x = 6$ (Ex 4) is a legitimate line.