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

Before we start, one picture to fix vocabulary so every reveal below is readable from line one.

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

Above: a line has a slope $m$ (how much $y$ rises per 1 step of $x$ ), a == $y$ -intercept== $c$ (height where it crosses the vertical axis, i.e. at $x = 0$ ), and an == $x$ -intercept== (where it crosses the horizontal axis, i.e. at $y = 0$ ). The angle $θ$ it makes with the positive $x$ -axis satisfies $m = tan θ$ . Keep this figure in mind — the traps all live in these four objects.

True or false — justify

Each item: decide, then the reveal gives the reasoning.

T/F: Every straight line can be written as

y = m x + c

False. A vertical line

x = k

has undefined slope, so no value of

m

works. Only

a x + b y + c = 0

covers all lines.

T/F: The line

a x + b y + c = 0

has slope

\frac{a}{b}

False. Solving for

y

gives

y = - \frac{a}{b} x - \frac{c}{b}

, so the slope is

- \frac{a}{b}

— the minus sign is essential.

T/F: If two lines have the same slope they are the same line.

False. Same slope means parallel; they coincide only if they also share a point (same intercept). See Parallel and perpendicular lines.

T/F: A line with a bigger

c

y = m x + c

is always steeper.

False. Steepness is controlled by

m

alone;

c

only shifts the line up or down without tilting it.

T/F: The two-point form

\frac{y - y _{1}}{x - x _{1}} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

works for any two distinct points.

False. If the two points share the same

x

(

x_{1} = x_{2}

), the right side divides by zero — that case is a vertical line

x = x_{1}

T/F: If a line passes through the origin, its equation has

c = 0

y = m x + c

True. At the origin

(0, 0)

, plugging in gives

0 = m \cdot 0 + c

, so

c = 0

; the line is

y = m x

T/F: In

a x + b y + c = 0

, we are always allowed to divide by

b

to find the slope.

False. If

b = 0

the line is vertical (

a x + c = 0 \Rightarrow x = - c / a

) and has no defined slope, so dividing by

b

is illegal.

T/F: A horizontal line has slope

0

, and a vertical line has slope

0

too.

False. Horizontal has slope

0

(no rise). Vertical has undefined slope (infinite steepness, run

= 0

) — the two are opposite extremes, not the same.

T/F: The equation

2 x + 4 y - 6 = 0

and

x + 2 y - 3 = 0

describe different lines.

False. The first is exactly

2

times the second; scaling every coefficient gives the same set of points, hence the same line.

T/F: Point–slope form requires the point to be the

y

-intercept.

False. It works for any known point

(x_{1}, y_{1})

on the line. Using the

y

-intercept

(0, c)

is just the special case that produces

y = m x + c

Spot the error

Each item states a flawed piece of work; the reveal names the mistake and fixes it.

"Slope between

(1, 2)

and

(4, 8)

\frac{4 - 1}{8 - 2} = \frac{3}{6} = \frac{1}{2}

Fraction is flipped. Slope is rise/run

= \frac{Δ y}{Δ x} = \frac{8 - 2}{4 - 1} = \frac{6}{3} = 2

, not

\frac{1}{2}

y - 3 = 4 (x - 2)

y = 4 x - 2

Forgot to distribute:

4 (x - 2) = 4 x - 8

, giving

y - 3 = 4 x - 8

, so

y = 4 x - 5

, not

4 x - 2

3 x + 4 y - 12 = 0

has slope

\frac{3}{4}

because it goes up."

Wrong sign and wrong direction. Slope

= - \frac{a}{b} = - \frac{3}{4}

; the line actually goes down to the right, so a negative slope is correct.

"The line

x = 5

has slope

m = 0

A vertical line does not have zero slope; its slope is undefined (run

= 0

). Slope

0

is a horizontal line like

y = 5

"To find the

x

-intercept of

2 x - 3 y + 6 = 0

, set

x = 0

Backwards. The

x

-intercept is where the line crosses the

x

-axis, meaning

y = 0

; setting

x = 0

gives the $y$ -intercept instead.

"Two points determine two possible lines depending on order."

Order doesn't matter. Swapping

(x_{1}, y_{1})

and

(x_{2}, y_{2})

negates both numerator and denominator of the slope, so the ratio is unchanged — one unique line.

a x + b y + c = 0

with

a = 0, b = 0, c = 5

is a horizontal line."

Not a line at all. The definition demands

(a, b) \neq = (0, 0)

; with

a = b = 0

the equation reads

5 = 0

, which is never true — no points, no line.

Why questions

Why does multiplying the point–slope equation by

(x - x_{1})

matter?

\frac{y - y _{1}}{x - x _{1}} = m

the point

(x_{1}, y_{1})

itself gives

\frac{0}{0}

, undefined. Multiplying to

y - y_{1} = m (x - x_{1})

clears the denominator so that point is legally included.

Why does the standard form

a x + b y + c = 0

exist if

y = m x + c

already gives lines?

Because

y = m x + c

cannot express a vertical line (undefined slope). Standard form can:

x - k = 0

works with

a = 1, b = 0

. It also treats distance formulas more symmetrically.

Why is slope defined as rise over run and not run over rise?

y = m x + c

m

multiplies

x

, so

m

answers "how much does

y

change per one unit of

x

?" That is

\frac{Δ y}{Δ x}

. Flipping it would answer a different question.

Why does

m = tan θ

rather than

sin θ

cos θ

On the right triangle formed by run (adjacent) and rise (opposite), tangent = opposite/adjacent = rise/run, which is exactly slope. Sine and cosine also involve the hypotenuse, which slope does not use. See Slope of a line.

Why do parallel lines share the same slope?

Same slope means same steepness/angle

θ

with the

x

-axis, so the lines tilt identically and never meet. Different intercepts just shift them apart. See Parallel and perpendicular lines.

Why can two different-looking standard equations be the same line?

Multiplying

a x + b y + c = 0

by any nonzero constant

k

gives

k a x + k b y + k c = 0

, the identical solution set. So a line's standard form is only unique up to scaling.

Edge cases

What is the equation of a vertical line through

(7, - 3)

, and its slope?

x = 7

(or

x - 7 = 0

). Slope is undefined — run is

0

, so rise/run divides by zero.

What is the equation of a horizontal line through

(7, - 3)

, and its slope?

y = - 3

. Slope is

0

— no rise for any run.

a x + b y + c = 0

, what does

c = 0

mean geometrically?

The line passes through the origin, because

(0, 0)

then satisfies

a \cdot 0 + b \cdot 0 + 0 = 0

a x + b y + c = 0

, what happens when

a = 0

(but

b \neq = 0

The equation becomes

b y + c = 0 \Rightarrow y = - \frac{c}{b}

, a horizontal line with slope

0

a x + b y + c = 0

, what happens when

b = 0

(but

a \neq = 0

It becomes

a x + c = 0 \Rightarrow x = - \frac{c}{a}

, a vertical line with undefined slope.

What line passes through

(2, 5)

and

(2, - 1)

Both points share

x = 2

, so it is the vertical line

x = 2

. The two-point/slope formula fails (division by zero) — recognise it as vertical instead.

What line passes through

(3, 4)

and

(- 1, 4)

Both share

y = 4

, so it is the horizontal line

y = 4

, slope

0

If a "line" is given as

0 x + 0 y + 3 = 0

, what is it?

Nothing — it reduces to

3 = 0

, false everywhere, so no points satisfy it. The condition

(a, b) \neq = (0, 0)

exists precisely to forbid this.

Recall One-line summary of every trap

Slope is rise/run (never flipped); standard-form slope is $- \frac{a}{b}$ (mind the minus, watch $b = 0$ ); vertical lines have undefined (not zero) slope and need $x = k$ ; scaling all coefficients keeps the same line; and $(a, b) \neq = (0, 0)$ keeps it a real line.

Connections

Parent topic (Hinglish)
Slope of a line — where $m = tan θ$ comes from.
Parallel and perpendicular lines — the same-slope and product $= - 1$ traps.
Angle between two lines — needs signed slopes handled correctly.
Distance of a point from a line — why standard form is convenient.
Intercept form of a line — another packaging of the same line.
Straight line as a first-degree equation — why degree 1 always means a line.