NDA Maths · Lines

Equations, Slope & Family of Lines

The slope of a line and the standard forms of its equation, the intercept form, the family of lines through a point or an intersection, and reflections in a line.

All Lines notes NDA Maths strategy

Why this matters

Every other tool in the chapter starts from a line's equation. Knowing which form to reach for — slope-intercept, point-slope, intercept, or the family L₁+λL₂ — turns most questions into one substitution.

Concept 1 of 4

Slope and the forms of a line

Intuition

A line is fixed by a point and a direction (its slope). The slope is the tangent of the angle it makes with the x-axis. Pick the form that matches what you're given — a point and slope, two points, or intercepts.

Definition

Slope $m=\tan\theta=\dfrac{y_2-y_1}{x_2-x_1}$ . Forms:

Slope-intercept: $y=mx+c$ . Point-slope: $y-y_1=m(x-x_1)$ .
Two-point: $\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}$ .
General: $ax+by+c=0$ has slope $-a/b$ . Normal form: $x\cos\theta+y\sin\theta=p$ ( $p$ = distance from origin).

Slope and forms of a line

m=\dfrac{y_2-y_1}{x_2-x_1}\qquad m_{ax+by+c=0}=-\dfrac{a}{b}\qquad y-y_1=m(x-x_1)\qquad y=mx+c

Worked example

Find the equation of the line through

(2,3)

with slope

4

Practice this conceptself-check · 4 quick reps

Slope is $\Delta y/\Delta x$ , not $\Delta x/\Delta y$ — and a vertical line has undefined slope

Two slips. First, slope is rise over run:

m=\dfrac{y_2-y_1}{x_2-x_1}

, not

\dfrac{x_2-x_1}{y_2-y_1}

— keep the

y

-difference on top. Second, a vertical line

x=k

has *undefined* slope (the run is

0

), not slope

0

— that's a *horizontal* line

y=k

. Likewise, the slope of

ax+by+c=0

-a/b

, with the minus sign — dropping it flips the line.

Drill 9 more on slope and the forms of a line

Concept 2 of 4

Intercept form and intercepts

Intuition

When a line's x- and y-intercepts matter, the intercept form

\tfrac{x}{a}+\tfrac{y}{b}=1

reads them off directly. Many questions give a relation between the intercepts (their sum, or a midpoint) and ask for the line.

Definition

Intercept form: $\dfrac{x}{a}+\dfrac{y}{b}=1$ , where $a$ is the x-intercept and $b$ the y-intercept. From $ax+by+c=0$ : x-intercept $=-c/a$ , y-intercept $=-c/b$ . If $(h,k)$ is the midpoint of the intercept segment, then $a=2h$ , $b=2k$ .

Intercept form and intercepts

\dfrac{x}{a}+\dfrac{y}{b}=1\qquad x\text{-intercept}=-\dfrac{c}{a}\qquad y\text{-intercept}=-\dfrac{c}{b}

Worked example

A line has x-intercept

4

and y-intercept

2

. Find its equation and the sum of intercepts.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2LinesMODERATE

What is the sum of the intercepts of the line

\dfrac{x}{a^2}+\dfrac{y}{b^2}=\dfrac{2}{a^2+b^2}

on the coordinate axes?

[Q60 · Sep · 2025]

Intercept form needs the constant on the **RHS as $1$ ** — $a,b$ are the intercepts only then

You can read the intercepts straight off

\dfrac{x}{a}+\dfrac{y}{b}=1

**only when the right side is exactly

1

**. From

ax+by=c

the x-intercept is

c/a

, not

a

: divide through by

c

first to reach

\dfrac{x}{c/a}+\dfrac{y}{c/b}=1

. Grabbing the coefficients before normalising to

1

is the classic error.

Drill 5 more on intercept form and intercepts

Concept 3 of 4

Family of lines and concurrency

Intuition

Any line through the intersection of

L_1=0

and

L_2=0

can be written

L_1+\lambda L_2=0

— without ever finding the intersection point. Choose

\lambda

from one extra condition. Three lines are concurrent when their intersection is shared.

Definition

Family (pencil): through $L_1\cap L_2$ , every line is $L_1+\lambda L_2=0$ ; fix $\lambda$ from a point or a slope condition. Parallel/perpendicular through a point: keep the same (or negative-reciprocal) slope. Concurrency: three lines are concurrent iff $\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0$ . If coefficients $A,B,C$ are in AP, $Ax+2By+C=0$ passes through the fixed point $(1,-1)$ (since $C=2B-A$ gives $A(x-1)+2B(y+1)=0$ ).

Family of lines and concurrency

L_1+\lambda L_2=0\qquad \begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0

Worked example

Find the line through the intersection of

x+y-1=0

and

2x-y-2=0

that passes through

(1,2)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3LinesMODERATE

If the lines

3y+4x=1

y=x+5

and

5y+bx=3

are concurrent, then what is the value of b?

[Q57 · Apr · 2019]

The pencil is $L_1+\lambda L_2=0$ — keep each $L_i$ in the form $=0$ first

The family through

L_1\cap L_2

L_1+\lambda L_2=0

, where **each

L_i

is the whole expression

a_ix+b_iy+c_i

** moved to one side so the line reads

L_i=0

. Combining

2x+3y=5

and

x-y=1

means using

L_1=2x+3y-5

and

L_2=x-y-1

— forgetting to move the constants over (using

2x+3y

and

x-y

) silently shifts the pencil off the intersection.

Drill 8 more on family of lines and concurrency

Concept 4 of 4

Image of a point and reflections

Intuition

The image of a point in a line is its mirror reflection: the line is the perpendicular bisector of the segment joining the point and its image. Use 'midpoint lies on the line' plus 'segment ⟂ line' to find the image, or recover the mirror from a point–image pair.

Definition

If $P'$ is the image of $P$ in line $L$ : the midpoint of $PP'$ lies on $L$ , and $PP'\perp L$ . These two conditions pin down $P'$ (or the mirror line). Foot of perpendicular from $P$ to $L$ is the midpoint of $PP'$ .

Worked example

Find the image of

(1,2)

in the line

y=x

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4LinesMODERATE

If the image of the point

(-4, 2)

by a line mirror is

(4, -2)

, then what is the equation of the line mirror?

[Q60 · Apr · 2021]

The foot of the perpendicular is the *midpoint* of $PP'$ , not the image itself

The foot of the perpendicular

F

from

P

to the line is halfway to the image:

F

is the *midpoint* of

P

and

P'

. So the image is

P'=2F-P

— you must double the displacement from

P

F

. Reporting the foot

F

as the reflected image gives a point only half as far across the line.

Drill 2 more on image of a point and reflections

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Slope and the forms of a line
Slope and forms of a line
$m=\dfrac{y_2-y_1}{x_2-x_1}\qquad m_{ax+by+c=0}=-\dfrac{a}{b}\qquad y-y_1=m(x-x_1)\qquad y=mx+c$
Intercept form and intercepts
Intercept form and intercepts
$\dfrac{x}{a}+\dfrac{y}{b}=1\qquad x\text{-intercept}=-\dfrac{c}{a}\qquad y\text{-intercept}=-\dfrac{c}{b}$
Family of lines and concurrency
Family of lines and concurrency
$L_1+\lambda L_2=0\qquad \begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0$

Watch out for (4)

Slope is $\Delta y/\Delta x$ , not $\Delta x/\Delta y$ — and a vertical line has undefined slope
→ Slope and the forms of a line
Intercept form needs the constant on the **RHS as $1$ ** — $a,b$ are the intercepts only then
→ Intercept form and intercepts
The pencil is $L_1+\lambda L_2=0$ — keep each $L_i$ in the form $=0$ first
→ Family of lines and concurrency
The foot of the perpendicular is the *midpoint* of $PP'$ , not the image itself
→ Image of a point and reflections

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