Lines — NDA Mathematics Formulas and Traps

Equations, Slope & Family of Lines

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

$m$ slope of the line
$\theta$ angle the LINE makes with the positive x-axis
$\alpha$ angle the PERPENDICULAR from the origin makes — normal form only
$c$ y-intercept in $y=mx+c$ ; the constant term in $ax+by+c=0$
$p$ distance from the origin to the line

Intercept form and intercepts

\dfrac{x}{a}+\dfrac{y}{b}=1\qquad\text{from }px+qy+r=0:\quad x\text{-int}=-\dfrac{r}{p},\quad y\text{-int}=-\dfrac{r}{q}

$a,b$ the x- and y-INTERCEPTS (numbers), in intercept form only
$p,q,r$ the COEFFICIENTS of a general line $px+qy+r=0$
$(h,k)$ midpoint of the segment cut between the axes

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

Image of a point and reflections

Image of a point in a line

\text{midpoint}(PP')\in L\qquad PP'\perp L\qquad P'=2F-P

$P$ the original point
$P'$ its image (mirror reflection) in the line
$F$ foot of the perpendicular from $P$ to the line — the MIDPOINT of $PP'$

Common traps

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

is

-a/b

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

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.

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

is

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.

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

to

F

. Reporting the foot

F

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

Angle Between Lines, Parallel & Perpendicular

Learn this subtopic in the notes

Angle between two lines

\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1 m_2}\right|

Parallel and perpendicular conditions

\text{Parallel: } m_1=m_2\qquad \text{Perpendicular: } m_1 m_2=-1\qquad a_1a_2+b_1b_2=0

Common traps

The difference of slopes is on top: $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$

The angle formula puts the difference

m_1-m_2

in the numerator and

1+m_1m_2

in the denominator — students often invert it to

\dfrac{1+m_1m_2}{m_1-m_2}

. Also watch the denominator's plus sign (

1+m_1m_2

, not

1-m_1m_2

); when

1+m_1m_2=0

the tangent blows up, correctly signalling

\theta=90^\circ

.

Perpendicular slope is the negative reciprocal: $m_2=-\dfrac{1}{m_1}$ , not $\dfrac{1}{m_1}$

Parallel ⇒ equal slopes (

m_1=m_2

); perpendicular ⇒ the product is

-1

(

m_1m_2=-1

), so the second slope is the negative reciprocal

-1/m_1

. The two classic slips: forgetting the minus (using

1/m_1

, the plain reciprocal), and swapping the two rules — "perpendicular means equal slopes" is wrong. If

m_1=\tfrac23

, a perpendicular line has slope

-\tfrac32

, not

\tfrac32

.

Distance, Section & Locus

Learn this subtopic in the notes

Distance: point-point, point-line, parallel lines

Distance formulas

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\qquad \dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\qquad \dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}

Section formula and ratios

Section formula and midpoint

\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)\qquad \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

Common traps

Put the line in $ax+by+c=0$ form first, and divide by $\sqrt{a^2+b^2}$ — not by $a+b$

Two things go wrong with point-to-line distance. (1) The line must be written as

ax+by+c=0

(everything on one side) before substituting — applying the formula to

y=mx+c

or

ax+by=c

as-is gives the wrong numerator. (2) The denominator is

\sqrt{a^2+b^2}

, not

a^2+b^2

or

a+b

. For parallel lines, the

a,b

coefficients of both lines must be made identical first, or

|c_1-c_2|

is meaningless.

In $m:n$ the weight $m$ multiplies the far endpoint $x_2$ — mind the cross-pairing

For a point dividing

P_1P_2

in ratio

m:n

(with

m

the part nearer

P_2

), the x-coordinate is

\dfrac{mx_2+nx_1}{m+n}

: the ratio's first number

m

pairs with the second point's coordinate

x_2

. Writing

\dfrac{mx_1+nx_2}{m+n}

swaps the weights and divides in ratio

n:m

instead — the point lands on the wrong side. Also remember the

m+n

in the denominator.

Equidistant from two points gives a line (perp. bisector); equidistant from two lines gives the angle bisectors

Don't conflate the two 'equidistant' loci. Equidistant from two points

A,B

is the perpendicular bisector of

AB

— a single straight line. Equidistant from two lines is the pair of angle bisectors between them. Setting up

PA^2=PB^2

(distances to points) when the question means distance-to-lines, or vice versa, produces the wrong locus entirely.

Triangles, Quadrilaterals & Polygons

Learn this subtopic in the notes

Area of a triangle and collinearity

Area of a triangle from vertices

\text{Area}=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Centroid, incentre, circumcentre

Centroid and incentre

G=\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)\qquad I=\dfrac{a\,A+b\,B+c\,C}{a+b+c}

Constructing a triangle: vertices, medians, altitudes

Vertex from a midpoint

B+C=2M\qquad C=2M-B

Parallelograms, squares and diagonals

Parallelogram: fourth vertex and area

A+C=B+D\qquad D=A+C-B\qquad \text{Area}=|u_1v_2-u_2v_1|

$A,B,C,D$ the vertices, labelled in order around the figure
$(u_1,v_1)$ components of the side vector $\vec{AB}=B-A$
$(u_2,v_2)$ components of the side vector $\vec{AD}=D-A$

Common traps

Don't forget the $\tfrac12$ and the absolute value — and collinearity is area $=0$

Two routine slips on the area formula

\tfrac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|

: dropping the leading $\tfrac12$ (which doubles the answer), and omitting the modulus (a clockwise vertex order makes the bare determinant negative — area can't be negative). When the expression equals

0

, the three points are collinear, so the 'area' test and the collinearity test are the same computation.

Centroid is the plain average; the incentre weights by the opposite side lengths $a,b,c$

The centroid

G

is the unweighted average of the three vertices,

\left(\tfrac{x_1+x_2+x_3}{3},\tfrac{y_1+y_2+y_3}{3}\right)

. The incentre is not that average — it is the side-length-weighted average

\dfrac{aA+bB+cC}{a+b+c}

, where

a,b,c

are the lengths of the sides opposite vertices

A,B,C

. Using equal weights for the incentre (or pairing a side with its adjacent vertex) is the usual mistake; the two centres coincide only for an equilateral triangle.

An altitude is perpendicular to the opposite side — use the negative-reciprocal slope

The altitude from a vertex is perpendicular to the opposite side, so its slope is the negative reciprocal of that side's slope (not the same slope, which would be parallel, and not the side's own slope). To recover a vertex from a midpoint, use

B+C=2M\Rightarrow C=2M-B

— i.e.

2M-B

, not

M-B

; the factor of

2

is essential because

M

is the average of

B

and

C

.

In parallelogram $ABCD$ the diagonals are $AC$ and $BD$ : $A+C=B+D$

The diagonals of

ABCD

join opposite vertices —

AC

and

BD

— and they bisect each other, so the midpoints match:

A+C=B+D

, giving

D=A+C-B

. The slip is pairing adjacent vertices (e.g. computing

A+B

); the order of the labels around the parallelogram tells you which pairs are diagonals. Get the pairing wrong and the 'fourth vertex' is misplaced.

Lines formulas

Equations, Slope & Family of Lines

Slope and the forms of a line

Image of a point and reflections

Common traps

Angle Between Lines, Parallel & Perpendicular

Common traps

Distance, Section & Locus

Distance: point-point, point-line, parallel lines

Section formula and ratios

Common traps

Triangles, Quadrilaterals & Polygons

Area of a triangle and collinearity

Centroid, incentre, circumcentre

Constructing a triangle: vertices, medians, altitudes

Parallelograms, squares and diagonals

Common traps

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