What is the condition for a homogeneous second-degree equation to represent perpendicular lines?

The condition is a + b = 0 - the sum of the coefficients of x squared and y squared vanishes. This is a useful test that requires no factoring.

What is the difference between the homogeneous form and the general second-degree equation?

The homogeneous form always represents a pair of lines through the origin. The general form represents a pair only if the determinant condition abc + 2fgh - af squared - bg squared - ch squared = 0 holds, and then the two lines can meet anywhere.

Introduction to Straight Lines

Q: When is the slope of a line undefined?

When the line is vertical (parallel to the y-axis). Its inclination is 90 degrees and tan 90 is undefined. Vertical lines have the equation x = k for some constant k.

Q: Why is p always taken positive in the normal form?

Because p represents a length - the perpendicular distance from the origin to the line - and lengths are non-negative by definition. The direction information is carried by the angle alpha.

Q: What is the parametric form used for?

It gives every point on a line as (x1 + r cos theta, y1 + r sin theta) where the absolute value of r is the distance from a fixed point. This makes it ideal for finding points on a line at a given distance from another point.

Q: What does the homogenisation trick achieve?

Given a line meeting a curve at two points A and B, homogenisation produces a single second-degree equation whose two lines are exactly OA and OB. This lets you find the angle at the origin or test perpendicularity without solving for A and B explicitly.

MathsStraight LinesFor JEE aspirants

A straight line in the coordinate plane is completely determined by any two of its properties - two points on it, a point and a slope, two intercepts, or the perpendicular from the origin. The equation of a straight line is a first-degree relation in $x$ and $y$ , and every first-degree equation represents a straight line. This concept covers the eight standard forms of the equation of a line, the angle between two lines, parallel and perpendicular conditions, the family of concurrent lines, and the theory of a pair of straight lines given by a second-degree equation.

Key Formulas - Quick Reference

Slope: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = tan θ$ , where $θ$ is the inclination.
Point-slope: $y - y_{1} = m (x - x_{1})$ .
Slope-intercept: $y = m x + c$ .
Two-point: $y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$ .
Intercept: $\frac{x}{a} + \frac{y}{b} = 1$ .
Normal: $x cos α + y sin α = p$ , $p > 0$ .
Parametric: $\frac{x - x _{1}}{cos θ} = \frac{y - y _{1}}{sin θ} = r$ .
General: $a x + b y + c = 0$ ; slope $= - \frac{a}{b}$ .
Angle between two lines: $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ .
Distance between parallel lines $a x + b y + c_{1} = 0$ and $a x + b y + c_{2} = 0$ : $\frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}}$ .
Pair through origin $a x^{2} + 2 h x y + b y^{2} = 0$ : slopes have $m_{1} + m_{2} = - \frac{2 h}{b}$ , $m_{1} m_{2} = \frac{a}{b}$ ; angle $tan θ = \frac{2 h ^{2} - ab}{a + b}$ .

1. Slope (Gradient) of a Line

Let $θ$ be the angle a straight line makes with the positive direction of the $x$ -axis, measured anti-clockwise, with $0^{\circ} \leq θ < 18 0^{\circ}$ . Then $θ$ is called the inclination of the line and

m = tan θ

is its slope (or gradient). Special cases:

$θ = 0^{\circ}$ : line is parallel to the $x$ -axis, $m = 0$ .
$θ = 9 0^{\circ}$ : line is parallel to the $y$ -axis (vertical); $m$ is undefined.
$0^{\circ} < θ < 9 0^{\circ}$ : line rises left-to-right, $m > 0$ .
$9 0^{\circ} < θ < 18 0^{\circ}$ : line falls left-to-right, $m < 0$ .

Figure 1. The four cases.

m = tan θ

is zero for a horizontal line, positive for

0^{\circ} < θ < 9 0^{\circ}

, undefined at

θ = 9 0^{\circ}

, and negative for

9 0^{\circ} < θ < 18 0^{\circ}

Slope from two points

If $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ are two points on a line and $x_{1} \neq = x_{2}$ , then

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

Figure 2. Slope is the tangent of the inclination:

m = tan θ = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

, the vertical step divided by the horizontal step between any two points of the line.

Solved Example 1

Find the slope of the line whose inclination is (i)

12 0^{\circ}

, (ii)

15 0^{\circ}

Solution:

(i) $m = tan 12 0^{\circ} = tan (18 0^{\circ} - 6 0^{\circ}) = - tan 6 0^{\circ} = - 3$ .

(ii) $m = tan 15 0^{\circ} = - tan 3 0^{\circ} = - \frac{1}{3}$ .

Solved Example 2

Find the slope of the line through

(1, 6)

and

(- 4, 2)

Solution:

$m = \frac{2 - 6}{- 4 - 1} = \frac{- 4}{- 5} = \frac{4}{5}$ .

2. Various Forms of the Equation of a Line

2.1 Point-slope form

Line with slope $m$ passing through $(x_{1}, y_{1})$ :

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

Solved Example 3

Find the equation of the line through

(2, - 3)

inclined at

13 5^{\circ}

with the

x

-axis.

Solution:

$m = tan 13 5^{\circ} = - 1$ . So $y - (- 3) = - 1 (x - 2)$ , i.e., $y + 3 = - x + 2$ or $x + y + 1 = 0$ .

2.2 Slope-intercept form

Line with slope $m$ and $y$ -intercept $c$ (i.e., cutting the $y$ -axis at $(0, c)$ ):

y = m x + c .

Solved Example 4

Find the equation of the line with slope

- 1

and

y

-intercept

- 4

Solution:

$y = - x - 4$ or equivalently $x + y + 4 = 0$ .

2.3 Two-point form

Line through two given points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :

y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1}) (x_{1} \neq = x_{2}) .

2.4 Determinant form

Same idea rewritten as a $3 \times 3$ determinant:

x y 1 x_{1} y_{1} 1 x_{2} y_{2} 1 = 0.

Expanding this determinant gives the two-point form directly.

Solved Example 5

Find the equation of the line joining

(- 1, 3)

and

(4, - 2)

Solution:

Slope $= \frac{- 2 - 3}{4 - ( - 1 )} = \frac{- 5}{5} = - 1$ .

Using point-slope with $(- 1, 3)$ : $y - 3 = - 1 (x + 1) \Rightarrow x + y - 2 = 0$ .

2.5 Intercept form

Line with $x$ -intercept $a$ and $y$ -intercept $b$ (i.e., meeting the axes at $(a, 0)$ and $(0, b)$ , both non-zero):

\frac{x}{a} + \frac{y}{b} = 1.

Figure 3. Intercept form. A line meeting the axes at

(a, 0)

and

(0, b)

has equation

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

, and cuts off a triangle of area

\frac{1}{2} ∣ ab ∣

with the axes.

Solved Example 6

A line passes through

(3, 4)

and the sum of its intercepts on the axes is

14

. Find its equation.

Solution:

Let intercepts be $a$ and $b$ with $a + b = 14$ , so $b = 14 - a$ . The line is $\frac{x}{a} + \frac{y}{14 - a} = 1$ .

Passing through $(3, 4)$ : $\frac{3}{a} + \frac{4}{14 - a} = 1 \Rightarrow 3 (14 - a) + 4 a = a (14 - a)$ .

$42 + a = 14 a - a^{2} \Rightarrow a^{2} - 13 a + 42 = 0 \Rightarrow (a - 6) (a - 7) = 0$ .

So $(a, b) = (6, 8)$ giving $\frac{x}{6} + \frac{y}{8} = 1$ or $4 x + 3 y = 24$ ; or $(a, b) = (7, 7)$ giving $x + y = 7$ .

2.6 Normal (perpendicular) form

If the perpendicular from the origin to the line has length $p$ and makes an angle $α$ with the positive $x$ -axis, then

x cos α + y sin α = p, p > 0, 0 \leq α < 2 π .

The constant $p$ is always taken as positive; the angle $α$ carries the direction information.

Figure 4. Normal form. If the perpendicular from the origin has length

p

and makes angle

α

with the positive

x

-axis, the line is

x cos α + y sin α = p

with

p > 0

Solved Example 7

A line is at distance

3

from the origin and the perpendicular from the origin makes an angle of

3 0^{\circ}

with the positive

x

-axis. Find its equation.

Solution:

$p = 3$ , $α = 3 0^{\circ}$ . Equation: $x cos 3 0^{\circ} + y sin 3 0^{\circ} = 3$ , i.e., $\frac{3}{2} x + \frac{1}{2} y = 3$ or $3 x + y = 6$ .

2.7 Parametric (distance) form

Line through $P (x_{1}, y_{1})$ inclined at angle $θ$ with the positive $x$ -axis:

\frac{x - x _{1}}{cos θ} = \frac{y - y _{1}}{sin θ} = r,

where $∣ r ∣$ is the distance from $P$ to the variable point $(x, y)$ on the line. So any point on the line is

(x, y) = (x_{1} + r cos θ, y_{1} + r sin θ) .

Figure 5. Parametric form. Every point of the line through

P (x_{1}, y_{1})

with inclination

θ

(x_{1} + r cos θ, y_{1} + r sin θ)

, where

r

is the signed distance from

P

This form is invaluable for "find points on this line at a given distance from a fixed point" problems.

Solved Example 8

Find the equation of the line through

A (2, 3)

making an angle

4 5^{\circ}

with the

x

-axis, and the length of its intercept between

A

and the line

x + y + 1 = 0

Solution:

Line: $\frac{x - 2}{cos 4 5 ^{\circ}} = \frac{y - 3}{sin 4 5 ^{\circ}}$ , i.e., $x - y + 1 = 0$ .

Any point on it at distance $r$ from $A$ : $(2 + \frac{r}{2}, 3 + \frac{r}{2})$ . Substituting into $x + y + 1 = 0$ : $2 + \frac{r}{2} + 3 + \frac{r}{2} + 1 = 0 \Rightarrow 2 r = - 6 \Rightarrow r = - 32$ .

Distance $= ∣ r ∣ = 32$ .

2.8 General form

Every straight line can be written as

a x + b y + c = 0, (a, b) \neq = (0, 0) .

From this general form:

Slope $= - \frac{a}{b}$ (undefined if $b = 0$ ; the line is then vertical).
$x$ -intercept $= - \frac{c}{a}$ ; $y$ -intercept $= - \frac{c}{b}$ .
Distance from origin $= \frac{∣ c ∣}{a ^{2} + b ^{2}}$ .

Solved Example 9

Find the slope,

x

-intercept, and

y

-intercept of

2 x - 3 y + 5 = 0

Solution:

Slope $= - \frac{2}{- 3} = \frac{2}{3}$ . $x$ -intercept $= - \frac{5}{2}$ . $y$ -intercept $= - \frac{5}{- 3} = \frac{5}{3}$ .

3. Angle Between Two Lines

Let $m_{1}$ and $m_{2}$ be the slopes of two intersecting lines with $m_{1} m_{2} \neq = - 1$ . The acute angle $θ$ between them satisfies

tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} .

Figure 6. The angle between two lines is the difference of their inclinations, so

tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}

. The lines are parallel when

m_{1} = m_{2}

and perpendicular when

m_{1} m_{2} = - 1

Two special cases:

Parallel lines: $m_{1} = m_{2}$ (both defined) or both undefined.
Perpendicular lines: $m_{1} m_{2} = - 1$ (both defined); alternatively one is horizontal ( $m = 0$ ) and the other is vertical.

Equivalently, for lines in general form:

$a_{1} x + b_{1} y + c_{1} = 0$ is parallel to $a_{2} x + b_{2} y + c_{2} = 0$ iff $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ .
The two lines are perpendicular iff $a_{1} a_{2} + b_{1} b_{2} = 0$ .

Line through a point making a given angle with a given line

The two lines through $(x_{1}, y_{1})$ making angle $α$ with the line $y = m x + c$ are

y - y_{1} = tan (θ - α) (x - x_{1}) and y - y_{1} = tan (θ + α) (x - x_{1}),

where $tan θ = m$ .

Solved Example 10

The acute angle between two lines is

\frac{π}{4}

and the slope of one of them is

\frac{1}{2}

. Find the slope of the other.

Solution:

Let the other slope be $m$ . Then $tan \frac{π}{4} = \frac{\frac{1}{2} - m}{1 + \frac{m}{2}} \Rightarrow 1 = \frac{1 - 2 m}{2 + m}$ .

So $1 - 2 m = \pm (2 + m)$ . Positive: $1 - 2 m = 2 + m \Rightarrow m = - \frac{1}{3}$ . Negative: $1 - 2 m = - 2 - m \Rightarrow m = 3$ .

The slope of the other line is $- \frac{1}{3}$ or $3$ .

Solved Example 11

A vertex of an equilateral triangle is

(2, 3)

and the opposite side lies on

x + y = 2

. Find the equations of the other two sides.

Solution:

Slope of $x + y = 2$ is $- 1$ . The other two sides pass through $(2, 3)$ and make $6 0^{\circ}$ with slope $- 1$ .

$tan 6 0^{\circ} = \frac{- 1 - m}{1 - m} \Rightarrow 3 (1 - m) = \pm (1 + m)$ .

Solving both signs: $m = \frac{3 - 1}{3 + 1} = 2 - 3$ or $m = 2 + 3$ .

Equations: $y - 3 = (2 - 3) (x - 2)$ and $y - 3 = (2 + 3) (x - 2)$ .

4. Distance Between Two Parallel Lines

The distance between the parallel lines $a x + b y + c_{1} = 0$ and $a x + b y + c_{2} = 0$ is

d = \frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}} .

Figure 7. Distance between parallels. With identical

a

and

b

in both equations,

d = \frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}}

; rescale one equation first if the coefficients do not already match.

Warning: the coefficients of $x$ and $y$ in both equations must be identical before applying the formula. If they aren't, multiply one equation through to match them.

Solved Example 12

Two sides of a square lie on

x + y = 1

and

x + y + 2 = 0

. Find its area.

Solution:

Side of the square $=$ distance between the two parallel lines $= \frac{∣1 - ( - 2 ) ∣}{1 ^{2} + 1 ^{2}} = \frac{3}{2}$ .

Area $= (\frac{3}{2})^{2} = \frac{9}{2}$ .

Area of a parallelogram formed by four lines

If the four sides are $y = m_{1} x + c_{1}, y = m_{1} x + c_{2}, y = m_{2} x + d_{1}, y = m_{2} x + d_{2}$ , then

Area = \frac{( c _{1} - c _{2} ) ( d _{1} - d _{2} )}{m _{1} - m _{2}} .

5. Family of Lines and Concurrency

Family through the intersection of two lines

If $L_{1} \equiv a_{1} x + b_{1} y + c_{1} = 0$ and $L_{2} \equiv a_{2} x + b_{2} y + c_{2} = 0$ meet at a point $P$ , then every line through $P$ (except $L_{2}$ itself) can be written as

L_{1} + λ L_{2} = 0, λ \in R .

Figure 8. The family of lines. If

L_{1} = 0

and

L_{2} = 0

meet at

P

, then

L_{1} + λ L_{2} = 0

passes through

P

for every real

λ

, and every line through

P

except

L_{2}

itself arises this way.

Choose $λ$ to make the resulting line satisfy the extra condition of the problem.

Concurrency of three lines

Three lines $a_{i} x + b_{i} y + c_{i} = 0$ ( $i = 1, 2, 3$ ) are concurrent iff

a_{1} b_{1} c_{1} a_{2} b_{2} c_{2} a_{3} b_{3} c_{3} = 0.

Equivalently, three lines are concurrent if constants $A, B, C$ (not all zero) exist with $A L_{1} + B L_{2} + C L_{3} \equiv 0$ (identically).

Solved Example 13

Find the equation of the line passing through

(2, - 3)

and through the intersection of

x + y + 4 = 0

and

3 x - y - 8 = 0

Solution:

Any line through the intersection: $(x + y + 4) + λ (3 x - y - 8) = 0$ . Passing through $(2, - 3)$ : $(2 - 3 + 4) + λ (6 + 3 - 8) = 0 \Rightarrow 3 + λ = 0 \Rightarrow λ = - 3$ .

Substituting: $(x + y + 4) - 3 (3 x - y - 8) = 0 \Rightarrow - 8 x + 4 y + 28 = 0 \Rightarrow 2 x - y - 7 = 0$ .

Solved Example 14

Show that

4 x + 7 y - 9 = 0

5 x - 8 y + 15 = 0

9 x - y + 6 = 0

are concurrent.

Solution:

$47 - 9 5 - 815 9 - 16 = 4 (- 48 + 15) - 7 (30 - 135) - 9 (- 5 + 72) = - 132 + 735 - 603 = 0$ . Concurrent.

6. Pair of Straight Lines Through the Origin

A homogeneous equation of degree two in $x$ and $y$ ,

a x^{2} + 2 h x y + b y^{2} = 0,

always represents a pair of straight lines passing through the origin. Dividing by $x^{2}$ and setting $\frac{y}{x} = m$ gives the quadratic $b m^{2} + 2 hm + a = 0$ whose two roots $m_{1}, m_{2}$ are the slopes of the two lines.

Slopes of the pair

m_{1} + m_{2} = - \frac{2 h}{b}, m_{1} m_{2} = \frac{a}{b} .

Figure 9. A homogeneous second-degree equation

a x^{2} + 2 h x y + b y^{2} = 0

always represents two lines through the origin, with

m_{1} + m_{2} = - \frac{2 h}{b}

m_{1} m_{2} = \frac{a}{b}

and

tan θ = \frac{2 h ^{2} - ab}{a + b}

Nature of the pair

$h^{2} > ab$ : two distinct real lines.
$h^{2} = ab$ : two coincident real lines.
$h^{2} < ab$ : no real lines - only the origin (imaginary pair, real point of intersection at $(0, 0)$ ).

Angle between the pair

The acute angle $θ$ between the two lines represented by $a x^{2} + 2 h x y + b y^{2} = 0$ is

tan θ = \frac{2 h ^{2} - ab}{a + b} .

Special cases

Perpendicular lines: $a + b = 0$ (coefficient of $x^{2}$ + coefficient of $y^{2}$ = 0).
Coincident lines: $h^{2} = ab$ .
Equally inclined to the $x$ -axis: $h = 0$ (the coefficient of $x y$ vanishes).

Extension. A homogeneous equation of degree

n

x, y

represents

n

straight lines through the origin (some possibly imaginary).

Angle bisectors of the pair

The equation of the pair of angle bisectors of $a x^{2} + 2 h x y + b y^{2} = 0$ is

\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h} .

Note: the bisectors are always perpendicular to each other (the two bisectors of any angle pair are).

Solved Example 15

Show that

6 x^{2} - 5 x y + y^{2} = 0

represents two distinct lines through the origin and find them.

Solution:

Here $a = 6, b = 1, h = - \frac{5}{2}$ . So $h^{2} - ab = \frac{25}{4} - 6 = \frac{1}{4} > 0$ : two distinct real lines.

Factorise: $6 x^{2} - 5 x y + y^{2} = (3 x - y) (2 x - y) = 0$ . Lines: $y = 3 x$ and $y = 2 x$ .

Solved Example 16

Find the angle between the pair of lines

4 x^{2} + 24 x y + 11 y^{2} = 0

Solution:

$a = 4, b = 11, 2 h = 24 \Rightarrow h = 12$ . $tan θ = \frac{2 144 - 44}{4 + 11} = \frac{2 \cdot 10}{15} = \frac{4}{3}$ .

Acute angle $= tan^{- 1} \frac{4}{3}$ .

Solved Example 17

Find the equation of the bisectors of the angles between the lines

3 x^{2} - 5 x y + 4 y^{2} = 0

Solution:

Here $a = 3, b = 4, h = - \frac{5}{2}$ . Bisector pair: $\frac{x ^{2} - y ^{2}}{3 - 4} = \frac{x y}{- 5/2}$ , i.e., $\frac{x ^{2} - y ^{2}}{- 1} = - \frac{2 x y}{5}$ .

Cross-multiplying: $- 5 (x^{2} - y^{2}) = - 2 x y \Rightarrow 5 x^{2} - 2 x y - 5 y^{2} = 0$ .

7. General Second-Degree Equation as a Pair of Lines

The general second-degree equation

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

represents a pair of straight lines if and only if

Δ \equiv ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0, i.e., ah g hb f g f c = 0,

along with $h^{2} \geq ab$ (so that the pair is real).

Angle between the pair

The angle between the two lines depends only on the second-degree part - i.e., it is the same as for $a x^{2} + 2 h x y + b y^{2} = 0$ :

tan θ = \frac{2 h ^{2} - ab}{a + b} .

Point of intersection of the pair

Assuming $h^{2} \neq = ab$ , the two lines meet at

(\frac{h f - b g}{ab - h ^{2}}, \frac{g h - a f}{ab - h ^{2}}) .

An easy way to remember: partially differentiate the LHS with respect to $x$ and $y$ , set both to zero, and solve. That gives the intersection point directly.

Condition for the pair to be parallel lines

The pair represents two parallel lines iff $h^{2} = ab$ and $a f^{2} = b g^{2}$ (or equivalently $b g^{2} = c h^{2}$ , $a f^{2} = c h^{2}$ ). The distance between the two parallel lines is

d = 2 \frac{g ^{2} - a c}{a ( a + b )} .

Solved Example 18

Show that

2 x^{2} + 5 x y + 3 y^{2} + 6 x + 7 y + 4 = 0

represents a pair of straight lines and find their point of intersection.

Solution:

Here $a = 2, b = 3, c = 4, h = \frac{5}{2}, g = 3, f = \frac{7}{2}$ .

$Δ = ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 2 \cdot 3 \cdot 4 + 2 \cdot \frac{7}{2} \cdot 3 \cdot \frac{5}{2} - 2 \cdot \frac{49}{4} - 3 \cdot 9 - 4 \cdot \frac{25}{4}$ .

$= 24 + \frac{105}{2} - \frac{49}{2} - 27 - 25 = - 28 + \frac{56}{2} = - 28 + 28 = 0$ . Pair confirmed.

Treating as a quadratic in $x$ : $2 x^{2} + (5 y + 6) x + (3 y^{2} + 7 y + 4) = 0$ . Discriminant $= (5 y + 6)^{2} - 8 (3 y^{2} + 7 y + 4) = y^{2} + 4 y + 4 = (y + 2)^{2}$ .

$x = \frac{- ( 5 y + 6 ) \pm ( y + 2 )}{4}$ , giving $x + y + 1 = 0$ and $2 x + 3 y + 4 = 0$ .

Solving: $y = - 2, x = 1$ . Point of intersection: $(1, - 2)$ .

8. Homogenisation - Joint Equation from the Origin

Given a curve $S \equiv a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ and a line $L \equiv ℓ x + m y + n = 0$ ( $n \neq = 0$ ), let the line cut the curve at $A$ and $B$ . The joint equation of the two straight lines $O A$ and $O B$ (from the origin to the intersection points) is obtained by making $S$ homogeneous of degree two using $L$ :

a x^{2} + 2 h x y + b y^{2} + 2 (g x + f y) (\frac{ℓ x + m y}{- n}) + c (\frac{ℓ x + m y}{- n})^{2} = 0.

Every term is now of degree two, so the equation represents a pair of straight lines through the origin - exactly $O A$ and $O B$ .

Figure 10. Homogenisation. Writing the line as

L = 1

and using it to raise every term of

S = 0

to degree two gives the joint equation of

O A

and

O B

, the pair of lines from the origin to the points where the line cuts the curve.

Solved Example 19

If the lines joining the origin to the points of intersection of

2 x^{2} + 3 y^{2} + m = 0

with

y = 2 x + 3

are perpendicular, find

m

Solution:

Write the line as $\frac{y - 2 x}{3} = 1$ . Homogenising $2 x^{2} + 3 y^{2} + m \cdot 1^{2} = 0$ :

$2 x^{2} + 3 y^{2} + m (\frac{y - 2 x}{3})^{2} = 0$ .

Coefficient of $x^{2}$ : $2 + \frac{4 m}{9}$ . Coefficient of $y^{2}$ : $3 + \frac{m}{9}$ .

For perpendicular lines the two coefficients sum to zero: $2 + \frac{4 m}{9} + 3 + \frac{m}{9} = 0 \Rightarrow 5 + \frac{5 m}{9} = 0 \Rightarrow m = - 9$ .

Solved Example 20

Find the equation of the lines joining the origin to the points of intersection of

3 x + 4 y - 5 = 0

with the curve

2 x^{2} + 3 y^{2} = 5

Solution:

Write $\frac{3 x + 4 y}{5} = 1$ . Homogenise: $2 x^{2} + 3 y^{2} = 5 \cdot 1^{2} = 5 \cdot (\frac{3 x + 4 y}{5})^{2} = \frac{( 3 x + 4 y ) ^{2}}{5}$ .

Multiply by $5$ : $10 x^{2} + 15 y^{2} = 9 x^{2} + 24 x y + 16 y^{2} \Rightarrow x^{2} - 24 x y - y^{2} = 0$ .

Common Mistakes to Avoid

Watch out

Slope of a vertical line: $tan 9 0^{\circ}$ is undefined - the line has no slope, not "infinite slope". Write vertical lines as $x = k$ , not as $y = m x + c$ .
Angle between lines when $1 + m_{1} m_{2} = 0$ : the formula gives division by zero because the lines are perpendicular; simply state $θ = 9 0^{\circ}$ instead of trying to plug in.
Distance between parallel lines: the coefficients of $x$ and $y$ must match. $2 x + 3 y + 5 = 0$ and $4 x + 6 y - 1 = 0$ - divide the second by $2$ first, then apply the formula.
Homogeneous eqn $a x^{2} + 2 h x y + b y^{2} = 0$ : the coefficient of $x y$ is $2 h$ , not $h$ . A student who reads $6 x y$ as $h = 6$ (instead of $h = 3$ ) will get the angle wrong.
General second-degree pair: the determinant condition $Δ = 0$ is necessary but you also need $h^{2} \geq ab$ for the lines to be real (otherwise you get an imaginary pair).
Homogenisation: the line must be written in the form $L = 1$ (i.e., $\frac{ℓ x + m y}{- n} = 1$ ), never $L = 0$ - otherwise you'd be multiplying by zero and lose information.
Family of lines: $L_{1} + λ L_{2} = 0$ generates every line through the intersection except $L_{2}$ itself. If the required line is $L_{2}$ , no finite $λ$ works.

Frequently Asked Questions

Q1. When is the slope of a line undefined?

When the line is vertical - i.e., parallel to the $y$ -axis. Its inclination is $9 0^{\circ}$ and $tan 9 0^{\circ}$ is undefined. Vertical lines have the equation $x = k$ for some constant $k$ .

Q2. Why is $p$ always taken positive in the normal form $x cos α + y sin α = p$ ?

Because $p$ represents a length - the perpendicular distance from the origin to the line - and lengths are non-negative by definition. The direction information is carried by the angle $α$ (which can range over $[0, 2 π)$ ).

Q3. What is the parametric form used for?

It gives every point on a line as $(x_{1} + r cos θ, y_{1} + r sin θ)$ where $∣ r ∣$ is the distance from a fixed point $(x_{1}, y_{1})$ . This makes it ideal for problems that ask "find the point on this line at distance $d$ from a given point" or "find the length of intercept between two curves along a given line".

Q4. How many lines can pass through the intersection of two given lines?

Infinitely many - the entire pencil of lines through that point. Algebraically, the family $L_{1} + λ L_{2} = 0$ (for varying real $λ$ ) covers every line through the point of intersection except $L_{2}$ itself.

Q5. What is the condition for $a x^{2} + 2 h x y + b y^{2} = 0$ to represent perpendicular lines?

The condition is $a + b = 0$ - i.e., the sum of the coefficients of $x^{2}$ and $y^{2}$ vanishes. This is a very useful test that requires no factoring.

Q6. What is the difference between $a x^{2} + 2 h x y + b y^{2} = 0$ and the general second-degree equation?

The homogeneous form (no first-degree or constant terms) always represents a pair of lines through the origin. The general form with all six terms represents a pair only if the extra condition $ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$ holds, and then the two lines can meet anywhere - not necessarily at the origin.

Q7. What does the homogenisation trick achieve?

Given a line $L = 0$ meeting a curve $S = 0$ at two points $A$ and $B$ , homogenisation produces a single second-degree equation whose two lines are exactly $O A$ and $O B$ . This lets you find the angle at the origin, test perpendicularity, or write the joint equation - all without solving for $A$ and $B$ explicitly.

Q8. Are the two angle bisectors of a pair of lines always perpendicular?

Yes. The two bisectors of the two angles between any pair of intersecting lines bisect supplementary angles, and supplementary bisectors are always perpendicular to each other. That's why $\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h}$ (the pair-of-bisectors equation) has coefficient of $x^{2}$ equal to negative of coefficient of $y^{2}$ - the perpendicularity condition.

Previous year questions on Introduction to Straight Lines

4 questions from past papers, each with a step-by-step solution.

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1. Slope (Gradient) of a Line

Slope from two points

2. Various Forms of the Equation of a Line

2.1 Point-slope form

2.2 Slope-intercept form

2.3 Two-point form

2.4 Determinant form

2.5 Intercept form

2.6 Normal (perpendicular) form

2.7 Parametric (distance) form

2.8 General form

3. Angle Between Two Lines

Line through a point making a given angle with a given line

4. Distance Between Two Parallel Lines

Area of a parallelogram formed by four lines

5. Family of Lines and Concurrency

Family through the intersection of two lines

Concurrency of three lines

6. Pair of Straight Lines Through the Origin

Slopes of the pair

Nature of the pair

Angle between the pair

Special cases

Angle bisectors of the pair

7. General Second-Degree Equation as a Pair of Lines

Angle between the pair

Point of intersection of the pair

Condition for the pair to be parallel lines

8. Homogenisation - Joint Equation from the Origin

Common Mistakes to Avoid

Frequently Asked Questions

Q1. When is the slope of a line undefined?

Q2. Why is p always taken positive in the normal form xcosα+ysinα=p?

Q3. What is the parametric form used for?

Q4. How many lines can pass through the intersection of two given lines?

Q5. What is the condition for ax2+2hxy+by2=0 to represent perpendicular lines?

Q6. What is the difference between ax2+2hxy+by2=0 and the general second-degree equation?

Q7. What does the homogenisation trick achieve?

Q8. Are the two angle bisectors of a pair of lines always perpendicular?

Previous year questions on Introduction to Straight Lines

Ready to master Straight Lines?

Q2. Why is $p$ always taken positive in the normal form $x cos α + y sin α = p$ ?

Q5. What is the condition for $a x^{2} + 2 h x y + b y^{2} = 0$ to represent perpendicular lines?

Q6. What is the difference between $a x^{2} + 2 h x y + b y^{2} = 0$ and the general second-degree equation?