Family of Lines: Definition, Types, Properties, Equation and Formula

In this article, we will cover the concept of Family of lines. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of fifteen questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2022.

What is Family of Lines?

A family of lines are lines that have similar features, properties, and characteristics. A family of lines has the same common intersection point. A common characteristic of a family of lines is they have the same slope.

Let us get the general equation of the straight line passing through a given fixed point in terms of two particular straight lines through the same point.

Equation of Family of Lines

Any equation of the line through the point of intersection of the lines $\mathrm{L}_1=\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{L}_2=\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$ can be represented as
$
\begin{aligned}
& \mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1+\lambda\left(\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2\right)=0 \\
& \text { or, } \mathrm{L}_1+\lambda \mathrm{L}_2=0
\end{aligned}
$

Where $\lambda$ is a parameter.

Note:

• The equation $\mathrm{L}_1+\lambda \mathrm{L}_2=0$ or $\mu \mathrm{L}_1+\mathrm{vL} \mathrm{L}_2=0$ represents a line passing through the intersection of the lines $L_1=0$ and $L_2=0$ which is a fixed point. And $\lambda, \mu, \mathrm{v}$ are constants.

Types of Families of Lines

Family of lines are of the following three types:

• Family of Intersecting Straight Lines
• Family of Parallel Straight Lines
• Family of Perpendicular Straight Lines

Family of Intersecting Straight Lines

If the y-intercept of a family of lines is the same, they form a family of intersecting straight lines. A family of Intersecting Lines passes through a common point. The slope for each line may vary, keeping the y-intercept constant for all.

The general equation of this family can be given as $\mathrm{y}-\mathrm{y}_1=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)$

Family of Parallel Straight Lines

The family of Parallel Straight Lines consists of lines whose slope remains the same for all, but the $y$-intercept varies.

If there is a line $a x+b y+c=0$, then the line parallel to this line is given by $\mathrm{ax}+\mathrm{by}+\mathrm{k}=0$, where k is a parameter.

Family of Perpendicular Straight Lines

A family of perpendicular straight lines refers to a set of straight lines in a plane such that each line in the family is perpendicular to each other. We can say that any two lines chosen from this family will always intersect at right angles.

If a family of lines is being represented as ax + by $+\mathrm{c}=0$ then, any two lines from this family having slopes $m_1$ and $m_2$ respectively, will be perpendicular if
$
m_1 \times m_2=-1
$

If there is a line $a x+b y+c=0$, then the line perpendicular to this line is given by $bx -ay +k=0$, where k is a parameter.

Properties of Family of Lines

The properties of Family of Lines are:

• The members of the family of lines share common properties such as all members will be parallel or intersecting or perpendicular.
• For an intersecting straight line, the sum of angles formed by adjacent crossing lines will be equal to $180^{\circ}$.
• The opposite angles at each intersection will be equal if the family of lines has the same slope.
• In the family of perpendicular lines, the angle between any two lines is $90^{\circ}$.
• A Family of Lines can have infinite members having common properties

Important points

1) Consider variable straight line $a x+b y+c=0$, where $a, b, c$ belong to $R$. For randomly chosen values of $a, b$, and c, lines obtained are not necessarily concurrent. But if $a, b, c$ are related by equation $a l+b m+c n=0$, where $I, m, n$ are constants, then lines are concurrent for different values of $a, b, c$ which satisfy the above equation

2) A straight line is such that the algebraic sum of the perpendicular drawn on it from the number of fixed points is zero. Then the line always passes through a fixed point which is the mean of the given points.

Solved Examples Based on Family of Lines

Example 1: Let a line $L$ pass through the point of intersection of the lines $\mathrm{b} x+10 y-8=0$ and $2 x-3 y=0, b \in \mathbf{R}-\left\{\frac{4}{3}\right\}$. If the line L also passes through the point $(1,1)$ and touches the circle $17\left(\mathrm{x}^2+\mathrm{y}^2\right)=16$, then the eccentricity of the ellipse $\frac{\mathrm{x}^2}{5}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ is [JEE MAINS 2022]

Solution

Line is passing through intersection of $b x+10 y-8=0$ and $2 x-3 y=0$ is $(b x+10 y-8)+\lambda(2 x-3 y)=0$.
As line is Passing through $(1,1)$ so $\lambda=\mathrm{b}+2$
$
\text { so } \frac{8}{\sqrt{(3 b+4)^2+(3 b-4)^2}}=\frac{4}{\sqrt{17}}
$

Now line $(3 b+4) x-(3 b-4) y-8=0$ is tangent to circle $17\left(x^2+y^2\right)=16^{b^2}=2 \Rightarrow e=\sqrt{\frac{3}{5}}$
Hence, the answer is $\sqrt{\frac{3}{5}}$
Example 2: Given the family of lines, $a(3 x+4 y+6)+b(x+y+2)=0$. The line of the family situated at the greatest distance from the point $\mathrm{P}(2,3)$ has the equation.
Solution: Two-point form of a straight line
$
y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)
$
wherein
The lines pass through $\left(x_1 y_1\right)$ and $\left(x_2 y_2\right)$
Equation of a line perpendicular to a given line -
$B x-A y+\lambda=0$ is the line perpendicular to $A x+B y+C=0$.

wherein

$\lambda$ is some other constant than $C$.
point of intersection is $\mathrm{A}(-2,0)$. The required line will be one which passes through $(-2,0)$ and is perpendicular to the line joining $\mathrm{A}(-2,0)$ and $\mathrm{B}(2,3)$ slope of $A B=3 / 4$

Hence, the slope of the required line will be $-4 / 3$
Hence, the required line will be $4 x+3 y+k=0$ which passes through $(-2,0)$.
Hence, the required line is $4 x+3 y+8=0$

Example 3 : If the lines $3 y+4 x=1, y=x+5$ and $5 y+b x=3$ are concurrent, then the value of $b$ is
Solution: Substituting the coordinates of the point of intersection of the first two lines, i.e. $(-2,3)$ in the third equation, we have $5.3-2 b=3 \Rightarrow b=6$.
Hence, the answer is 6
Example 4: The number of possible straight lines, passing through $(2,3)$ and forming a triangle with coordinate axes, whose area is $12 \mathrm{sq} . \mathrm{units}$, is

Solution: The equation of any line through $(2,3)$ is $\mathrm{y}-3=\mathrm{m}(\mathrm{x}-2)$
$
\mathrm{y}=\mathrm{mx}-2 \mathrm{~m}+3
$
with the help of the fig. area of $\triangle \mathrm{OAB}= \pm 12$
ie. $\frac{1}{2}\left(\frac{2 \mathrm{~m}-3}{\mathrm{~m}}\right)(3-2 \mathrm{~m})= \pm 12$
taking + sign me get $(2 m+3)^2=0$
this gives one value of $\mathrm{m}=-3 / 2$
taking negative signs we get
$
4 \mathrm{~m}^2-36 \mathrm{~m}+9=0 \quad(\mathrm{D}>0)
$
quadratic in m gives 2 values of m
$\Rightarrow \quad 3$ st. lines are possible.
Hence, the answer is three

Example 5: A line $4 \mathrm{x}+\mathrm{y}=1$ through the point $\mathrm{A}(2,-7)$ meets the line BC , whose equation is $3 \mathrm{x}-4 \mathrm{y}+1=0$ at the point B . Find the equation to the line AC, so that $\mathrm{AB}=\mathrm{AC}$.

Solution: Equation of line AB is $4 \mathrm{x}+\mathrm{y}=1$.
Its slope $=-4$.
Slope of line $3 x-4 y+1=0$ is $3 / 4$.
If $\alpha$ is angle ABC, then
$
\tan \alpha=\frac{-4-3 / 4}{1+(-4) \cdot(3 / 4)}=\frac{19}{8}
$

Given that $\mathrm{AB}=\mathrm{AC}$

Hence the given line passes through the point $(\mathrm{a} / \mathrm{c}, \mathrm{b} / \mathrm{c})$, which is a fixed point since $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are constants.
$
\therefore \angle \mathrm{ACB}=\angle \mathrm{ABC}=\alpha
$

If the slope of AC is m , then
$
\begin{aligned}
& \qquad \tan \alpha=\frac{19}{8}= \pm \frac{\mathrm{m}-3 / 4}{1+\mathrm{m} \cdot(3 / 4)} \\
& \text { If the slope of } \mathrm{AC} \text { is } \mathrm{m} \text {, then } \\
& \text { or } 19(4+3 \mathrm{~m})= \pm 8(4 \mathrm{~m}-3) \Rightarrow \mathrm{m}=-4 \text { or }-52 / 89
\end{aligned}
$

But -4 is the slope of line AB , the slope of AC
$
=\mathrm{m}=-52 / 89
$

Hence the equation of the line that passes through $\mathrm{A}(2,-7)$ and has a slope
$
\begin{aligned}
& \mathrm{m}=-52 / 89 \text { is } \quad \mathrm{y}+7=\left(-\frac{52}{89}\right)(\mathrm{x}-2) \\
& \text { or } \quad 52 \mathrm{x}+89 \mathrm{y}+519=0
\end{aligned}
$

Hence, the answer is $52 x+89 y+519=0$

Summary

A family of lines refers to a group of straight lines that share a common geometric property or mathematical relationship, typically characterized by a parameter or set of parameters. This parameterization allows for a versatile representation of lines within the family, accommodating various orientations, slopes, and intercepts while maintaining a unifying characteristic.