Parallel and Perpendicular Lines - Definition, Properties and Examples

In this article, we will cover the concept of Parallel and Perpendicular 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 ten questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2014, one in 2015, one in 2019, two in 2020, two in 2021, and one in 2022.

What are parallel and perpendicular lines?

If two straight lines lie in the same plane, and if they never intersect each other, they are called parallel lines. They are always the same distance apart and are equidistant lines. The symbol || is used to represent parallel lines. On the other hand, when two lines intersect each other at an angle of 90°, they are known as perpendicular lines. Perpendicular lines are denoted by the symbol ⊥.

Properties of Parallel Lines

• Parallel lines are always equidistant from each other.
• They never meet at any common point.
• They lie in the same plane.

Properties of Perpendicular Lines

• Perpendicular lines always intersect at 90°.
• All perpendicular lines can be termed intersecting lines, but all intersecting lines cannot be called perpendicular because they need to intersect at right angles

Difference Between Parallel and Perpendicular Lines

The following table shows the difference between parallel and perpendicular lines.

Line parallel and perpendicular to a given line

Equation of line parallel to a given line

If the angle between the two lines is $0^{\circ}$ or $\pi$ then the lines are parallel to each other. In this case, $m_1=m_2$ where $m_1$ and $m_2$ are slopes of two lines.

The equation of the line parallel to $a x+b y+c=0$ is given as $a x+b y+\lambda=$ 0 , where $\lambda$ is some constant.

Equation of the given line is $a x+b y+c=0$
Its slope is $(-\mathrm{a} / \mathrm{b})$
So, any equation of line parallel to $a x+b y+c=0$ is
$
\begin{aligned}
& y=\left(-\frac{a}{b}\right) x+c_1 \\
& a x+b y-b c_1=0 \\
& a x+b y+\lambda=0
\end{aligned}
$

The inclination of a line parallel to the x-axis is 0 degrees. Thus, the slope of a horizontal line is tan(0 degree0) = 0.

Equation of lines perpendicular to each other

If the angle between the two line is $\frac{\pi}{2}$ or $-\frac{\pi}{2}$ then lines are perpendicular to each other. Then in this case $m_1 \cdot m_2=-1$ where $m_1$ and $m_2$ are slopes of two lines.

The equation of the line perpendicular to $a x+b y+c=0$ is given as $b x-a y+$ $\lambda=0$, where $\lambda$ is some constant.

Equation of the given line is $a x+b y+c=0$
Its slope is $(-\mathrm{a} / \mathrm{b})$
The slope of the perpendicular line will be (b/a)
So, any equation of line perpendicular to $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ is
$
\begin{aligned}
& y=\left(\frac{b}{a}\right) x+c_1 \\
& b x-b y-a c_1=0 \\
& b x-a y+\lambda=0
\end{aligned}
$

Solved Examples Based on Parallel and Perpendicular Lines

Example 1: The equations of the sides $\mathrm{AB}, \mathrm{BC}$ and CA of a triangle ABC are $2 \mathrm{x}+\mathrm{y}=0, \mathrm{x}+\mathrm{py}=15 \mathrm{a}$ and $\mathrm{x}-\mathrm{y}=3$ respectively. If its orthocentre is $(2, \mathrm{a}),-\frac{1}{2}<\mathrm{a}<2$, then P is equal to [JEE MAINS 2022]

Solution:
Coordinates of $\mathrm{A}(1,-2), \mathrm{B}\left(\frac{15 \mathrm{a}}{1-2 \mathrm{p}} ; \frac{-30 \mathrm{a}}{1-2 \mathrm{P}}\right)$ and orthocentre $\mathrm{H}(2, \mathrm{a})$
Slope of $\mathrm{AH}=\mathrm{p}$
$
\mathrm{a}+2=\mathrm{P}
$

Slope of $\mathrm{BH}=-1$
$
31 \mathrm{a}-2 \mathrm{ab}=15 \mathrm{a}+4 \mathrm{p}-2
$
from(1) and(2)
$
a=1 \& P=3
$

Hence, the answer is 3 .

Example 2: A square ABCD has all its vertices on the curve $x^2 y^2=1$. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is ____________. [JEE MAINS 2021]

Solution

$\begin{aligned} & \mathrm{OA} \perp \mathrm{OB} \\ & \Rightarrow\left(\frac{0-\left(-\frac{1}{q}\right)}{0-q}\right)\left(\frac{0-\frac{1}{p}}{0-p}\right)=-1 \\ & \Rightarrow\left(\frac{1}{\mathrm{p}^2}\right)\left(-\frac{1}{\mathrm{q}^2}\right)=-1 \\ & \Rightarrow x^2 y^2=1 \\ & \Rightarrow(p+q)^2\left(\frac{1}{p}-\frac{1}{q}\right)^2=16 \\ & \Rightarrow(p+q)^2(p-q)^2=16 \\ & \Rightarrow\left(p^2-q^2\right)^2=16 \\ & \Rightarrow p^2-\frac{1}{p^2}= \pm 4 \\ & \Rightarrow \mathrm{p}^4 \pm 4 \mathrm{p}^2-1=0 \\ & \Rightarrow \mathrm{p}^2=\frac{ \pm 4 \pm \sqrt{20}}{2}= \pm 2 \pm \sqrt{5} \\ & \Rightarrow \mathrm{p}^2=2+\sqrt{5} \text { or }-2+\sqrt{5}\end{aligned}$

$\begin{aligned} & \mathrm{OB}^2=\mathrm{p}^2+\frac{1}{\mathrm{p}^2}=2+\sqrt{5}+\frac{1}{2+\sqrt{5}} \text { or }-2+\sqrt{5}+\frac{1}{-2+\sqrt{5}}=2 \sqrt{5} \\ & \text { Area }=4\left(\frac{1}{2}\right)(\mathrm{OA})(\mathrm{OB})=2(\mathrm{OB})^2=4 \sqrt{5}\end{aligned}$

Hence, the answer is 80.

Example 3: The locus of the mid-points of the perpendiculars drawn from points on the line, $x=2 y$ to the line $x=y$. [JEE MAINS 2020]

Solution: The slope of the line joining two points

If $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ are two points on a straight line then the slope of the line is
$
\tan \theta=\frac{B C}{A C}=\frac{y_2-y_1}{x_2-x_1}
$

The equation of the line parallel to $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ is given as $\mathrm{ax}+\mathrm{by}+\lambda=0$, where $\lambda$ is some constant.
Equation of the given line is $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$
Its slope is $(-a / b)$
So, any equation of line parallel to $a x+b y+c=0$ is $\mathrm{ax}+\mathrm{by}+\mathrm{k}=0$ where k is some constant whose value can be obtained by given conditions.

The equation of the line perpendicular to $a x+b y+c=0$ is given as $b x-a y+\lambda=0$, where $\lambda$ is some constant.
$
\begin{aligned}
& \text { slope of } P Q=\frac{x-a}{y-2 a}=-1 \\
& \Rightarrow x-a=-y+2 a \\
& \Rightarrow a=\frac{x+y}{3}
\end{aligned}
$

Using midpoint

$
\begin{aligned}
& 2 x=2 a+b \\
& 2 y=a+b \\
& a=2 x-2 y \\
& \frac{\mathrm{x}+\mathrm{y}}{3}=2(\mathrm{x}-\mathrm{y})
\end{aligned}
$
so locus is $6 x-6 y=x+y$
$
5 \mathrm{x}=7 \mathrm{y}
$

Hence, the answer is $5 x-7 y=0$

Example 4: In a triangle $P Q R$, the coordinates of the points $P$ and $Q$ are $(-2,4)$ and $(4,-2)$ respectively. If the equation of the perpendicular bisector of PR is $2 x-y+2=0$, then the centre of the circumcircle of the $\triangle P Q R$ is : [JEE MAINS 2021]

Solution

The equation of line PQ is

$
\begin{aligned}
& y+2=\frac{4-(-2)}{-2-4}(x-4) \\
& x+y=2
\end{aligned}
$

The slope of the perpendicular bisector of PQ is 1 and passes through the midpoint of P and Q.
The equation of the perpendicular bisector of $P Q$ is
$
y=x
$

Solving with $2 x-y+2=0$ will give the circumcenter of triangle PQR
Hence, the Circumcentre of the triangle is $(-2,-2)$.

Example 5: If a $\triangle A B C$ has vertices $A(-1,7), B(-7,1)$ and $C(5,-5)$, then its orthocentre has coordinates:
[JEE MAINS 2020]

Solution: Let orthocentre is $\mathrm{H}\left(\mathrm{x}_0, \mathrm{y}_0\right)$

$\begin{aligned} & m_{\mathrm{AH}} \cdot \mathrm{m}_{\mathrm{BC}}=-1 \\ \Rightarrow & \left(\frac{\mathrm{y}_0-7}{\mathrm{x}_0+1}\right)\left(\frac{1+5}{-7-5}\right)=-1 \\ \Rightarrow & 2 \mathrm{x}_0-\mathrm{y}_0+9=0\end{aligned}$
$\begin{aligned} & \text { and } m_{B H} \cdot m_{A C}=-1 \\ & \Rightarrow\left(\frac{\mathrm{y}_0-1}{\mathrm{x}_0+7}\right)\left(\frac{7-(-5)}{-1-5}\right)=-1 \\ & \Rightarrow x_0-2 y_0+9=0\end{aligned}$

Solving equations (1) and (2)

we get

$\left(\mathrm{x}_0, \mathrm{y}_0\right) \equiv(-3,3)$

Hence, the answer is (-3, 3).

Summary

It can be said that if the slope of two lines is the same, they are identified as parallel lines, whereas, if the slope of two given lines are negative reciprocals of each other, they are identified as perpendicular lines.