
Equation of a line through the origin and perpendicular to, the line joining $(a,0)$ and $(-a,0)$ is
A. \[y=0\]
B. \[x=0\]
C. \[x=-a\]
D. \[y=-a\]
Answer
217.2k+ views
Hint: In this question, we are to find the equation of the line passing through the origin and perpendicular to the other line equation. To find this, we have a standard equation for the line in the coordinate geometry. By applying the given values in it, we get the required line equation.
Formula Used:The equation of the line passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
The equation of the line perpendicular to $ax+by+c=0$ and passing through $({{x}_{1}},{{y}_{1}})$ is
$b(x-{{x}_{1}})-a(y-{{y}_{1}})=0$
Complete step by step solution:It is given that,
The required equation of the line is perpendicular to the line joining the given points. Those points are:
\[\begin{align}
& ({{x}_{1}},{{y}_{1}})=(a,0) \\
& ({{x}_{2}},{{y}_{2}})=(-a,0) \\
\end{align}\]
Then, the equation of the line that passes through these points is obtained by using the formula we have
The equation of the line passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
On substituting,
\[\begin{align}
& y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}}) \\
& \text{ }y-0=\dfrac{0-0}{-a-a}(x-a) \\
& \text{ }\therefore y=0\text{ }...(1) \\
\end{align}\]
Then, the equation of the line perpendicular to the line (1) and passing through the origin is,
Here \[({{x}_{1}},{{y}_{1}})=(0,0);a=0;b=1;c=0\]
\[\begin{align}
& b(x-{{x}_{1}})-a(y-{{y}_{1}})=0 \\
& \Rightarrow 1(x-0)-0(y-0)=0 \\
& \Rightarrow x=0 \\
\end{align}\]
Option ‘B’ is correct
Note: Here we may go wrong with the condition that the required line should be perpendicular to the given line. Firstly, we need to find the line passing through the given points and then we can find the perpendicular line equation using the obtained line by the formula we have in the coordinate geometry of straight lines.
Formula Used:The equation of the line passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
The equation of the line perpendicular to $ax+by+c=0$ and passing through $({{x}_{1}},{{y}_{1}})$ is
$b(x-{{x}_{1}})-a(y-{{y}_{1}})=0$
Complete step by step solution:It is given that,
The required equation of the line is perpendicular to the line joining the given points. Those points are:
\[\begin{align}
& ({{x}_{1}},{{y}_{1}})=(a,0) \\
& ({{x}_{2}},{{y}_{2}})=(-a,0) \\
\end{align}\]
Then, the equation of the line that passes through these points is obtained by using the formula we have
The equation of the line passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
On substituting,
\[\begin{align}
& y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}}) \\
& \text{ }y-0=\dfrac{0-0}{-a-a}(x-a) \\
& \text{ }\therefore y=0\text{ }...(1) \\
\end{align}\]
Then, the equation of the line perpendicular to the line (1) and passing through the origin is,
Here \[({{x}_{1}},{{y}_{1}})=(0,0);a=0;b=1;c=0\]
\[\begin{align}
& b(x-{{x}_{1}})-a(y-{{y}_{1}})=0 \\
& \Rightarrow 1(x-0)-0(y-0)=0 \\
& \Rightarrow x=0 \\
\end{align}\]
Option ‘B’ is correct
Note: Here we may go wrong with the condition that the required line should be perpendicular to the given line. Firstly, we need to find the line passing through the given points and then we can find the perpendicular line equation using the obtained line by the formula we have in the coordinate geometry of straight lines.
Recently Updated Pages
Area vs Volume: Key Differences Explained for Students

Mutually Exclusive vs Independent Events: Key Differences Explained

Addition of Three Vectors: Methods & Examples

Addition of Vectors: Simple Guide for Students

Algebra Made Easy: Step-by-Step Guide for Students

Relations and Functions: Complete Guide for Students

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding Atomic Structure for Beginners

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

