Answer
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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.
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