A line perpendicular to the line $ax+by+c=0$ and passes through $(a,b)$. The equation of the line is
A. $bx-ay+({{a}^{2}}-{{b}^{2}})=0$
B. $bx-ay-({{a}^{2}}-{{b}^{2}})=0$
C. $bx-ay=0$
D. None of these

Hint: In this question, we are to find the equation of the line that is perpendicular to the given line and passing through the given point. To do this, the slope of the given line is evaluated and used it for the required line in the point-slope formula. Since we know that the product of the slopes of the perpendicular lines is $-1$.



Formula Used:The equation of the with a slope $m$ and passing through a point $({{x}_{1}},{{y}_{1}})$ is
$y-{{y}_{1}}=m(x-{{x}_{1}})$
The slope of the line with the equation $ax+by+c=0$ is $m=\dfrac{-a}{b}$
The product of the slopes of the two lines that are perpendicular to one another is $-1$. Hence, the slope of the perpendicular line is $\dfrac{-1}{m}$.
The equation of the line that is perpendicular to the line $ax+by+c=0$ and passing through a point $({{x}_{1}},{{y}_{1}})$ is $b(x-{{x}_{1}})-a(y-{{y}_{1}})=0$



Complete step by step solution:Here it is given that,
The equation of the line is $ax+by+c=0$
We are to find its perpendicular line’s equation.
The slope of the given line is ${{m}_{1}}=\dfrac{-a}{b}$
Thus, the slope of the perpendicular line is
 $\begin{align}
  & {{m}_{2}}=\dfrac{-1}{{{m}_{1}}} \\
 & \Rightarrow {{m}_{2}}=\dfrac{-1}{\left( \dfrac{-a}{b} \right)} \\
 & \Rightarrow {{m}_{2}}=\dfrac{b}{a} \\
\end{align}$
Thus, the equation of the required perpendicular line with the slope ${{m}_{2}}=\dfrac{b}{a}$ and the point $({{x}_{1}},{{y}_{1}})=(a,b)$ is
$\begin{align}
  & y-{{y}_{1}}={{m}_{2}}(x-{{x}_{1}}) \\
 & y-b=\dfrac{b}{a}(x-a) \\
 & a(y-b)=b(x-a) \\
 & ay-ab=bx-ab \\
 & \Rightarrow bx-ay=0 \\
\end{align}$



Option ‘C’ is correct



Note: Here we may go wrong with the calculation of the slope of the perpendicular line. We need to remember that the product of the slopes of the perpendicular lines is $-1$. To find this type of question, we can also use the direct formula mentioned above.