Question

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

Answer 1

Hint: In this question, we are to find the equation of the perpendicular line for the given equation. This is obtained by using the property of the straight line which states that the product of slopes of two perpendicular lines is $-1$.

Formula used: The standard equation for a line is $ax+by+c=0$.
The equation of a line that has a slope $m$ and passing through the point $({{x}_{1}},{{y}_{1}})$ is
$y-{{y}_{1}}=m(x-{{x}_{1}})$
The slope of a line $ax+by+c=0$ is $m=\dfrac{-a}{b}$
The slopes of two lines that are perpendicular lines are given by the condition,
${{m}_{1}}\times {{m}_{2}}=-1$

Complete step by step solution: Given line is $ax+by+c=0\text{ }...(1)$
The slope of line (1) is
${{m}_{1}}=\dfrac{-a}{b}$
Then, the slope of the line that is perpendicular to (1) is
$\begin{align}
  & {{m}_{1}}\times {{m}_{2}}=-1 \\
 & \Rightarrow {{m}_{2}}=\dfrac{-1}{{{m}_{1}}} \\
 & \Rightarrow {{m}_{2}}=\dfrac{-1}{\dfrac{-a}{b}}=\dfrac{b}{a} \\
\end{align}$
Then, the equation with slope $m=\dfrac{b}{a}$ and passes through a point $({{x}_{1}},{{y}_{1}})=(a,b)$ is
$\begin{align}
  & y-{{y}_{1}}=m(x-{{x}_{1}}) \\
 & \Rightarrow y-b=\dfrac{b}{a}(x-a) \\
 & \Rightarrow ay-ab=bx-ab \\
 & \therefore bx-ay=0 \\
\end{align}$

Thus, Option (A) is correct.

Note: Here we need to remember that, the required line is perpendicular to the given line. Here we can also solve this by applying a direct equation i.e., $bx-ay+k=0$ where $k$ is a constant. By substituting the given point through the line passes, we get the $k$ value. Then, the required equation is obtained.