Question

Equation of the line passing through the point \[\left( {1,2} \right)\]and perpendicular to the line \[y = 3x - 1\] is
A. \[x + 3y + 7 = 0\]
B. \[x + 3y - 7 = 0\]
C. \[x + 3y = 0\]
D. \[x - 3y = 0\]

Answer 1

Hint: Two lines are said to be in perpendicular if \[{m_1}{m_2} = - 1\] where \[{m_1},{m_2}\] are the slopes of the given two lines. This statement is also known as the condition of perpendicularity of two straight lines.

Complete step-by-step answer:
Given line is \[y = 3x - 1\] but , we know that the general equation of the line is \[y = mx + c\] where \[m\] is the slope of the equation.
So, slope of the line \[y = 3x - 1\] is \[{m_1} = 3\] and
Let \[{m_2}\] be the slope of the required line
By the condition of perpendicularity,\[{m_1}{m_2} = - 1\]
i.e. \[3 \times {m_2} = - 1\]
 \[\therefore {m_2} = \dfrac{{ - 1}}{3}\]
The line perpendicular to the line \[y = {m_1}x + c\] is given by\[y = {m_2}x + c\].
i.e. \[y = \dfrac{{ - 1}}{3}x + c\]
But this line is passing through \[\left( {1,2} \right)\]. So, it must satisfy the equation if we put the point \[\left( {1,2} \right)\]in the line.
\[
   \Rightarrow 2 = \dfrac{{ - 1}}{3}\left( 1 \right) + c \\
   \Rightarrow 6 = - 1 + 3c \\
   \Rightarrow 3c = 7 \\
  \therefore c = \dfrac{7}{3} \\
\]
So, the required line is
\[
  y = \dfrac{{ - 1}}{3}x + \dfrac{7}{3} \\
  y = \dfrac{{ - x + 7}}{3} \\
  3y = - x + 7 \\
  \therefore x + 3y - 7 = 0 \\
\]
Thus, the line passing through the point \[\left( {1,2} \right)\]and perpendicular to the line \[y = 3x - 1\] is \[x + 3y - 7 = 0\].
Therefore, the answer is option B. \[x + 3y - 7 = 0\].

Note: The above problem can be done by a using a direct formula i.e. equation of line passing through the point \[\left( {{x_1},{y_1}} \right)\] and perpendicular to the line \[y = mx + c\] is given by the equation\[x - {x_1} + m\left( {y - {y_1}} \right) = 0\].