# 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. \[x - 3y = 0\]

Last updated date: 17th Mar 2023

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Answer

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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\[x - {x_1} + m\left( {y - {y_1}} \right) = 0\].

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\[x - {x_1} + m\left( {y - {y_1}} \right) = 0\].

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