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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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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.

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