Question

How do you find the equation of a line, L which passes through the point $\left( {3, - 1} \right)$ and perpendicular to the line with equation $9x - 3y = 2$?

Answer 1

In this question we have to find the equation of the line which passes through the given point and perpendicular to the given line, first we will have to find the slope of the given line by converting the given line into slope-intercept form, which is $y = mx + b$ and then we will get the slope of the given line, then by using the fact which is, if the slopes of the lines that are perpendicular are ${m_1}$ and ${m_2}$, then product of the slopes will be -1, i.e, ${m_1} \times {m_2} = - 1$, by using the formula we will get the slope of the required line and then by using slope-point formula which is given by, - {y_1} = m\left( {x - {x_1}} \right)$, by substituting the given point and the slope and further simplification we will get the required equation of the line.

Complete step by step solution:
Given that the required equation which passes through the point $\left( {3, - 1} \right)$ and perpendicular to the line with equation $9x - 3y = 2$,
First convert the given equation of the line $9x - 3y = 2$, into slope intercept form, i.e, $y = mx + b$,
$\Rightarrow 9x - 3y = 2$,
Now add 3y to both sides of the equation we get,
$\Rightarrow 9x - 3y + 3y = 2 + 3y$,
Now simplifying we get,
$\Rightarrow 9x = 2 + 3y$,
Now subtract 2 from both sides of the equations we get,
$\Rightarrow 9x - 2 = 2 + 3y - 2$,
Now simplifying we get,
$\Rightarrow 9x - 2 = 3y$,
Now dividing both sides with 3, we get,
$\Rightarrow \dfrac{{3y}}{3} = \dfrac{{9x}}{3} - \dfrac{2}{3}$,
Now simplifying we get,
$\Rightarrow y = 3x - \dfrac{2}{3}$,
This is in form of slope intercept form $y = mx + b$, so, here $m = 3$ and $b = - \dfrac{2}{3}$,
So, slope of the given line is 3,
Now using the fact that slope of the given line, then by using the fact which is, if the slopes of the lines that are perpendicular are ${m_1}$ and ${m_2}$, then product of the slopes will be -1, i.e, ${m_1} \times {m_2} = - 1$,
So, here ${m_1} = 3$and the slope of the required line is ${m_2}$, by substituting the values in the formula we get,
$\Rightarrow 3 \times {m_2} = - 1$,
Now dividing both sides with -3, we get,
$\Rightarrow \dfrac{{3 \times {m_2}}}{3} = \dfrac{{ - 1}}{3}$,
Now simplifying we get,
$\Rightarrow {m_2} = \dfrac{{ - 1}}{3}$,
So, the slope of the required line is $\dfrac{{ - 1}}{3}$,
Now by using slope-point formula which is given by, $y - {y_1} = m\left( {x - {x_1}} \right)$, so here, point is $\left( {3, - 1} \right)$ and the slope is $\dfrac{{ - 1}}{3}$,
So here ${x_1} = 3$ and ${y_1} = - 1$ and $m = \dfrac{{ - 1}}{3}$, by substituting the values in the formula we get,
$\Rightarrow \left( {y - \left( { - 1} \right)} \right) = \dfrac{{ - 1}}{3}\left( {x - 3} \right)$,
Now simplifying we get,
$\Rightarrow \left( {y + 1} \right) = \dfrac{{ - 1}}{3}\left( {x - 3} \right)$,
Now taking 3 to the left hand and multiplying we get,
$\Rightarrow 3\left( {y + 1} \right) = - 1\left( {x - 3} \right)$,
Now multiplying we get,
$\Rightarrow 3y + 3 = - x + 3$,
Now simplifying we get,
$\Rightarrow 3y + 3 + x - 3 = 0$,
Now eliminating like terms we get,
$\Rightarrow x + 3y = 0$,
So, the required line is $x + 3y = 0$.

**
$\therefore$The equation of a line which passes through the point $\left( {3, - 1} \right)$ and perpendicular to the line with equation will be $x + 3y = 0$.**

Note:
Remember that if the slope of a line is equal to zero then it is parallel to x-axis and if the slope tends to infinity then it is perpendicular to x-axis. Also, we can remember that if the x-coordinates of the two points through which the line passes are same it must be perpendicular to the x-axis and y-coordinates of the two points through which the line passes are same it must be parallel to the x-axis.