Question

How do you find the slope that is perpendicular to the line $- 9x = - 6y + 18?$

Answer 1

First find the slope of the tangent of the given line by finding the derivative of the line with respect to $x$ and then use the fact that product of slopes of a tangent and a normal or perpendicular equals negative one, to find the respective slope of the perpendicular to the given line.

Complete step by step solution:
In order to find the slope that is perpendicular to the given line $- 9x = - 6y + 18$ we will first find slope of the line itself by calculating its derivative with respect to $x$
$\Rightarrow - 9x = - 6y + 18$
Differentiating both sides of the equation with respect to $x$, we will get

$$\Rightarrow \dfrac{{d( - 9x)}}{{dx}} = \dfrac{{d\left( { - 6y + 18} \right)}}{{dx}} \\\ \Rightarrow - 9\dfrac{{d(x)}}{{dx}} = \dfrac{{d\left( { - 6y + 18} \right)}}{{dx}} \\\ \Rightarrow - 9 = - 6\dfrac{{dy}}{{dx}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 9}}{{ - 6}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{3}{2} \\\$$

Now, we know that product of slopes of two perpendicular lines perpendicular to each other, is equals to $- 1$, that is if we consider the slope of a line ${m_1}$ and slope of another line which is perpendicular to the first line, to be ${m_2}$ then it can be written as
${m_1}{m_2} = - 1$
So, let us consider the slope of required perpendicular to be $m$, then we can write,
$\Rightarrow \dfrac{3}{2} \times m = - 1 \\\ \Rightarrow m = - \dfrac{2}{3} \\\$ Therefore $- \dfrac{2}{3}$ is the required slope of the perpendicular to the given line.

Note: Sometimes when line is parallel to x-axis then the slope of its perpendicular is equals to infinity and when the line is parallel to y-axis then slope of the perpendicular will become zero.
This problem can be directly solved by differentiating the given equation of the line with respect to $y$ and then put a negative sign in the derivative, you will get the slope of perpendicular.