Question

How do you find the slope that is perpendicular to the line $5x - 3y = 2?$

Answer 1

First, by finding the derivative of the line with respect to the slope of the tangent of the given line, and then using the fact that the product of the slopes of the tangent and the normal or perpendicular equals the negative one, to find the corresponding 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 $5x - 3y = 2$ we will first find slope of the line itself by calculating its derivative with respect to $x$
$\Rightarrow 5x - 3y = 2$
Differentiating both sides of the equation with respect to $x$, we will get

$$\Rightarrow \dfrac{{d(5x - 3y)}}{{dx}} = \dfrac{{d(2)}}{{dx}} \\\ \Rightarrow 5 - 3\dfrac{{dy}}{{dx}} = 0 \\\ \Rightarrow 3\dfrac{{dy}}{{dx}} = 5 \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{5}{3} \\\$$

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{5}{3} \times m = - 1 \\\ \Rightarrow m = - \dfrac{3}{5} \\\$
Therefore $- \dfrac{3}{5}$ is the required slope of the perpendicular to the given line.

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