Question

What is the slope of a line perpendicular to the graph of the equation $5x-3y=27$?

Answer 1

In this problem we have to find the slope of a line perpendicular to the graph of the given equation. Here we can find the slope by the slope intercept form, $y=mx+b$ where m in the slope intercept form is the slope of line perpendicular to the graph of the given equation.

Complete step-by-step solution:
We know that the slope intercept form is,
$y=mx+b$
We know that the given equation is,
$5x-3y=27$
We can now bring the given equation to the slope intercept form therefore,
$\Rightarrow -3y=27-5x$
Then
$\Rightarrow -y=\dfrac{27}{3}-\dfrac{5x}{3}$
We can now multiply the whole term by negative sign (-) we get,
$\Rightarrow y=\dfrac{5x}{3}-\dfrac{27}{3}$
Now this term is in the slope intercept form therefore by equating
$\Rightarrow m=\dfrac{5}{3}$
and
$\Rightarrow b=-\dfrac{27}{3}$
To find the slope of the perpendicular line, there is a condition where,
$\Rightarrow {{m}_{1}}{{m}_{2}}=-1$ here
$\Rightarrow {{m}_{1}}=\dfrac{5}{3}$
Now we can substitute this in the perpendicular condition to get the slope of a line perpendicular to the graph of the given equation.
$\Rightarrow \dfrac{5}{3}{{m}_{2}}=-1$
By simplifying we get,
$\Rightarrow {{m}_{2}}=-\dfrac{3}{5}$
Therefore the slope of the perpendicular line to the graph of the given equation is
$\Rightarrow {{m}_{2}}=-\dfrac{3}{5}$

Therefore, the slope of a line perpendicular to the graph of the equation $5x-3y=27$ is $-\dfrac{3}{5}$.

Note: In this condition we have to remember the slope intercept form. And we have to apply the correct condition for the slope for the perpendicular line. Students will make mistakes in applying the condition either they may apply parallel conditions instead of perpendicularly. Here in the slope intercept form, m stands for slope and b is the y-intercept and will appear in the coordinate form.