Answer: The required equation of the line is [tex]y=4x-29.[/tex]
Step-by-step explanation: We are given to write the equation of the line that is perpendicular to the following line and passes through the point (8, 3).
[tex]5x+20y=10~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We know that
the slope-intercept form of a straight line is given by
[tex]y=mx+c,[/tex]
where m is the slope and c is the y-intercept of the line.
From equation (i), we have
[tex]5x+20y=10\\\\\Rightarrow 20y=-5x+10\\\\\\\Rightarrow y=-\dfrac{5}{20}x+\dfrac{10}{20}\\\\\\\Rightarrow y=-\dfrac{1}{4}x+\dfrac{1}{2}.[/tex]
So,
[tex]\textup{slope, }m=-\dfrac{1}{4}~~\textup{and}~~\textup{y-intercept, }c=\dfrac{1}{2}[/tex]
Since the product of the slopes of two perpendicular lines is - 1.
Let, m' be the slope of the line perpendicular to line (i).
Then, we must have
[tex]m\times m'=-1\\\\\\\Rightarrow -\dfrac{1}{4}\times m'=-1\\\\\Rightarrow m'=4.[/tex]
Since the line passes through the point (8, 3), so its equation will be
[tex]y-3=m'(x-8)\\\\\Rightarrow y-3=4(x-8)\\\\\Rightarrow y=4x-32+3\\\\\Rightarrow y=4x-29.[/tex]
Thus, the required equation of the line is [tex]y=4x-29.[/tex]
