Given:
y = 6x + 4
The slope of a perpendicular line, is the negative reciprocal of the slope of the original line.
Using the slope intercept form:
y = mx + b
Where m is the slope and b is the y-intercept
The slope of the origi line: y = 6x + 4 is = 6
The negative reciprocal of 6 is:
[tex]-\frac{1}{6}[/tex]Thus, the slope of the perpendicular line is
[tex]-\frac{1}{6}[/tex]To find the equation of the perpendicular line that passes through the point (5, 10), use the slope-intercept form:
y = mx + b
Substitute -1/6 for m, 5 for x and 10 for y to find b.
We have:
[tex]\begin{gathered} y=mx+b \\ \\ 10=-\frac{1}{6}\ast5+b \\ \\ 10=-\frac{5}{6}+b \\ \\ \text{Multiply through by 6:} \\ 10\ast6=-\frac{5}{6}\ast6+6b \\ \\ 60=-5+6b \\ \\ 60+5=-5+5+6b \\ \\ 65=6b \\ \\ \frac{65}{6}=b \end{gathered}[/tex]Therefore, the equation of the perperndicular line in slope intercept form is:
[tex]y=-\frac{1}{6}x+\frac{65}{6}[/tex]ANSWER:
[tex]y=-\frac{1}{6}x+\frac{65}{6}[/tex]