Answer:
The equation of a line that is perpendicular to [tex]y=\frac{7}{5}x+6[/tex] and that passes through the point (2, −6) is [tex]y=-\frac{5}{7}x-\frac{32}{7}[/tex].
Step-by-step explanation:
If a linear equation is written in this form [tex]y=mx+b[/tex], m represents the slope and b represents the y-intercept.
When one line has a slope of [tex]m[/tex], a perpendicular line has a slope of [tex]\frac{-1}{m}[/tex].
We know the slope/intercept equation of the line [tex]y=\frac{7}{5}x+6[/tex], the slope of this equation is [tex]\frac{7}{5}[/tex] and the negative reciprocal of that slope is:
[tex]m=\frac{-1}{\frac{7}{5} } \\m=-\frac{5}{7}[/tex]
The perpendicular line will have a slope of:
[tex]y-y_1=(-\frac{5}{7})(x-x_1)[/tex]
we use the point given (2, -6) to find the equation of the perpendicular line
[tex]y-(-6)=(-\frac{5}{7})(x-2)[/tex] and we solve for y
[tex]y+6-6=\left(-\frac{5}{7}\right)\left(x-2\right)-6\\\\y=-\frac{5}{7}x+\frac{10}{7}-6\\\\y=-\frac{5}{7}x-\frac{32}{7}[/tex]
The equation of a line that is perpendicular to [tex]y=\frac{7}{5}x+6[/tex] and that passes through the point (2, −6) is [tex]y=-\frac{5}{7}x-\frac{32}{7}[/tex].
We can check our work with the graph of the two lines.
