Answer:
The point slope form is: [tex]y-7=-\frac{1}{3} \,(x+2)[/tex]
Step-by-step explanation:
Recall the general point-slope form of a line of slope "m" and going through a point [tex](x_0,y_0)[/tex], is given by: [tex]y-y_0=m(x-x_0)[/tex]
Notice therefore, that the given line is expressed in "point-slope" form
[tex]y-8=3(x-10)[/tex]
where the slope is given by the value: "3".
Recall that the slope of a line perpendicular to another given line of known slope "m" is the "opposite of the reciprocal" of that "m". That is:
[tex]m_{perp}=-\frac{1}{m}[/tex]
Then for our case, the line perpendicular to the one given, must have slope given by:
[tex]m_{perp}=-\frac{1}{m}=-\frac{1}{3}[/tex]
So now we can create easily the point-slope form of the requested perpendicular line that goes through the point (-2,7) using the general point-slope form mentioned above:
[tex]y-y_0=m(x-x_0)\\y-7=-\frac{1}{3} \,(x-(-2))\\y-7=-\frac{1}{3} \,(x+2)[/tex]
