Answer :
Given point = (2, -12)
We know, the slope of a line perpendicular to another line is the opposite reciprocal.
So slope of the new line will be [tex]- \frac{1}{3} [/tex]
We know, point slope form is,
[tex]y-y_1=m(x-x_1)\\\\ y-(-12)= -\frac{1}{3}(x-2)\\\\ \boxed{y+12= -\frac{1}{3}(x-2)}\\\\y+12= -\frac{x}{3} + \frac{2}{3}\\\\y= -\frac{x}{3} + \frac{2}{3}-12\\\\y= -\frac{x}{3} + \frac{2-36}{3}\\\\ \boxed{y= -\frac{x}{3} - \frac{34}{3}}[/tex]
The first box is your answer in point-slope form. The second box is your answer in slope-intercept form.
We know, the slope of a line perpendicular to another line is the opposite reciprocal.
So slope of the new line will be [tex]- \frac{1}{3} [/tex]
We know, point slope form is,
[tex]y-y_1=m(x-x_1)\\\\ y-(-12)= -\frac{1}{3}(x-2)\\\\ \boxed{y+12= -\frac{1}{3}(x-2)}\\\\y+12= -\frac{x}{3} + \frac{2}{3}\\\\y= -\frac{x}{3} + \frac{2}{3}-12\\\\y= -\frac{x}{3} + \frac{2-36}{3}\\\\ \boxed{y= -\frac{x}{3} - \frac{34}{3}}[/tex]
The first box is your answer in point-slope form. The second box is your answer in slope-intercept form.