Answer:
[tex]\textsf{Slope of line q}: \quad 3[/tex]
[tex]\textsf{Slope of line p}:\quad -\dfrac{1}{3}[/tex]
[tex]\textsf{Equation of line p in slope-point form}: \quad y+5=-\dfrac{1}{3}(x-6)[/tex]
[tex]\textsf{Equation of line p in slope-intercept form}: \quad y=-\dfrac{1}{3}x-3[/tex]
Step-by-step explanation:
Slope-intercept form of a linear equation: [tex]y = mx + b[/tex]
(where m is the slope and b is the y-intercept)
Given:
Therefore, the slope of line q is 3.
As line p is perpendicular to line q, the slope of line p is the negative reciprocal of the slope of line q.
Therefore, the slope of line p is -1/3
Equation of line p, using the point-slope form, the slope of -1/3 and the point (6, -5):
[tex]\begin{aligned}y-y_1 &=m(x-x_1)\\\implies y-(-5) &=-\dfrac{1}{3}(x-6)\\y+5 &=-\dfrac{1}{3}(x-6)\end{aligned}[/tex]
Simplify to slope-intercept form:
[tex]\implies y=-\dfrac{1}{3}x-3[/tex]
