Q:

URGENT PLEASE HELPThe equation of line CD is y = 3x βˆ’ 3. Write an equation of a line perpendicular to line CD in slope-intercept form that contains point (3, 1). y = negative 1 over 3x + 2 y = negative 1 over 3x + 0

Accepted Solution

A:
Answer:y = - [tex]\frac{1}{3}[/tex] x + 2Step-by-step explanation:The equation of a line in slope- intercept form isy = mx + c ( m is the slope and c the y- intercept )y = 3x - 3 ← is in slope- intercept formwith slope m = 3Given a line with slope m then the slope of a line perpendicular to it is[tex]m_{perpendicular}[/tex] = - [tex]\frac{1}{m}[/tex] = - [tex]\frac{1}{3}[/tex], hencey = - [tex]\frac{1}{3}[/tex] x + c ← is the partial equation of the perpendicular lineTo find c substitute (3, 1) into the partial equation1 = - 1 + c β‡’ c = 1 + 1 = 2y = - [tex]\frac{1}{3}[/tex] x + 2 ← equation of perpendicular line