Question 427670
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Hi
Using the standard slope-intercept form for an equation of a line y = mx + b
  where m is the slope and b the y-intercept. 
 y=1/3x-4  |Slope m = (1/3) AND the slope of a line perpendicular to it is -3/1
 Perpendicular Lines have slopes that are negative reciproclas of one another.
New Line:
 y = -3x + b    |using Pt(0,1) to solve for b
 1 = -3*0 + b
 1 = b   EQ of this line is y = -3x+1
{{{drawing(300,300,   -6, 6, -6, 6,  grid(1),
circle(0, -4,0.3),
circle(0, 1,0.3),
graph( 300, 300, -6, 6, -6, 6,0,-3x+1,(1/3)x-4))}}}