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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.
line that is passing through the point of (-2,5)
and is perpendicular to the equation y=3x+4. slope m = 3/1
perpendicular have slopes that are negative reciproclas of one another:
New Line
y = (-1/3)x + b |Using ordered pair pt(-2,5) to solve for b
5 = (-1/3)*-2 + b
13/3 = b
y = (-1/3)x + 13/3 OR x+ 3y = 13