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A line perpendicular to x + 3y = 5 passes through (1, -1). What is the equation of the line in standard form?
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Thanks but how did you get it?
Alright...First, write the equation of the first line in the standard form as 3y=5-x y=(5/3)-(x/3) Therefore, the slope of the above line is -1/3 Now, you can simply get the slope of the perpendicular line as
m2=-(1/m1)=-(1/-(1/3))=3 So, the equation of the new line is y=m2x+b=3x+b but this line passes through (1,-1)...substitute the point into the above equation you get -1=3(1)+b --> b=-4 Therefore, y=3x-4
Oh wow. Thank you so much
No problem :)
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