Posted by ann on .
Could you please help me??
pointslope form linear equation
y3 = 3(x+1)
what is the equation in standard form of a perpendicular line that passes through (5,1). i think you take 1/3 as slope, reciprocal, I am confused about how to proceed. I know you put variables on one side and constants on other side in standard form.
what is the xintercept of the perpendicular line?
thanks

algebra 
Henry,
I assume the line is perpendicular to
the given line.
y3 = 3(x+1).
y3 = 3x+3,
3x+y = 6,
3xy = 6.
m1 = A/B = 3/1 = 3.
m2 = 1/3 = Negative reciprocal.
P(5,1).
y = mx + b,
y = (1/3)*5 + b = 1,
b = 1 + 5/3 = 2 2/3 = 8/3.
y = (1/3)x+8/3,
Convert to STD form:
x/3 + y = 8/3,
X + 3Y = 8. 
algebra 
Henry,
ADDITION: Let Y = 0 to find Xint.
X + 3Y = 8,
X + 3*0 = 8,
X = 8 = xint. 
algebra 
ann,
thank you so much for your help Henry, I really appreciate it. Ann
I understand now where I went wrong. first mistake i made was using positive 1/3 and then it was downhill from there.