Posted by ann on Thursday, January 12, 2012 at 10:22pm.
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, Saturday, January 14, 2012 at 8:21pm
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, Saturday, January 14, 2012 at 8:26pm
ADDITION: Let Y = 0 to find Xint.
X + 3Y = 8,
X + 3*0 = 8,
X = 8 = xint.

algebra  ann, Saturday, January 14, 2012 at 9:00pm
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.
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