Posted by ann on Thursday, January 12, 2012 at 10:22pm.
Could you please help me??
point-slope form linear equation
y-3 = 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 x-intercept of the perpendicular line?
- algebra - Henry, Saturday, January 14, 2012 at 8:21pm
I assume the line is perpendicular to
the given line.
y-3 = 3(x+1).
y-3 = 3x+3,
-3x+y = 6,
3x-y = -6.
m1 = -A/B = -3/-1 = 3.
m2 = -1/3 = Negative reciprocal.
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 X-int.
X + 3Y = 8,
X + 3*0 = 8,
X = 8 = x-int.
- 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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