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April 20, 2014

April 20, 2014

Posted by **ann** on Thursday, January 12, 2012 at 10:22pm.

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?

thanks

- algebra -
**Henry**, Saturday, January 14, 2012 at 8:21pmI 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.

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 8:26pmADDITION: 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 9:00pmthank 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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