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November 29, 2014

November 29, 2014

Posted by **Kelsie** on Thursday, February 4, 2010 at 10:15pm.

I have the given points (-1,3) and (2,-4).

Also, how do you write an equation in standard form when perpendicular through another?

[the question is "through (-1,2) and perpendicular to 2x-3y=-5]

- math -
**lelia**, Thursday, February 4, 2010 at 10:30pmfirst use y2-y1/x2-x1 to find the slope

-4-3/2-(-1)=-7/3

then(assuming the equation is parellel)use point slope form y-y=slope(x-x)

y-3=-7/3(x-(-1))

y-3=-7/3x-7/3

y=-7/3x+2/3

then to put it in standard form it must be in the Ax+By=C form with no fractions and the leading coefficient cannot be negative.

y=-7/3x+2/3

7/3x+y=2/3

3(7/3x+y=2/3)

7x+y=2

- math -
**lelia**, Thursday, February 4, 2010 at 10:36pmthe way i would work it is first move the equation to standard form because i find it easier to work it that way.

2x-3y=-5

-3y=-2x-5

y=2/3x+5/3

an equation perpendicular to this equation would have a slope of -3/2 because perpendicular lines have opposite recirocal slopes. now use point slope.

y-2=-3/2(x-(-1))

y-2=-3/2x-3/2

y=-3/2x+1/2

now move the equation back to standard form.

y=-3/2x+1/2

3/2x+y=1/2

2(3/2x+y=1/2)

3x+y=1

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