first 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
How do I put an equation into standard form, Ax+Bx=C?
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]
2 answers
the 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
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