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Sagot :
A line that is parallel has the same slop where a line that is perpendicular has a slop that is negative and reciprocal.
so for the parallel one you don't need to worry about the slop because it will be 2/3x. But yous the point slope equation form
y-y1=m(x-x1)
y+5=2/3(x+2)
y+5=2/3x+ 4/3
y=2/3x-11/3
-2/3x+y=-11/3
multiple by -1 so A inst negative
2/3-y=11/3
For a line that is perpendicular you just need to flip the original 2/3x slope and make it negative.
y+5=-3/2(x+2)
y+5=-3/2x-3
y=-3/2x-8
3/2x+y=-8
so for the parallel one you don't need to worry about the slop because it will be 2/3x. But yous the point slope equation form
y-y1=m(x-x1)
y+5=2/3(x+2)
y+5=2/3x+ 4/3
y=2/3x-11/3
-2/3x+y=-11/3
multiple by -1 so A inst negative
2/3-y=11/3
For a line that is perpendicular you just need to flip the original 2/3x slope and make it negative.
y+5=-3/2(x+2)
y+5=-3/2x-3
y=-3/2x-8
3/2x+y=-8
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