i have one more question, and i hope someone can help with this, because im absolutely lost for this one.:
write the equation of a line that is perpendicular to the given line and passes through the given point.
y+2 = 1/3(x-5) ; (-4, 3)
i am so confused plz help
yummm ... tacos
perpendicular lines have slopes that are negative-reciprocals
the negative-reciprocal of 1/3 is -3
use point-slope to write the equation of the new line
so is it y= -3x +15?
or is it y= -1/3x -11
point-slope ... y - 3 = -3 (x - -4)
but it doesnt say to use point slope?
okay ... so which of your answers is correct?
y+2 = 1/3(x-5), (-4, 3).
3y+6 = x-5
Eq1: -x+3y = -11. slope = -A/B = 1/3.
Eq2 slope = -3.
y = mx+b.
3 = -3(-4) + b
b = -9.
Eq2: Y = -3x - 9.