Would this be B?
3(-1) - 2(-2) = 1
(-3) - (-4) =1
1 = 1
What is the standard form of the equation of the line passing through the point (-1,-2) and perpendicular to the line y= - 2/3x - 1?
a. 3x-2y=-1
b.3x-2y=1
c.3x+2y=1
d.3x+2y=-1
I don't understand, can anyone help me please?
2 answers
Yes.
However, I don't think you understand why. Just checking to see which of the lines contains your point is not enough.
The line y = -2/3x - 1 has slope -2/3.
Any line perpendicular to that line has slope -1/(-2/3) = 3/2.
So, now you have a slope and a point.
Using the point-slope form a line,
y-(-2) = 3/2 (x-(-1))
y+2 = 3/2 (x+1)
2(y+2) = 3(x+1)
2y + 4 = 3x + 3
3x - 2y = 1
However, I don't think you understand why. Just checking to see which of the lines contains your point is not enough.
The line y = -2/3x - 1 has slope -2/3.
Any line perpendicular to that line has slope -1/(-2/3) = 3/2.
So, now you have a slope and a point.
Using the point-slope form a line,
y-(-2) = 3/2 (x-(-1))
y+2 = 3/2 (x+1)
2(y+2) = 3(x+1)
2y + 4 = 3x + 3
3x - 2y = 1
1 answer