First I would suggest expanding your equation for the line from C to D, then get y =
and see what is connected to x. The piece connected to x is the original slope. In this case -3, so you need the perpendicular slope (that is when you multiply them together they equal 1 ... as they are negative reciprocals : )
So your new slope is 1/3, now sub in the point and solve for b : )
Line CD passes through point E.
Line CD: y−5=−3(x+2)
Point E: (6,−1)
What is the equation of the line that is perpendicular to line CD?
y=1/3x−4
y=−1/3x−19/3
y=−1/3x+5
y=−1/3x+19/3
y=1/3x+4
3 answers
Line CD: y-5 = -3(x+2), E(6, -1).
y-5 = -3x-6,
3x + y = -1.
m1 = -A/B = -3/1 = -3 = slope of CD.
m2 = 1/3 = slope of line perpendicular to CD.
Y = mx+b.
-1 = (1/3)6 + b,
b = -3.
y-5 = -3x-6,
3x + y = -1.
m1 = -A/B = -3/1 = -3 = slope of CD.
m2 = 1/3 = slope of line perpendicular to CD.
Y = mx+b.
-1 = (1/3)6 + b,
b = -3.
Note: Point E(6, -1) does not satisfy the given Eq.
There is an error somewhere.
There is an error somewhere.
1 answer