perpendicular to 3y+x=2, through the pt. (2,-5). This is exactly as it appears on my take home assignment. Can someone walk me through the solution? I would really appreciate it, as I am completely lost.
2006-09-27 04:19:42 · 5 answers · asked by poohntao 2 in Science & Mathematics ➔ Mathematics
3y+x=2
3y= (-1)x +2
y=(-1/3)x+(2/3)
So according to y=mx+c slope of the line =(-1/3)
Now when two lines are perpendicular
m1*m2= -1 , m2 is slope of given line
So m1*(-1/3)= -1
so m1=3 , m1 is the slope of line perpendicular to the line you gave .
The line perpendicular to the given one passes (2,-5) and has slope 3
So using the equation (y-y1)=m1(x-x1)
where (x1,y1)=(2,-5)
(y-(-5))=3(x-2)
y+5=3x-6
y-3x+11=0
2006-09-27 04:38:59 · answer #1 · answered by Anonymous · 0⤊ 0⤋
First, put 3y+x=2 in slope intercept form: y=-x/3+2/3. Therefore, the slope is equal to -1/3, so the slope of the new line must be positive three.
2006-09-27 04:22:49 · answer #2 · answered by bruinfan 7 · 0⤊ 0⤋
3y + x = 2
3y = -x + 2
y = (-1/3)x + (2/3)
Perpendicular Lines have opposite reciprocal slopes
(2,-5), m = 3
-5 = 3(2) + b
-5 = 6 + b
b = -11
ANS : 3x - 11
2006-09-27 04:33:34 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
3y+x=2
y=-x/3+2/3
slope of this eqn m=-1/3
slope of a line perpendicular to it=3
(since product of slopes of per. lines=-1)
So for the required eqn. we have
slope=3
point=2,-5
y=mx+c
substituting
-5=3*2+c
c=-11
the required eqn is
y=3x-11
2006-09-27 05:59:55 · answer #4 · answered by openpsychy 6 · 0⤊ 0⤋
3y + x = 2 .... y = -1/3x + 2/3 m = -1/3
"M" perpendicular to -1/3 is +3
fit into equation (y-y1) = m (x-x1) where (x1,y1) = (2,-5)
y-(-5) = 3(x-2) .... y+5 = 3x - 6
y = 3x - 11
2006-09-27 05:39:28 · answer #5 · answered by Brian D 5 · 0⤊ 0⤋