What this means is that you want two equations that do not intersect:
For 2x+3y=5, you need to put it in slope intercept form: y=mx+b.
First, you need to subtract 2x on both sides:
3y=-2x+5
Divide 3 on both sides:
y=-2/3x+5/3
Now try this formula for points (-4,-1):
(y-y1)=m(x-x1)
y-(-1)=(-2/3)(x-(-4))
y+1=(-2/3)(x+4)
y+1=-2/3x-8/3
y=-2/3x-8/3-3/3
y=-2/3x-11/3
Now you should be able to graph them on graph paper.
2006-10-29 03:17:34
·
answer #1
·
answered by Anonymous
·
5⤊
0⤋
the equation paralell to the above line can be written as....
2x+3y=k (k is a constant)
since the line passes through (-4,-1) ...this must satisfy the above equation so ...
2(-4)+3(-1)=k
which implies....k =-11
there fore 2x+3y+11=0 is the eqn
2006-10-29 11:10:35
·
answer #2
·
answered by indurti karthik 2
·
1⤊
0⤋
Normal Equation for line pass from x0,y0,z0 and parallel with a,b,c Vector (P.D. of line) is: (x-x0)/a = (y-y0)/b = (z-z0)/c
That: a,b,c <>0.
(only if numerator equal zero, denominator must equal zero)
2x+3y=5 ==> x/3+y/2 = 5/6 ==> y/2 - (5/3)/2 = -(x/3) ==>
(y-5/3)/2 = x/-3 ===>> a = -3 , b = 2
Your: x0 = -4 , y0 = -1
Then, Your line is: (x-(-4))/-3 = (y-(-1))/2 ==> (x+4)/-3 = (y+1)/2
==> 2x+3y+11 = 0.
2006-10-30 07:36:45
·
answer #3
·
answered by Saeed M 1
·
0⤊
0⤋
2x + 3y = -11 or 2x + 3y + 11 = 0
2006-10-29 11:27:56
·
answer #4
·
answered by sikar 2
·
1⤊
0⤋
Sorry, you're wrong about the slope ...
Rearrange to the form y= mx +b. You'll find the slope is not 3. Plug in the slope for m. Plug in the point (-4,1) for x and y and solve for b for the new equation.
2006-10-29 11:10:46
·
answer #5
·
answered by Gene 7
·
1⤊
0⤋
if lines are parellel
a1x+b1y+c1=0
a2x+b2y+c2=0
therefore
a1=a2
b1=b2
2x+3y-5=0
if parallel
2x+3y+c=0
put value (-4,-1)
instead of x and y
2*-4+3*-1+c=0
c=11
therefore eq. is
2x+3y+11=0
2006-10-29 11:13:11
·
answer #6
·
answered by Anonymous
·
1⤊
0⤋