SOLUTION: Hi! I need some help here :). I need to find the equations of lines (i) parallel and (ii) perpendicular to the following and passing through the given point. Here is the the equat

Algebra ->  Finance -> SOLUTION: Hi! I need some help here :). I need to find the equations of lines (i) parallel and (ii) perpendicular to the following and passing through the given point. Here is the the equat      Log On


   

Question 844453: Hi! I need some help here :). I need to find the equations of lines (i) parallel and (ii) perpendicular to the following and passing through the given point.
Here is the the equation and given point.
"3x - 2y + 4 = 0" and "(3,4)"
Thanks a bunch!
Found 2 solutions by josgarithmetic, fcabanski:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
(i)
If parallel, then the coefficients on x and y are unchanged; but the constant term will be different.
3x-2y%2Bp=0 is a parallel line for any p%3C%3E4.
Solving for p and then using (3,4),
p=2y-3x
p=2%2A4-3%2A3
p=-1
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The parallel line is highlight%283x-2y-1=0%29.

(ii)
Sufficient understanding of the standard form of the equation, know that the slope is 3%2F2, and then the slope of a line perpendicular to the given line must be -%282%2F3%29. The equation we want is 2x%2B3y%2Br=0 and we want to use the given point to solve for r. (One should examine standard form and compare to slope-intercept form to learn how you can use standard form).
-
First solve symbolically for r.
r=-2x-3y.
Substitute the point that it should contain.
r=-2%2A3-3%2A4
r=-6-12
r=-18
-
Now knowing r, the equation containing (3,4) and perpendicular to 3x-2y+4=0 is highlight%282x%2B3y-18=0%29.

Answer by fcabanski(1391) About Me  (Show Source):
You can put this solution on YOUR website!
Manipulate the equation into the slope intercept form. y = mx + b where m is the slope.


3x-2y + 4=0


-2y=-3x + -4
y = (3/2)x + +2. The slope is 3/2.


Parallel Lines have the same slope. Perpendicular lines have slopes that are negative reciprocals. y-y1 = m(x-x1) is the equation of a line passing through the point (x1,y1)


Parallel: slope is 3/2: y-4 = 3/2 (x-3)


y-4 = (3/2)x - 9/2


y = (3/2)x -1/2


Perpendicular: slope is -2/3


y-4 = -2/3 (x-3)


y-4 = (-2/3)x +2


y= (-2/3)x +6