SOLUTION: Write an equation of the line in standard form that Satisfies the stated conditions 1) perpendicular to the line 3x-2y=6 at the point where it crosses the y axis. 2) parallel to

Algebra ->  Rational-functions -> SOLUTION: Write an equation of the line in standard form that Satisfies the stated conditions 1) perpendicular to the line 3x-2y=6 at the point where it crosses the y axis. 2) parallel to       Log On


   

Question 902881: Write an equation of the line in standard form that Satisfies the stated conditions
1) perpendicular to the line 3x-2y=6 at the point where it crosses the y axis.
2) parallel to the line 3x + 4y=6; passing through the x-intercept of the graph of the line x + 2y= 6
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Number 2:

Any 3x+4y=c will be a parallel line to the first equation.

The x-intercept of x%2B2y=6, is x%2B2%2A0=6, x%2B0=6, highlight_green%28x=6%29.
This is the point (6,0).

You want 3x+4y=c to pass through or contain point (6,0).
c=3%2A6%2B4%2A0
highlight_green%28c=18%29.

The line you are looking for is highlight%283x%2B4y=18%29.


Question Number 1:
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You should study what I showed for Number 2, and then solve Number 1 yourself.
In number 2, you wanted a line parallel to another. In Number 1 you want a line PERPENDICULAR to another. Either use some understanding of standard form, or convert the given first equation into slope-intercept form, and identify the slope. You want the product of the slopes to be -1. for lines to be perpendicular. Use the needed point to find the value for c in Ax+By=C, along with the needed values for A and B in the expression Ax+By.
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LESS WORDY WAY TO THINK OF THIS,
3x-2y=6
-2y=-3x%2B6
y=%28-3%2F-2%29x%2B6%2F%28-2%29
What is the slope?
You want to use slope of m%28-3%2F-2%29=-1 for the line you want.
NOW, known slope m, known given point, determine what is C.