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Question 79648: Hello,I'm having problems with some slope-intercept problems that I really need help with,The first one is,Write an equation in slope-intercept form of a line with the following characteristics:perpendicular to the graph of:2x-3y=12 and passes through (4,3).The next one is,Write an equation in slope-intercept form of a line with the following characteristics.Perpendicular to the graph of:y=3/2x-3 with a y-intercept of 5.The third one is,Write an equation in slope-intercept form of a line with the following characteristics:parallel to the graph of:-4x+3y=12 and passes through the origin.And the last one is,Change y=x-3 so that the graph of the new equation has a steeper,negative slope and the same y-intercept,thanks to whoever can help.
Answer by checkley75(3666)   (Show Source):
You can put this solution on YOUR website! 2X-3Y=12
-3Y=-2X+12
Y=-2X/-3+12/-3
Y=2X/3-4 THIS (RED) LINE HAS A SLOPE(m) OF 2/3.
FOR A LINE TO BE PERPENDICULAR TO THIS LINE THE SLOPE MUST BE THE NEGATIVE RECIPRICAL OF THIS SLOPE. THUS THE NEW SLOPE IS -3/2.
NOW WE SUBSTITUTE THE POINTS (4,3) FOR X & Y AND USING THE NEW SLOPE (-3/2) WE SOLVE THE LINE EQUATION [Y=mX+b] FOR b THE Y INTERCEPT.
3=-3/2*3+b
3=-9/2+b
b=3+9/2
b=3+4.5
b=7.5 THE Y INTERCEPT.
THUS THE EQUATION IS
Y=-3X/2+7.5 (THE GREEN LINE).
 (graph 300x300 pixels, x from -10 to 10, y from -10 to 10, of TWO functions y = 2x/3 -4 and y = -3x/2 +7.5).
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solved the first one. now see if you can handle the rest. :)
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