## MSFFL 3 years ago For the equation find the equation of the graph that is parallel and perpendicular that passes through the specified point. Then graph the equations The equation is 3x - y = 9 So i have the equations y = 3x + 9 y = 3x - 2 y = 1/3x + 4/3 but I am confused on how to graph it

1. myko

3x - y = 9 so y = 3x-9 perpendicular: y =-1/3x -9

2. 2bornot2b

@MSFFL the equation for perpendicular line, that you have prepared is incorrect, to the best of my knowledge.

3. MSFFL

I got the 3 equation by doing the following The specified points are (1,1) 1. 3x - y = 9 2. -y = 3x + 9 3. y = 3x -9 (Slope = 3) So now 1. y = mx+b 2. 1 = 3(1) + b 3. b = -2 Equation is y = 3x - 2 for perpendicular Slope = -1/3 1. y = mx + b 2. 1 = -1/3(1) + b 3. b = 4/3 Equation is y = -1/3x + 4/3

4. myko

looks ok

5. MSFFL

Now how do I graph it?

6. myko

slope m means : for 1 horizontal unit you go up m vertical units the last term in y = mx + b equation, b, is the point where you intersect y axis

7. 2bornot2b

Here are the graphs,

8. 2bornot2b