SOLUTION: Can you explain and show how to graph the point-slope form equation y + 5 = -(x - 2). I understand that the (x,y)is my "starting" point and the slope is actually -1/1; however my

Algebra ->  Linear-equations -> SOLUTION: Can you explain and show how to graph the point-slope form equation y + 5 = -(x - 2). I understand that the (x,y)is my "starting" point and the slope is actually -1/1; however my       Log On


   



Question 176886: Can you explain and show how to graph the point-slope form equation y + 5 = -(x - 2). I understand that the (x,y)is my "starting" point and the slope is actually -1/1; however my line is in quadrants 1 and 3 instead of Q 1 & 3. . .what is correct (x,y?) Why is it not (2,5). I.E, if the equation was y - 5 = -2/3 (x + 2),then (2,5) would be correct, right? Thanks in advance for your assistance.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Can you explain and show how to graph the point-slope form equation:
y + 5 = -(x - 2)
Arrange the equation in the y = form
removing the brackets changes the signs to
y + 5 = -x + 2
y = -x + 2 - 5
y = -x - 3; the correct values for x,y are which ever satisfy this equation
:
choose two x values to substitute in this equation, find y
x = -6 then y = +3
x = +3 then y = -6
join these two points with a straight line
:
should look like this
+graph%28+300%2C+200%2C+-8%2C+8%2C+-8%2C+8%2C+-x-3%29+