Since it is parallel, the new equation must look like
y = -x + b, (it would differ only in the constant)
but (2,-2) lies on this line, so
-2 = -(2) + b
b = 0
new equation :
y = -x
write an equation in slope intercept form of the line that passses through the given point and is parallel to the graph of the given equation
(2,-2) ;y=-x-2
2 answers
Parallel lines have the same slope.
So, you want a line with slope = -1
Now you have a point and a slope. Use that to get
y+2 = -1(x-2)
y+2 = -x+2
y = -x
Or, given y = -x-2, you know that any parallel line will be y = -x + b
So, plug in x=2 into your equation, and you have
y = -2-2 = -4
But, you want y = -2, which is 2 units above that line. So, your final line is
y = -x-2+2 = -2x
So, you want a line with slope = -1
Now you have a point and a slope. Use that to get
y+2 = -1(x-2)
y+2 = -x+2
y = -x
Or, given y = -x-2, you know that any parallel line will be y = -x + b
So, plug in x=2 into your equation, and you have
y = -2-2 = -4
But, you want y = -2, which is 2 units above that line. So, your final line is
y = -x-2+2 = -2x
1 answer