If l is a line whose equation is y=5x-3, find the equation of the image of l under each of the following translations a. (x, y) --> (x, y-6) b. (x, y) --> (x+2,y)

Consider the line y=5x-3. Two solutions (x, y) of this equation, are points on the line.

For example, let x=0, then y=-3. So (0, -3) is a point on the line.

If x=1, then y=5*1-3=5-3=2, so (1, 2) is a point on the line.

a) 

Consider the transformations of these two points under (x,y) --> (x, y-6):

(0,-3) --> (0, -3-6)=(0, -9),
(1, 2) --> (1, 2-6)=(1, -4).

So (0, -3) and (1, 2) are translated to respectively (0, -9) and (1, -4).

The whole line is translated to the line containing (0, -9) and (1, -4).

The slope of the new line is . The equation of the line is:

                                   , 

that is,                               y=5x-9.

b) 

Consider the transformations of these two points under (x,y) --> (x+2, y):

(0,-3) --> (0+2, -3)=(2, -3),
(1, 2) --> (1+2, 2)=(3, 2).

The slope of the new line is .

The equation of the line is
                 
                                   , 
which can be written as
                                        y=5x-13.

a)  y=5x-9

b)  y=5x-13

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4