Find the direct variation equation of the graph through the points (0, 0) and (1, -2). write in y=kx form.
a. y = 2x
b. y = -2x
c. y = 1/2 x
d. y = -1/2 x

find the direct variation equation which passes through (0, 0) and (4, 1).
a. y = 4x
b. y = -4x
c. y = 1/4 x
d. y = -1/4 x

1. ( y - 0)/( -2 - 0) = (x - 0)/(1 - 0);
Then, y/(-2) = x/1;
finalyy, y = -2x;
The correct answer is B.

2. In the same way, we obtain y =4x; the correct answer is A.

The direct variation between x and y may be expressed as y = kx where k is the constant of variation. Substitute the values of the abscissa to the x and ordinates to the y's to get the k's

1. 0 = k(0)     ; -2 = k(1)    ; k = -2.  
  Therefore, the direct variation is y = -2x. The answer is letter B.

2. 0 = k(0)     ; 1 = k(4)     ; k = 1/4
  Therefore, the direct variation is y = (1/4) x. The answer is letter C.