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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

User Bogdan
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2 Answers

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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.
User Russell Giddings
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8.6k points
2 votes
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.
User Kingoleg
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