Question

For which values of k is the set of order pairs (2,3),(k, 6),(4,0) a function? ​

Answer 1

Slope1 = (6–4)/(k-2) = 2/(k-2)

Slope2 = (6–0)/(k-4) = 6/(k-4)

Setting them equal to the known slope of (4–0)/(2-4) = 4/(-2) we will find k

2/(k-2) = -2

2 = (-2)*(k-2)

-1 = k-2

1 = k

checking with Slope2 we have

6/(k-4) = 6/(1–4) = 6/(-3) = -2.