Question 1010002
Find two values of k such that the points (-3,4), (0,k), and (k,10) are collinear. 
----
Look at the slopes::
1st point to 3rd point::
(k-4)/(0+3) = (10-4)/(k+3)
-----
Cross multiply::
k^2 - k - 12 = 3*6
---
k^2 - k - 30 = 0
-----
(k-6)(k+5) = 0
k = 6 or k = -5
-----
Cheers,
Stan H.
-----------