Question

for what value K the points A(1,5), B(K,1) and C (4,11) are collinear

Answer 1

for the points to be collinear,
x1(y2-y3) + x2(y3-y1) + x3(y1-y2)=0... (1)
now the points are A(1,5) B(K,1) and C(4,11) i.e.
x1=1 y1=5
x2=K y2=1
x3=4 y3=11
put these values in equation 1..
1(1-11) + K(11-5) + 4(5-1) =0
-10+ 6K +16 =0
6K+6=0.... 6K=-6
i.e. K=-1...
hope it helps you..