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    				| 1. | For what value of `k`are thepoints `(k ,2-2k),(-k+1,2k)a n d(-4-k ,6-2k)`collinear? | 
| Answer» Let the three given points be `A-=(x_1,y_1)-=(k,2-2k),B-=(x_2,y_2)-=(-k+1,2k)`, and `C-=(x_3,y_3)-=(-4-k,6-2k)`. If the given points are collinear, then `Delta=0`, i.e., `x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0` or `k(2k-6+2k)+(-k+1)(6-2k-2+2k)+ (-4-k)(2-2k) =0` or `k(4k-6)-4(k-1) +(4+k)(4k-2)=0` or `4k^2 -6k-4k+4+4k^2+14k-8=0` or `8k^2+4k-4=0` or `2k^2+k-1=0` or `(2k^2-1)(k+1)=0` i.e., `k=(1)/(2)` or `-1` But for `t=1//2`, the first two points are coincident. Hence ,`K=-1`. | |