Find the value of K for which the points (3K-1,K-2),(K,K-7)and (K-1,-K-2) are collinear.

Find the value of K for which the points (3K-1,K-2),(K,K-7)and (K-1,-K

1.	Find the value of K for which the points (3K-1,K-2),(K,K-7)and (K-1,-K-2) are collinear.
Answer» Given points are A(3k -1, k - 2), B(k, k - 7) and C(k-1, -k - 2)We know that points A, B, C will be collinear, if the area of the ΔABC =0Area of ΔABC={tex}\\frac{1}{2}{/tex}\|x1(y2 -\xa0y3)+x2(y3\xa0- y1)+x3(y1\xa0- y2)\|here, x1\xa0=3k-1, x2\xa0= k, x3= k-1, y1= k-2, y2= k-7, y3= -k-2Area of ΔABC = 0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2}{/tex}\|(3k−1)[(k−7)−(−k−2)]+k[(−k−2)−(k−2)]+(k−1)[(k−2)−(k−7)]\|=0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2}{/tex}\|(3k−1)(k−7+k+2)+k(−k−2−k+2)+(k−1)(k−2−k+7)\|=0{tex}\\Rightarrow{/tex}\xa0\|(3k−1)(2k−5)+k(−2k)+(k−1)(5)\|=0{tex}\\Rightarrow{/tex}\xa0\|3k(2k−5)−1(2k−5)−2k2+5k−5\|=0{tex}\\Rightarrow{/tex}\xa0\|6k2−15 k−2k+5−2k2+5k−5\|=0{tex}\\Rightarrow{/tex}\xa0\|4k2 - 10k - 2k\|=0{tex}\\Rightarrow{/tex}\xa04k2 - 12k = 0{tex}\\Rightarrow{/tex}\xa04k(k-3) = 0{tex}\\Rightarrow{/tex}\xa0k = 0 or k - 3 =0{tex}\\Rightarrow{/tex}\xa0k = 0 or k = 3