(no link)find the values of k for which the points (3k - 1,k- 22),(k,k-7) and (k-1,-k-2) are collinear Share with your friends Share 0 Neha Sethi answered this Dear student Let the three points be A3k-1,k-22,Bk,k-7 and Ck-1,-k-2Since the three points are collinear so area of the triangle formed by them is zeroie12x1y2-y3+x2y3-y1+x3y1-y2=0x1y2-y3+x2y3-y1+x3y1-y2=03k-1k-7+k+2+k-k-2-k+22+k-1k-22-k+7=03k-12k-5+k-2k+20+k-1-15=06k2-15k-2k+5-2k2+20k-15k+15=04k2-12k+20=0k2-3k+5=0k=3±9-202 Regards -4 View Full Answer Vishweshwargupta answered this substitute the points in area of triangle formula since colinear area = 0 and solve -2
