A point A divides the join of P−5,1 and Q3,5 in the ratio k:1, Then the integral value of k for which the area of ΔABC, where B is 1,5 and C is 7,−2, is equal to 2 units in magnitude is

Using section formula, A=3k−5k+1,5k+1k+1.Area of triangle ABC is 2 sq. units. Therefore,121516−713k−5k+15k+1k+11=±2Operating R2→R2;R3→R3−R1, we get 1516−703k−5k+15k+1k+10=±4or 65k+1−5k−5k+1+73k−5−k−1k+1=±4or −24+72k−6=±4k+1i,e., k=7 of k=319