 Hello and welcome to this session. Let us understand the following problem today. Find value of k if area of triangle is 4 square units and vertices are k,0,4,0,1,0,2. Now where is y? We know area of triangle is given by half into x1, y1,1, x2, y2,1, x3, y3,1 and given to us is area is equal to 4 square units. Now area is equal to half into, putting the vertices here we get k,0,1,4,0,1,0,2,1. Now solving this we get half into k into 0 minus 2 minus 0 into 4 minus 0 plus 1 into 8 minus 0, which is equal to half into minus 2k minus 0 plus 8, which is equal to half into minus 2k plus 8, which is equal to taking two common so we two and two get cancelled so we are left with 4 minus k. Now we have got the equation as delta is equal to, that is area is equal to 4 minus k and this is given to be 4. So it implies 4 is equal to 4 minus k, which implies k is equal to 0. Now this is a situation when k is equal to 4, now when k is equal to minus 4, then minus 4 is equal to 4 minus k, which implies k is equal to 8. Hence the required answer is 0,8. I hope you understood the problem, bye and have a nice day.