 Hello and welcome to the session. Let us understand the following problem today. If area of the triangle is 35 square units with vertices 2, minus 6, 5, 4 and k, 4 then k is a12, b minus 2, c minus 12, minus 2, d12, minus 2. Now that is like the solution. Now area of the triangle with the given vertices is given by half 2 minus 6, 1, 5, 4, 1, k, 4, 1 which is equal to plus minus 35 which implies 2 minus 6, 1, 5, 4, 1, k, 4, 1 is equal to plus minus 70. Now applying r2 tends to r2 minus r3. We will get it implies 2 minus 6, 1, 5 minus k, 0, 0, k, 4, 1 which is equal to plus minus 70 which implies now minus of 5 minus k into minus 6, minus 4 which implies minus of 5 minus k into minus 10 which implies 10 into 5 minus k is equal to plus minus 70 which implies 50 minus 10 k is equal to plus minus 70. Now we have two cases case 1 when 50 minus 10 k is equal to plus 70 it implies 50 minus 70 is equal to 10 k which implies 10 k is equal to minus 20 which implies k is equal to minus 2. Now case 2 when 50 minus 10 k is equal to minus 70 which implies 50 plus 70 is equal to 10 k which implies 10 k is equal to 120 which implies k is equal to 12. So the two values of k are k is equal to 12 comma minus 2 and the required values of k are 12 comma minus 2. I hope you understood the problem and here the correct answer is D. I hope you enjoyed the session bye and have a nice day.