Find the value of k where area is 24sq.km with vertices (1,-1),(-4,2k),(-k,-5)
Answers
Answered by
0
24=1/2|{1(2k+5)-4(-5+1)-k(-1-2k)}|
48=|2k+5+16+k+2k^2|
48=|2k^2+3k+21|
now break mod
2k^2+3k+21=48
2k^2+3k-27=0
k={-3+_√225}/4=(-3+_15)/4=3,-9/2
k=3,-9/2
again
2k^2+3k+21=-48
2k^2+3k+69=0
but this isn't possible because D<0
hence k=3 and -9/2 is possible
48=|2k+5+16+k+2k^2|
48=|2k^2+3k+21|
now break mod
2k^2+3k+21=48
2k^2+3k-27=0
k={-3+_√225}/4=(-3+_15)/4=3,-9/2
k=3,-9/2
again
2k^2+3k+21=-48
2k^2+3k+69=0
but this isn't possible because D<0
hence k=3 and -9/2 is possible
Similar questions