Find the values of K if Area of the triangle is 4 sq. units and vertices are (k,0) (4,0) (0,2) .

 Solution:- Given that Area of triangle =4 sq. units 

Then, 1/2|k(0-2) +4(2-0) +0(0-0) |=4

1/2|-2k+8+0|=4

|-2k+8|=8

-2K+8=±8

For +ve value 

-2k+8=8

-2k=8-8

-2k=0

k=0/-2

k=0

For -ve value 

-2k+8=-8

-2k=-8-8

-2k=-16

k=16/2

k=8

Hence the value of k is 0,8