Given vertices of a triangle is, (K, 0 ), (4, 0), (0, 2) and area of triangle is 4 sq. units. We know that, area of triangle with vertices (x1,y1),(x2,y2)(x3,y3) is given by A=|
x1
x2
x3
y1
y2
y3
z1
z2
z3
|
A=
1
2
∣(x1(y1−y3)+x2(y3−y2)+x3(y1−y2))
So,
A=
1
2
|K(0−2)+4(2−0)−0(0−0)|=4
⇒
1
2
|−2K+8|=4 ⇒8−2K=±8 For +8, we have 8−2K=8⇒K=0 For –8, we have 8−2K=−8⇒K=8