Find values of K, if the area of a triangle is 4 square units whose vertices are

(– 2,0), (0, 4) and (0, k)

Given: – Vertices of triangle are (– 2,0), (0, 4) and (0, k) and the area of the triangle is 4 sq. units.


Tip: – If vertices of a triangle are (x1,y1), (x2,y2) and (x3,y3), then the area of the triangle is given by:



Now,


Substituting given value in above formula



Removing modulus



Expanding along R1



[ – 2(4 – k) – 0(0) + 1(0 – 0)] = ±8


– 8 + 2k = ± 8


Taking + ve sign, we get


8 = – 8 + 2k


2k = 16


k = 8


Taking – ve sign, we get


– 8 = 2x – 8


2k = 0


k = 0


Thus k = 0, 8


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