Q. 13 A3.7( 3 Votes )

# Find values of K, if the area of a triangle is 4 square units whose vertices are(k,0), (4,0) and (0,2)

Given: – Vertices of triangle are (k, 0), (4, 0) and (0, 2) and area of 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 [k(– 2) – 0(4) + 1(8 – 0)] = ±8

[ – 2k + 8] = ± 8

Taking + ve sign, we get

+ 8 = – 2x + 8

– 2k = 0

k = 0

Taking – ve sign, we get

– 8 = – 2x + 8

– 2x = – 16

x = 8

Thus x = 0, 8

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