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)

Answer :

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 (x_{1},y_{1}), (x_{2},y_{2}) and (x_{3},y_{3}), then the area of the triangle is given by:

Now,

Substituting given value in above formula

⇒

Removing modulus

⇒

Expanding along R_{1}

⇒

⇒ [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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