Answer :

Vertices of triangle A (0, 0), B (4, a) and C (6, 4)

Area of triangle = 17 sq. units

Area of triangle =

x_{1} = 0, x_{2} = 4 and x_{3} = 6

y_{1} = 0, y_{2} = a and y_{3} = 4

⇒

⇒

⇒

⇒ 17 × 2 = 16 – 6a

⇒ 34 = 16 – 6a

⇒ 34 + 6a = 16

⇒ 6a = 16 – 34

⇒ 6a = – 18

⇒

⇒ a = –3

Therefore, the required vertices are (4, –3)

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