The area of a triangle is 5 sq. units and two of its vertices are (2,

1.	The area of a triangle is 5 sq. units and two of its vertices are (2, 1) and (3, – 2). If the third vertex is (x, y), where y = x + 3, then find the co-ordinates of the third vertex.
Answer» The area of a triangle with vertex (x₁, y₁), (x₂, y₂) and (x₃, y₃) is ∆ = \(\frac{1}{2}\)\|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)\| Here, x₁ = 2, x₂ = 3, x₃ = x y₁ = 1, y₂ = – 2, y₃ = x + 3. ∆ = \(\frac{1}{2}\)\|2(−2 − x − 3) + 3(x + 3 − 1) + x(1 + 2)\| ⇒ 5 = \(\frac{1}{2}\)\|2(−5 − x) + 3 (x + 2) + 3x\| ⇒ 10 = \|−10 − 2x + 3x + 6 + 3x\| ⇒ \|4x − 4\| = 10 ⇒ 4x − 4 = 10 or 4x – 4 = – 10 ⇒ x = \(\frac{7}{2}\) or = − \(\frac{3}{2}\) ⇒ y = \(\frac{13}{2}\) ⇒ y = \(\frac{3}{2}\) ∴ (x, y.) = (\(\frac{7}{2}\), \(\frac{13}{2}\)) or (−\(\frac{3}{2}\), \(\frac{3}{2}\)).

The area of a triangle is 5 sq. units and two of its vertices are (2, 1) and (3, – 2). If the third vertex is (x, y), where y = x + 3, then find the co-ordinates of the third vertex.

Answer»

The area of a triangle with vertex (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

∆ = \(\frac{1}{2}\)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

Here, x₁ = 2, x₂ = 3, x₃ = x

y₁ = 1, y₂ = – 2, y₃ = x + 3.

∆ = \(\frac{1}{2}\)|2(−2 − x − 3) + 3(x + 3 − 1) + x(1 + 2)|

⇒ 5 = \(\frac{1}{2}\)|2(−5 − x) + 3 (x + 2) + 3x|

⇒ 10 = |−10 − 2x + 3x + 6 + 3x|

⇒ |4x − 4| = 10

⇒ 4x − 4 = 10 or 4x – 4 = – 10

⇒ x = \(\frac{7}{2}\) or = − \(\frac{3}{2}\)

⇒ y = \(\frac{13}{2}\) ⇒ y = \(\frac{3}{2}\)

∴ (x, y.) = (\(\frac{7}{2}\), \(\frac{13}{2}\)) or (−\(\frac{3}{2}\), \(\frac{3}{2}\)).