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Q.

Two vertices of a triangle are (2, –1) and (3, 2) and third vertex lies on the line x + y = 5. If the area of the triangle is 4 units then third vertex is

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a

(0, 5) or (4, 1)

b

(5, 0) or (4, 1)

c

(0, 5) or (1, 4)

d

(5, 0) or (1, 4)

answer is B.

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Detailed Solution

Let A ≡ (2, –1) and B ≡ (3, 2).
Let the third vertex be Cα,β
Then, α+β=5 (given)                                          (1)
Area of  ABC=122-11321αβ1=±4 (given)
β-3α=1                                                        (2)
or  β-3α=-15                                                   (3)
Solving (1) and (2), we get, α = 1, β = 4
Solving (1) and (3), we get, α = 5, β = 0
Thus, the third vertex is either (5, 0) or (1, 4).

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