First slide

Cartesian plane

Question

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

Moderate

A

(0, 5) or (4, 1)
B

(5, 0) or (1, 4)
C

(5, 0) or (4, 1)
D

(0, 5) or (1, 4)

Solution

Let A ≡ (2, –1) and B ≡ (3, 2).
Let the third vertex be $C (α, β)$
Then, $α + β = 5$ (given) (1)
Area of $△ A B C = \frac{1}{2} |\begin{matrix} 2 & - 1 & 1 \\ 3 & 2 & 1 \\ α & β & 1 \end{matrix}| = \pm 4$ (given)
$\Rightarrow β - 3 α = 1$ (2)
$o r β - 3 α = - 15$ (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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