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Find the area of the $Δ A B C$ with $A (1, − 4)$ and mid-point of sides through $A$ being (2, -1) and $(0, − 1)$ .

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12 unit 2

10 unit 2

4 unit 2

9 unit 2

answer is A.

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Detailed Solution

It is given that,
In

Δ A B C

A 1, − 4

and mid-point of sides through A being

(2, − 1)

and

(0, − 1)

.
Let the other two vertices of the

Δ A B C

are

B (x, y) and C (m, n)

.
So the middle point of A and B is (2, -1) and the middle point of B and C is (0,-1).
Now,
We know that, mid-point Formula is given by,

(x m, y m) = x 1 + x 2 2, y 1 + y 2 2

.
Therefore, we get,

1 + x 2, − 4 + y 2 = (2, − 1) and 1 + m 2, − 4 + n 2 = (0, − 1)

So,

1 + x 2 = 2 ⇒ 1 + x = 4 ⇒ x = 3 and − 4 + y 2 = − 1 ⇒ − 4 + y = − 2 ⇒ y = 2

The coordinates of point B is (3,2).
Similarly,

1 + m 2 = 0 ⇒ 1 + m = 0 ⇒ m = − 1 and − 4 + n 2 = − 1 ⇒ − 4 + n = − 2 ⇒ n = 2

The coordinates of point C is (-1,2).
We know that the area of a triangle, whose vertices are

A x 1, y 1, B x 2, y 2, C x 3, y 3

is given by

Δ A B C = 12 x 1 y 2 − y 3 + x 2 y 3 − y 1 + x 3 y 1 − y 2 unit 2

.
In

Δ A B C

x 1 = 1, y 1 = − 4, x 2 = 3, y 2 = 2, x 3 = − 1, y 3 = 2

Then, a r (Δ A B C) = 12 [1 (2 − 2) + 3 (2 − (− 4)) + (− 1) (− 4 − 2)] ⇒ a r (Δ A B C) = 12 [0 + 18 + 6] ⇒ a r (Δ A B C) = 12 units 2

Hence, option (1) is correct.

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