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Find the area of ABC with A(1, -4) and midpoints of sides thr

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Find the area of $Δ A BC$ with $A (1, - 4)$ and midpoints of sides through A being $(2, - 1)$ and $(0, - 1)$ .

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Find the area of

Δ A BC

with

A (1, - 4)

and midpoints of sides through A being

(2, - 1)

and

(0, - 1)

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

Given: A

Δ A BC

with

A (1, - 4)

Let F and E are the midpoints of AB and AC respectively

Let Coordinates of F are

(2, - 1)

and Coordinates of E are

(0, - 1)

Let Coordinates of B be

(x_{1}, y_{1})

and Coordinates of C be

(x_{2}, y_{2})

Find coordinate of B:

using section formula:

2 = \frac{1 + x _{1}}{2} \Rightarrow x_{1} = 4 - 1 = 3

- 1 = \frac{4 + y _{1}}{2} \Rightarrow y_{1} = - 2 + 4 = 2

Coordinate of B are

(3, 2)

Find coordinate of C:

using section formula:

0 = \frac{1 + x _{2}}{2} \Rightarrow 1 + x_{2} = 0 \Rightarrow x_{2} = - 1

- 1 = \frac{- 4 + y _{2}}{2} \Rightarrow - 4 + y_{2} = - 2

\Rightarrow y_{2} = - 2 + 4 = 2

Coordinate of C are

(- 1, 2)

Now,

Area of

Δ A BC = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]

Area of triangle ABC:

= \frac{1}{2} [1 (2 - 2) + 3 (2 + 4) + (- 1) (- 4 - 2)]

= \frac{1}{2} [1 \times 0 + 3 \times 6 + (- 1) \times (- 6)]

= \frac{1}{2} [0 + 18 + 6] = \frac{24}{2} = 12

sq. units.

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Question Text	Find the area of $Δ A BC$ with $A (1, - 4)$ and midpoints of sides through A being $(2, - 1)$ and $(0, - 1)$ .
Answer Type	Text solution:1
Upvotes	150

Find the area of ΔABC with A(1,−4) and midpoints of sides through A being (2,−1) and (0,−1).

Text solutionVerified

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Find the area of $Δ A BC$ with $A (1, - 4)$ and midpoints of sides through A being $(2, - 1)$ and $(0, - 1)$ .