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$L, M$ and $N$ are the midpoints of the sides $BC, CA$ and $AB$ respectively of triangle $ABC$ . If the vertices are $A (3, - 4), B (5, - 2)$ and $C (1, 3)$ the area of $△ LMN$ is

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2.25 square units

2.5 square units

2.75 square units

3 square units

answer is A.

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Detailed Solution

It is given that

L, M

and

N

are the midpoints of the sides

BC, CA

and

AB

respectively of triangle

ABC

and vertices are

A (3, - 4), B (5, - 2)

and

C (1, 3)

.
Area of triangle having vertices

(x_{1}, y_{1}), (x_{2}, y_{2})

and

(x_{3}, y_{3})

is given by,

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

square unit.
Now let

(x_{1}, y_{1})

= (3, - 4)

(x_{2}, y_{2})

= (5, - 2)

and

(x_{3}, y_{3})

= (1, 3)

So, area of

ΔABC = \frac{1}{2} [3 (- 2 - 3) + 5 (3 + 4) + 1 (- 4 + 2)]

= \frac{1}{2} (- 15 + 35 - 2)

= 9

sq. unit

∴

Area of

ΔLMN = \frac{1}{4} \times ΔABC

= \frac{1}{4} \times 9

= 2.25

square units
So, the correct option is 1.

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