The area of the figure formed by the inner section of line $x = 0, y = 0, x = 3, y = 4$ will
be-

3 sq.unit

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4 sq.unit

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6 sq .unit

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12 sq.unit

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Solution

The correct option is D 12 sq.unit
Area o a polygon is given by the equation,
$a r e a = | (x_{1} y_{2} - y_{1} x_{2}) + (x_{2} y_{3} - y_{2} x_{3}) + . . . + (x_{n} y_{1} - y_{n} x_{1}) | a r e a = | [0 \times 0 - 3 \times 0] + [3 \times 4 - 0 \times 0] + [4 \times 0 - 0 \times 0] | = 12 square.unit$

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Similar questions

x = 0

y = 0

x = 3

y = 4

4 y - 3 x - a = 0, 3 y - 4 x + a = 0, 4 y - 3 x - 3 a = 0

3 y - 4 x + 2 a = 0

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2 x - 5 y + 7 = 0, 5 x + 2 y - 1 = 0, 2 x - 5 y + 2 = 0

5 x + 2 y + 4 = 0

4 x - 3 y - a = 0, 3 x - 4 y + a = 0

4 x - 3 y - 3 a = 0

3 x - 4 y + 2 a = 0

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