Question : What is the area (in unit squares) of the region enclosed by the graphs of the equations $2 x - 3 y + 6 = 0, 4 x + y = 16$ and $y = 0$ ?
Option 1: 12
Option 2: 10.5
Option 3: 14
Option 4: 11.5

Question

Question : What is the area (in sq. units) of the triangle formed by the graphs of the equations $2 x + 5 y - 12 = 0, x + y = 3,$ and $y = 0$ ?
Option 1: 3
Option 2: 2
Option 3: 5
Option 4: 6

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Answer 1

Correct Answer: 14

Solution : Given: $2 x - 3 y + 6 = 0$
$2 x - 3 y = - 6$ ---------(i)
$4 x + y = 16$ ---------(ii)
Multiplying equation (ii) by 3
$12 x + 3 y = 48$ ---------(iii)
Adding (i) and (iii),
$2 x + 12 x = 48 - 6$
⇒ $14 x = 42$
⇒ $x = 3$
Putting the value of $x$ in equation (i),
$2 \times 3 - 3 y = - 6$
⇒ $- 3 y = - 12$
⇒ $y = 4$
⇒ $(x_{1}, y_{1}) = (3, 4)$
Putting $y = 0$ in equation(ii),
$4 x = 16$
⇒ $x = 4$
⇒ $(x_{2}, y_{2}) = (4, 0)$
Putting $y = 0$ in equation (i),
$2 x = - 6$
⇒ $x = - 3$
⇒ $(x_{3}, y_{3}) = (- 3, 0)$
Area of triangle = $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$
= $\frac{1}{2} | 3 (0 - 0) + 4 (0 - 4) + - 3 (4 - 0) |$
= $\frac{1}{2} | - 16 - 12 |$
= $\frac{28}{2}$
= 14
Hence, the correct answer is 14.

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Question : What is the area (in unit squares) of the region enclosed by the graphs of the equations 2x−3y+6=0,4x+y=16 and y=0?Option 1: 12Option 2: 10.5Option 3: 14Option 4: 11.5

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Question : What is the area (in unit squares) of the region enclosed by the graphs of the equations $2 x - 3 y + 6 = 0, 4 x + y = 16$ and $y = 0$ ?
Option 1: 12
Option 2: 10.5
Option 3: 14
Option 4: 11.5