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: 12
Option 2: 10.5
Option 3: 14
Option 4: 11.5
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Correct Answer: 14
Solution :
Given: $2x - 3y + 6 = 0$
$2x - 3y = -6 $---------(i)
$4x + y = 16$ ---------(ii)
Multiplying equation (ii) by 3
$12x + 3y = 48$ ---------(iii)
Adding (i) and (iii),
$2x + 12x = 48 - 6$
⇒ $14x = 42$
⇒ $x = 3$
Putting the value of $x$ in equation (i),
$2 × 3 - 3y = -6$
⇒ $-3y = -12$
⇒ $y = 4$
⇒ $(x_1,y_1) = (3,4)$
Putting $y=0$ in equation(ii),
$4x = 16$
⇒ $x = 4$
⇒ $(x_2, y_2) = (4,0)$
Putting $y=0$ in equation (i),
$2x = -6$
⇒ $x = -3$
⇒ $(x_3,y_3) = (-3,0)$
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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