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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Question : The graphs of the equations $4 x+\frac{1}{3} y=\frac{8}{3}$ and $\frac{1}{2} x+\frac{3}{4} y+\frac{5}{2}=0$ intersect at a point P. The point P also lies on the graph of the equation:
Question : What is the equation of the line perpendicular to the line $2x+3y=-6$ and having y-intercept 3?
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