Question : The area of the triangle formed by the graph of the straight lines $x-y=0$, $x+y=2$, and the $x$-axis is:
Option 1: 1 sq. unit
Option 2: 2 sq. units
Option 3: 4 sq. units
Option 4: 5 sq. units
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Correct Answer: 1 sq. unit
Solution : Given: The straight lines $x-y=0$, $x+y=2$. By substituting $x=y$ in the equation $x+y=2$ , we get, $2y=2$ ⇒ $y=1$ and therefore, $x$ = 1 The point of intersection of two lines is $(1,1)$. By substituting $y=0$ in the equation $x+y=2$, we get, $x+0=2$ ⇒ $x=2$ The point of intersection on $x$–axis is $(2,0)$. So, the area of $\triangle OAC=\frac{1}{2} \times OA \times CD$. Or, the area of $\triangle OAC=\frac{1}{2} \times 2 \times1$ The area of $\triangle OAC=1$ sq. unit Hence, the correct answer is 1 sq. unit.
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Question : What is the area (in sq. units) of the triangle formed by the graphs of the equations $2x + 5y - 12=0, x + y = 3,$ and $y = 0$?
Option 1: 3
Option 2: 2
Option 3: 5
Option 4: 6
Question : The area of triangle with vertices A(0, 8), O(0, 0), and B(5, 0) is:
Option 1: 8 sq. units
Option 2: 13 sq. units
Option 3: 20 sq. units
Option 4: 40 sq. units
Question : What is the area (in unit squares) of the triangle enclosed by the graphs of $2 x+5 y=12, x+y=3$ and the x-axis?
Option 1: 2.5
Option 2: 3.5
Option 3: 3
Option 4: 4
Question : Find the coordinates of the points where the graph $57x – 19y = 399$ cuts the coordinate axes.
Option 1: x-axis at(–7, 0) and y-axis at (0, –21)
Option 2: x-axis at(–7, 0) and y-axis at (0, 21)
Option 3: x-axis at (7, 0) and y-axis at (0, –21)
Option 4: x-axis at (7, 0) and y-axis at (0, 21)
Question : The area of triangle formed by the straight line $3x + 2y = 6$ and the co-ordinate axes is:
Option 1: 3 square units
Option 2: 6 square units
Option 3: 4 square units
Option 4: 8 square units
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