We are tasked with solving the equations graphically and then finding the areas of triangles enclosed by these lines with the x-axis and the y-axis.

The two given equations are:

1. $x - 2y = 0$
2. $x + y = 9$

Step 1: Rearrange the equations

We need to express each equation in the form $y = mx + c$ (slope-intercept form) to plot them easily.

• For $x - 2y = 0$: $x = 2y \implies y = \frac{x}{2}$ So, the first line has the equation $y = \frac{x}{2}$.

• For $x + y = 9$: $y = 9 - x$ So, the second line has the equation $y = 9 - x$.

Step 2: Plot the lines and find the points of intersection with axes

1. Line 1: $y = \frac{x}{2}$:

   • When $x = 0$, $y = 0$ (origin).
   • When $y = 0$, $x = 0$ (also origin).
   • To find another point, let $x = 4$, then $y = \frac{4}{2} = 2$. So, one point on the line is $(4, 2)$.

2. Line 2: $y = 9 - x$:

   • When $x = 0$, $y = 9$. So, one point is $(0, 9)$.
   • When $y = 0$, $x = 9$. So, another point is $(9, 0)$.

Step 3: Find the point of intersection

To find where the two lines intersect, solve the system of equations $\frac{x}{2} = 9 - x$.

• Multiply both sides by 2: $x = 18 - 2x$
• Add $2x$ to both sides: $3x = 18 \implies x = 6$
• Substitute $x = 6$ into one of the original equations, say $y = 9 - x$: $y = 9 - 6 = 3$ So, the point of intersection is $(6, 3)$.

Step 4: Find the area of the triangles

(i) Triangle enclosed by the x-axis and the two lines

The triangle is formed by the x-axis and the two lines $y = \frac{x}{2}$ and $y = 9 - x$. The vertices are:

• $(0, 0)$
• $(9, 0)$
• $(6, 3)$

We can calculate the area of this triangle using the formula for the area of a triangle with given vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$: $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ Substituting the coordinates $(0, 0)$, $(9, 0)$, and $(6, 3)$: $\text{Area} = \frac{1}{2} \left| 0(0 - 3) + 9(3 - 0) + 6(0 - 0) \right| = \frac{1}{2} \times 27 = 13.5$ So, the area of the triangle enclosed by the x-axis and the two lines is $13.5 \, \text{square units}$.

(ii) Triangle enclosed by the y-axis and the two lines

The triangle is formed by the y-axis and the two lines. The vertices are:

• $(0, 0)$
• $(0, 9)$
• $(6, 3)$

Using the same formula for the area of a triangle, substituting the coordinates $(0, 0)$, $(0, 9)$, and $(6, 3)$: $\text{Area} = \frac{1}{2} \left| 0(9 - 3) + 0(3 - 0) + 6(0 - 9) \right| = \frac{1}{2} \times 54 = 27$ So, the area of the triangle enclosed by the y-axis and the two lines is $27 \, \text{square units}$.

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Summary:

• Area of the triangle enclosed by the x-axis and the two lines: $13.5 \, \text{square units}$.
• Area of the triangle enclosed by the y-axis and the two lines: $27 \, \text{square units}$.

Would you like further details or have any questions?

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Here are five related questions to explore:

1. What is the significance of the intersection point in these kinds of problems?
2. How can we verify the area of the triangles using integration?
3. How does the slope of a line affect the shape of the triangle formed?
4. Can we solve these equations algebraically and get the same result without graphing?
5. What happens to the triangle's area if we shift one of the lines?

Tip: When calculating the area of a triangle formed by lines, always double-check the vertices and ensure they are correctly substituted into the area formula!