 Solve the following equations graphically.
x – 2y = 0 and x + y = 9
(i) Find the area of the triangle enclosed by the x-axis and the two lines.
(ii) Find the area of the triangle enclosed by the y-axis and the two lines.


We are given two linear equations:

1. \( x - 2y = 0 \)
2. \( x + y = 9 \)

### (i) Finding the Area of the Triangle Enclosed by the X-axis and the Two Lines:

#### Step 1: Solve the system of equations to find the points of intersection.

**Solve \( x - 2y = 0 \) for \( x \):**

\[
x = 2y
\]

Substitute \( x = 2y \) into the second equation \( x + y = 9 \):

\[
2y + y = 9
\]
\[
3y = 9 \quad \Rightarrow \quad y = 3
\]

Substitute \( y = 3 \) into \( x = 2y \):

\[
x = 2(3) = 6
\]

So, the lines intersect at the point \( (6, 3) \).

#### Step 2: Find the x-intercepts of the lines.

- For \( x - 2y = 0 \), set \( y = 0 \):
  \[
  x = 0
  \]
  So, the x-intercept is \( (0, 0) \).

- For \( x + y = 9 \), set \( y = 0 \):
  \[
  x = 9
  \]
  So, the x-intercept is \( (9, 0) \).

#### Step 3: The vertices of the triangle are \( (0, 0) \), \( (9, 0) \), and \( (6, 3) \).

#### Step 4: Calculate the area using the formula for the area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substitute the coordinates \( (0, 0) \), \( (9, 0) \), and \( (6, 3) \):

\[
\text{Area} = \frac{1}{2} \left| 0(0 - 3) + 9(3 - 0) + 6(0 - 0) \right|
\]
\[
\text{Area} = \frac{1}{2} \left| 9 \times 3 \right| = \frac{1}{2} \times 27 = 13.5 \, \text{square units}
\]

### (ii) Finding the Area of the Triangle Enclosed by the Y-axis and the Two Lines:

#### Step 1: Find the y-intercepts of the lines.

- For \( x - 2y = 0 \), set \( x = 0 \):
  \[
  -2y = 0 \quad \Rightarrow \quad y = 0
  \]
  So, the y-intercept is \( (0, 0) \).

- For \( x + y = 9 \), set \( x = 0 \):
  \[
  y = 9
  \]
  So, the y-intercept is \( (0, 9) \).

#### Step 2: The vertices of the triangle are \( (0, 0) \), \( (0, 9) \), and \( (6, 3) \) (the point of intersection).

#### Step 3: Calculate the area using the same formula as before:

\[
\text{Area} = \frac{1}{2} \left| 0(9 - 3) + 0(3 - 0) + 6(0 - 9) \right|
\]
\[
\text{Area} = \frac{1}{2} \left| 6 \times (-9) \right| = \frac{1}{2} \times 54 = 27 \, \text{square units}
\]

### Final Answers:
1. The area of the triangle enclosed by the x-axis and the two lines is \( 13.5 \, \text{square units} \).
2. The area of the triangle enclosed by the y-axis and the two lines is \( 27 \, \text{square units} \).

Would you like further details or explanations?

### Related Questions:
1. How would the area calculation change if the lines had different slopes?
2. What happens to the area if one of the lines is shifted parallel to its current position?
3. How do you graphically determine the intersection points without solving the equations?
4. What is the significance of the slopes of the lines in determining the shape of the triangle?
5. How would the solution change if the lines formed a quadrilateral instead of a triangle?

### Tip: 
For faster calculations of triangle areas on the coordinate plane, using the determinant method or the shoelace formula can be very effective!

Solve the following equations graphically.
x – 2y = 0 and x + y = 9
(i) Find the area of the triangle enclosed by the x-axis and the two lines.
(ii) Find the area of the triangle enclosed by the y-axis and the two lines.

Learn how to find the area of triangles formed by two linear equations and the coordinate axes using graphical methods. This detailed solution includes solving x - 2y = 0 and x + y = 9 to find the points of intersection, followed by applying the triangle area formula. Suitable for Grades 9-12.

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