Graphical Solution and Area with Y-Axis
Video Explanation
Question
Solve the following system of linear equations graphically and find the area of the region bounded by these two lines and the y-axis:
\[ 2x – 3y + 6 = 0 \]
\[ 2x + 3y – 18 = 0 \]
Solution
Step 1: Write Both Equations in the Form \(y = mx + c\)
Equation (1):
\[ 2x – 3y + 6 = 0 \Rightarrow -3y = -2x – 6 \Rightarrow y = \frac{2}{3}x + 2 \]
Equation (2):
\[ 2x + 3y – 18 = 0 \Rightarrow 3y = 18 – 2x \Rightarrow y = 6 – \frac{2}{3}x \]
Step 2: Prepare Tables of Values
For Equation (1): \(y = \frac{2}{3}x + 2\)
| x | y |
|---|---|
| 0 | 2 |
| 3 | 4 |
For Equation (2): \(y = 6 – \frac{2}{3}x\)
| x | y |
|---|---|
| 0 | 6 |
| 3 | 4 |
Step 3: Graphical Representation
Plot the following points on the same Cartesian plane:
- Line 1: (0, 2) and (3, 4)
- Line 2: (0, 6) and (3, 4)
Join each pair of points to obtain two straight lines.
The two straight lines intersect at the point (3, 4).
Result
The graphical solution of the given system of equations is:
\[ x = 3,\quad y = 4 \]
Step 4: Region Bounded with the Y-Axis
The bounded region is formed by:
- The line \(2x – 3y + 6 = 0\)
- The line \(2x + 3y – 18 = 0\)
- The y-axis \((x = 0)\)
Shade the triangular region enclosed by these two lines and the y-axis.
Step 5: Area of the Bounded Region
Vertices of the triangle are:
- (0, 2)
- (0, 6)
- (3, 4)
Base of the triangle (along y-axis) = \(6 – 2 = 4\) units
Height of the triangle = horizontal distance of point (3, 4) from y-axis = 3 units
\[ \text{Area} = \frac{1}{2} \times 4 \times 3 = 6 \]
Answer
The graphical solution of the given system of equations is:
\[ (x, y) = (3, 4) \]
Area of the region bounded by the given lines and the y-axis = 6 square units.
Conclusion
The required triangular region is shaded and its area is 6 square units.