Graph of Linear Equations, Shaded Region and Area
Video Explanation
Question
Draw the graphs of the following equations and determine the coordinates of the vertices of the triangle formed by these lines and the x-axis. Shade the triangular region and find its area:
\[ x – y + 1 = 0 \]
\[ 3x + 2y – 12 = 0 \]
Solution
Step 1: Write Both Equations in the Form \(y = mx + c\)
Equation (1):
\[ x – y + 1 = 0 \Rightarrow y = x + 1 \]
Equation (2):
\[ 3x + 2y – 12 = 0 \Rightarrow 2y = 12 – 3x \Rightarrow y = 6 – \frac{3}{2}x \]
Step 2: Prepare Tables of Values
For Equation (1): \(y = x + 1\)
| x | y |
|---|---|
| -1 | 0 |
| 1 | 2 |
For Equation (2): \(y = 6 – \frac{3}{2}x\)
| x | y |
|---|---|
| 4 | 0 |
| 0 | 6 |
Step 3: Graphical Representation
Plot the following points on the same Cartesian plane:
- Line 1: (−1, 0) and (1, 2)
- Line 2: (4, 0) and (0, 6)
Join each pair of points to obtain two straight lines.
The two straight lines intersect at the point (2, 3).
Step 4: Vertices of the Triangle with the X-Axis
The triangle is formed by:
- Intersection of \(x – y + 1 = 0\) with x-axis → (−1, 0)
- Intersection of \(3x + 2y – 12 = 0\) with x-axis → (4, 0)
- Intersection of the two lines → (2, 3)
Step 5: Shading of the Required Region
Shade the triangular region enclosed by:
- The line \(x – y + 1 = 0\)
- The line \(3x + 2y – 12 = 0\)
- The x-axis \((y = 0)\)
Step 6: Area of the Triangle
Base of the triangle = distance between (−1, 0) and (4, 0) = 5 units
Height of the triangle = y-coordinate of the vertex (2, 3) = 3 units
\[ \text{Area} = \frac{1}{2} \times 5 \times 3 = \frac{15}{2} \]
Answer
Coordinates of the vertices of the triangle are:
- (−1, 0)
- (4, 0)
- (2, 3)
Area of the triangular region = \(\frac{15}{2}\) square units.
Conclusion
The triangle formed by the given lines and the x-axis is shaded and its area is \(\frac{15}{2}\) square units.