Question 7 – Graphical Solution and Triangle Area
Question: Draw the graphs of the equations \( x – y + 1 = 0 \) and \( 3x + 2y – 12 = 0 \). Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
📚 What This Question Asks
This is a comprehensive graphical problem that requires:
- Drawing graphs of two linear equations
- Finding intersection points:
- Where the two lines meet each other
- Where each line meets the x-axis
- Identifying vertices of the triangle formed
- Shading the triangular region
- Calculating area of the triangle (bonus)
Step 1: Rewrite Equations in Slope-Intercept Form
Equation (1): \( x – y + 1 = 0 \)
Rearrange to get y:
\[ y = x + 1 \]Slope: m = 1, y-intercept: c = 1
Equation (2): \( 3x + 2y – 12 = 0 \)
Rearrange to get y:
\[ 2y = -3x + 12 \] \[ y = -\frac{3}{2}x + 6 \]Slope: m = -3/2, y-intercept: c = 6
Step 2: Find Points for Each Line
For Line 1: \( y = x + 1 \)
| x | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|
| y = x + 1 | 0 | 1 | 2 | 3 | 4 |
| Points (x, y) | (-1, 0) | (0, 1) | (1, 2) | (2, 3) | (3, 4) |
x-intercept: When y = 0, x = -1 → Point: (-1, 0)
For Line 2: \( y = -\frac{3}{2}x + 6 \)
| x | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| y = -1.5x + 6 | 6 | 3 | 0 | -3 |
| Points (x, y) | (0, 6) | (2, 3) | (4, 0) | (6, -3) |
x-intercept: When y = 0, x = 4 → Point: (4, 0)
Step 3: Find Intersection Point of Two Lines
Solve the system:
From equation (1): \( y = x + 1 \)
Substitute in equation (2): \( 3x + 2y – 12 = 0 \)
\[ 3x + 2(x + 1) – 12 = 0 \] \[ 3x + 2x + 2 – 12 = 0 \] \[ 5x – 10 = 0 \] \[ x = 2 \]Substitute \( x = 2 \) in \( y = x + 1 \):
\[ y = 2 + 1 = 3 \]Intersection point: (2, 3)
Step 4: Identify Triangle Vertices
The three vertices of the triangle are:
- Vertex A: Intersection of Line 1 with x-axis = (-1, 0)
- Vertex B: Intersection of Line 2 with x-axis = (4, 0)
- Vertex C: Intersection of Line 1 and Line 2 = (2, 3)
Step 5: Visual Graph with Shaded Triangle
📊 Complete Graphical Solution
Graph showing triangle formed by two lines and x-axis (shaded region)
📊 Graph Analysis:
- Blue Line: \( x - y + 1 = 0 \) or \( y = x + 1 \)
- Red Line: \( 3x + 2y - 12 = 0 \) or \( y = -\frac{3}{2}x + 6 \)
- Triangle vertices: A(-1, 0), B(4, 0), C(2, 3)
- Shaded region: Purple/violet area represents the triangle
Step 6: Calculate Area of Triangle
📐 Area Formula Using Coordinates
For a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \):
\[ \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \]
Given vertices:
A(-1, 0), B(4, 0), C(2, 3)
Let: \( x_1 = -1, y_1 = 0 \)
\( x_2 = 4, y_2 = 0 \)
\( x_3 = 2, y_3 = 3 \)
Calculate area:
\[ \text{Area} = \frac{1}{2} |(-1)(0 - 3) + 4(3 - 0) + 2(0 - 0)| \]
\[ = \frac{1}{2} |(-1)(-3) + 4(3) + 2(0)| \]
\[ = \frac{1}{2} |3 + 12 + 0| \]
\[ = \frac{1}{2} |15| \]
\[ = \frac{15}{2} \]
\[ = 7.5 \text{ square units} \]
Alternative method (Base × Height):
Base AB: Distance from A(-1, 0) to B(4, 0)
\[ \text{Base} = 4 - (-1) = 5 \text{ units} \]Height: Perpendicular distance from C(2, 3) to x-axis
\[ \text{Height} = 3 \text{ units} \]Area:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \] \[ = \frac{1}{2} \times 5 \times 3 \] \[ = 7.5 \text{ square units} \]
✅ Final Answer:
Vertices of Triangle:
A = (-1, 0)
B = (4, 0)
C = (2, 3)
Area of Triangle = 7.5 square units
Vertices of Triangle:
A = (-1, 0)
B = (4, 0)
C = (2, 3)
Area of Triangle = 7.5 square units
📊 Summary Table
| Element | Details |
|---|---|
| Line 1 | \( x - y + 1 = 0 \) or \( y = x + 1 \) |
| Line 2 | \( 3x + 2y - 12 = 0 \) or \( y = -\frac{3}{2}x + 6 \) |
| Line 1 x-intercept | (-1, 0) |
| Line 2 x-intercept | (4, 0) |
| Intersection of lines | (2, 3) |
| Triangle vertices | A(-1, 0), B(4, 0), C(2, 3) |
| Base (AB) | 5 units |
| Height | 3 units |
| Area | 7.5 square units |
⚠️ Common Mistakes to Avoid
❌ Mistake 1: Not finding x-intercepts correctly.
✅ Correct: To find x-intercept, set y = 0 and solve for x.
✅ Correct: To find x-intercept, set y = 0 and solve for x.
❌ Mistake 2: Confusing which points form the triangle.
✅ Correct: Triangle is formed by: two x-intercepts + intersection of the two lines.
✅ Correct: Triangle is formed by: two x-intercepts + intersection of the two lines.
❌ Mistake 3: Plotting points incorrectly on graph.
✅ Correct: Always verify at least 3 points per line before drawing.
✅ Correct: Always verify at least 3 points per line before drawing.
❌ Mistake 4: Sign errors in area formula.
✅ Correct: Use absolute value in area formula to ensure positive result.
✅ Correct: Use absolute value in area formula to ensure positive result.
💡 Key Points to Remember
Graphing Linear Equations:
- Find at least 3 points for each line to ensure accuracy
- x-intercept: Set y = 0 and solve for x
- y-intercept: Set x = 0 and solve for y
- Slope-intercept form: \( y = mx + c \) makes plotting easier
Finding Triangle Vertices:
- Vertex 1: Where Line 1 meets x-axis
- Vertex 2: Where Line 2 meets x-axis
- Vertex 3: Where Line 1 and Line 2 intersect
Area Calculation:
- Method 1: Coordinate formula (works for any triangle)
- Method 2: \( \frac{1}{2} \times \text{Base} \times \text{Height} \) (when base is on x-axis)
- Always use absolute value to ensure positive area
🎉 CONGRATULATIONS! 🎉
You've completed ALL 7 questions of Exercise 3.2!
Chapter 3: Pair of Linear Equations in Two Variables
You've completed ALL 7 questions of Exercise 3.2!
Chapter 3: Pair of Linear Equations in Two Variables
📚 Related Questions - Complete Exercise 3.2
Farhan Mansuri
M.Sc. Mathematics, B.Ed.
15+ Years Teaching Experience
Passionate about making mathematics easy and enjoyable for students.