Question 7 – Find vertices of triangle formed by two lines and x-axis
Understanding the Problem
Given:
- Two linear equations: \( x + 3y = 6 \) and \( 2x – 3y = 12 \)
- We need to draw both lines on the same graph
To Find:
- The coordinates of the three vertices of the triangle formed by these two lines and the x-axis
- Shade the triangular region
📚 Key Concept
A triangle is formed when:
- Two lines intersect at one point (this gives us one vertex)
- Each line crosses an axis at different points (these give us two more vertices)
- Total: 3 vertices form a triangle
Step 1: Find Points for First Line
Rearranging: \( x = 6 – 3y \)
Let’s find points by choosing values of y:
| y | 0 | 2 | -2 | 1 |
|---|---|---|---|---|
| x = 6 – 3y | 6 | 0 | 12 | 3 |
| Points | (6, 0) | (0, 2) | (12, -2) | (3, 1) |
Important Points:
- Point (6, 0) is where this line crosses the x-axis
- Point (0, 2) is where this line crosses the y-axis
Step 2: Find Points for Second Line
Rearranging: \( x = \frac{12 + 3y}{2} \)
Let’s find points by choosing values of y:
| y | 0 | -4 | 2 | -2 |
|---|---|---|---|---|
| x = (12 + 3y)/2 | 6 | 0 | 9 | 3 |
| Points | (6, 0) | (0, -4) | (9, 2) | (3, -2) |
Important Points:
- Point (6, 0) is where this line crosses the x-axis
- Point (0, -4) is where this line crosses the y-axis
Step 3: Visual Graph Representation
📊 Graphical Solution
Interactive graph showing both lines and the shaded triangular region
- Blue Line: \( x + 3y = 6 \) passes through (0, 2), (3, 1), and (6, 0)
- Red Line: \( 2x - 3y = 12 \) passes through (0, -4), (3, -2), and (6, 0)
- Intersection Point: Both lines meet at A(6, 0) on the x-axis
- Shaded Region: The green shaded area represents triangle ABC
Step 4: Find the Intersection Point Algebraically
Given equations:
\[ x + 3y = 6 \quad \text{...(1)} \] \[ 2x - 3y = 12 \quad \text{...(2)} \]Method: Addition
Add equations (1) and (2):
\[ (x + 3y) + (2x - 3y) = 6 + 12 \] \[ x + 2x + 3y - 3y = 18 \] \[ 3x = 18 \] \[ x = 6 \]Substitute \( x = 6 \) in equation (1):
\[ 6 + 3y = 6 \] \[ 3y = 0 \] \[ y = 0 \]Therefore, the intersection point is A(6, 0) ✓
Step 5: Identify All Three Vertices
Finding the Three Vertices of the Triangle:
Vertex A: Where both lines intersect
From our calculation: A(6, 0)
This point lies on the x-axis.
Vertex B: Where first line \( x + 3y = 6 \) crosses the y-axis
Set x = 0:
\[ 0 + 3y = 6 \] \[ y = 2 \]Vertex B = (0, 2)
Vertex C: Where second line \( 2x - 3y = 12 \) crosses the y-axis
Set x = 0:
\[ 0 - 3y = 12 \] \[ -3y = 12 \] \[ y = -4 \]Vertex C = (0, -4)
⚠️ Important Observation:
Since both lines intersect at point (6, 0) which is on the x-axis, the triangle is actually formed by:
- The two given lines
- The y-axis (not the x-axis as the question suggests)
The three vertices are at the intersection of the two lines (point A on x-axis) and where each line crosses the y-axis (points B and C).
Step 6: Verify the Triangle
- A(6, 0) - Intersection of both lines (on x-axis)
- B(0, 2) - First line crosses y-axis
- C(0, -4) - Second line crosses y-axis
These three non-collinear points form triangle ABC.
Points B(0, 2) and C(0, -4) both lie on the y-axis (x = 0).
Point A(6, 0) lies on the x-axis (y = 0).
Since A is not on the y-axis and B, C are not on the x-axis, the three points are not collinear.
Therefore, they form a valid triangle ✓
The three vertices of the triangle are:
A(6, 0), B(0, 2), and C(0, -4)
Additional Calculations
Using the coordinate formula:
\[ \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \]Where A(6, 0), B(0, 2), C(0, -4):
\[ \text{Area} = \frac{1}{2} |6(2 - (-4)) + 0(-4 - 0) + 0(0 - 2)| \] \[ = \frac{1}{2} |6 \times 6 + 0 + 0| \] \[ = \frac{1}{2} \times 36 \] \[ = 18 \text{ square units} \]Side AB: Distance from A(6, 0) to B(0, 2)
\[ AB = \sqrt{(6-0)^2 + (0-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \approx 6.32 \text{ units} \]Side AC: Distance from A(6, 0) to C(0, -4)
\[ AC = \sqrt{(6-0)^2 + (0-(-4))^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \approx 7.21 \text{ units} \]Side BC: Distance from B(0, 2) to C(0, -4)
\[ BC = \sqrt{(0-0)^2 + (2-(-4))^2} = \sqrt{0 + 36} = 6 \text{ units} \]Perimeter:
\[ P = AB + AC + BC = 2\sqrt{10} + 2\sqrt{13} + 6 \approx 19.53 \text{ units} \]⚠️ Common Mistakes to Avoid
✅ Correct: Always find the intersection point first. If it lies on one axis, the triangle is formed with the other axis.
✅ Correct: Always solve the equations simultaneously to confirm the exact intersection coordinates.
✅ Correct: X-axis intercept: y = 0. Y-axis intercept: x = 0. Remember this clearly.
✅ Correct: Use a ruler and mark points carefully. Choose an appropriate scale for your graph.
💡 Key Points to Remember
- Triangle Formation: Requires three non-collinear points
- Intersection Point: Found by solving both equations simultaneously
- X-axis Intercept: Set y = 0 in the equation
- Y-axis Intercept: Set x = 0 in the equation
- Special Case: When lines intersect on one axis, triangle forms with the other axis
- Verification: Always verify graphical solutions algebraically
- Collinearity: Three points on the same line cannot form a triangle
- Shading: Shade only the interior region bounded by the three sides
- Labeling: Clearly label all vertices with coordinates
- Scale: Use consistent scale throughout the graph
📝 Practice Problems
Try These Similar Problems:
- Find vertices of triangle formed by \( x + y = 5 \), \( 2x - y = 4 \), and the y-axis
- Find vertices of triangle formed by \( 3x + 4y = 12 \), \( x - 2y = 2 \), and the x-axis
- Draw \( 2x + y = 8 \), \( x - y = 1 \) and find the triangle with x-axis
- Calculate area of triangle formed by \( x + 2y = 10 \), \( 3x - y = 6 \), and y-axis
🎉 Congratulations!
You have completed all 7 questions of
Class 10 Maths Chapter 3 Exercise 3.1
✅ Question 1 - Word Problems (2 parts)
✅ Question 2 - Graphical Representation (3 parts)
✅ Question 3 - Consistency Check (5 parts)
✅ Question 4 - Consistency with Graphical Solutions (4 parts)
✅ Question 5 - Rectangular Garden Problem
✅ Question 6 - Drawing Line Pairs (3 parts)
✅ Question 7 - Triangle Vertices with Visual Graph
Keep practicing and mastering linear equations!
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Farhan Mansuri
M.Sc. Mathematics, B.Ed.
15+ Years Teaching Experience
Passionate about making mathematics easy and enjoyable for students.