Graphical Solution and Ratio of Areas of Triangles
Video Explanation
Question
Graphically solve the following pair of equations:
\[ 2x + y = 6 \]
\[ 2x – y + 2 = 0 \]
Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and with the y-axis.
Solution
Step 1: Write Both Equations in the Form \(y = mx + c\)
Equation (1):
\[ 2x + y = 6 \Rightarrow y = 6 – 2x \]
Equation (2):
\[ 2x – y + 2 = 0 \Rightarrow y = 2x + 2 \]
Step 2: Prepare Tables of Values
For Equation (1): \(y = 6 – 2x\)
| x | y |
|---|---|
| 0 | 6 |
| 3 | 0 |
For Equation (2): \(y = 2x + 2\)
| x | y |
|---|---|
| 0 | 2 |
| -1 | 0 |
Step 3: Graphical Representation
Plot the above points on the same Cartesian plane and draw the two straight lines.
The two lines intersect at the point:
\[ 6 – 2x = 2x + 2 \Rightarrow x = 1,\; y = 4 \]
Intersection point = \((1, 4)\)
Triangle Formed with the X-Axis
Vertices are:
- (3, 0)
- (-1, 0)
- (1, 4)
Base = distance between (3, 0) and (−1, 0) = 4 units
Height = y-coordinate of (1, 4) = 4 units
\[ \text{Area}_1 = \frac{1}{2} \times 4 \times 4 = 8 \]
Triangle Formed with the Y-Axis
Vertices are:
- (0, 6)
- (0, 2)
- (1, 4)
Base = distance between (0, 6) and (0, 2) = 4 units
Height = x-coordinate of (1, 4) = 1 unit
\[ \text{Area}_2 = \frac{1}{2} \times 4 \times 1 = 2 \]
Ratio of Areas
\[ \text{Area}_1 : \text{Area}_2 = 8 : 2 = 4 : 1 \]
Answer
The graphical solution of the given equations is:
\[ (x, y) = (1, 4) \]
The ratio of the areas of the triangles formed with the x-axis and the y-axis is 4 : 1.
Conclusion
Thus, the required ratio of the areas of the two triangles is 4 : 1.