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प्रश्न
Draw the graphs of the equations x + 2y = 4 and 3x – 2y = 4. Find the area of triangle formed by the lines and x-axis.
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उत्तर
1. Find the vertices of the triangle
The triangle is formed by the intersection of the two given lines and their respective intersections with the x-axis, where y = 0.
Intersection of the two lines:
To find where x + 2y = 4 and 3x – 2y = 4 meet, add the two equations together:
(x + 2y) + (3x – 2y) = 4 + 4
4x = 8
⇒ x = 2
Substitute x = 2 into x + 2y = 4:
2 + 2y = 4
⇒ 2y = 2
⇒ y = 1
The intersection point apex of the triangle is (2, 1).
Intersection with the x-axis (y = 0):
For x + 2y = 4:
If y = 0, then x = 4.
Point: (4, 0)
For 3x – 2y = 4:
If y = 0, then 3x = 4
⇒ `x = 4/3 ≈ 1.33`
Point: `(4/3, 0)`
2. Draw the graphs
To draw the lines, plot the intercepts and the intersection point found above:
Line 1 (x + 2y = 4):
Passes through (0, 2) and (4, 0).
Line 2 (3x – 2y = 4):
Passes through (0, –2) and (2, 1).

3. Calculate the area
The area of a triangle is given by the formula:
Area = `1/2` × base × height
Base: The distance between the two x-intercepts on the x-axis.
Base = `4 - 4/3`
= `(12 - 4)/3`
= `8/3 ≈ 2.67` units
Height: The y-coordinate of the intersection point (2, 1).
Height = 1 unit
Area calculation:
Area = `1/2 xx 8/3 xx 1`
= `4/3 ≈ 1.33` sq units
The area of the triangle formed by the lines and the x-axis is 1.33 square units or `4/3` sq units.
