Graphical Solution and Area of the Region Bounded by Lines and X-Axis

Video Explanation

Question

Solve graphically the following system of linear equations and find the area bounded by these lines and the x-axis:

\[ 4x – 3y + 4 = 0 \]

\[ 4x + 3y – 20 = 0 \]

Solution

Step 1: Write Both Equations in the Form \(y = mx + c\)

Equation (1):

\[ 4x – 3y + 4 = 0 \Rightarrow -3y = -4x – 4 \Rightarrow y = \frac{4}{3}x + \frac{4}{3} \]

Equation (2):

\[ 4x + 3y – 20 = 0 \Rightarrow 3y = 20 – 4x \Rightarrow y = \frac{20}{3} – \frac{4}{3}x \]

Step 2: Prepare Tables of Values

For Equation (1): \(y = \frac{4}{3}x + \frac{4}{3}\)

x y
-1 0
2 4

For Equation (2): \(y = \frac{20}{3} – \frac{4}{3}x\)

x y
5 0
2 4

Step 3: Graphical Representation

Plot the following points on the same Cartesian plane:

  • Line 1: (−1, 0) and (2, 4)
  • Line 2: (5, 0) and (2, 4)

Join each pair of points to obtain two straight lines.

The two straight lines intersect at the point (2, 4).

Step 4: Triangle Formed with the X-Axis

The triangle is formed by:

  • Intersection of \(4x – 3y + 4 = 0\) with x-axis → (−1, 0)
  • Intersection of \(4x + 3y – 20 = 0\) with x-axis → (5, 0)
  • Intersection point of the two lines → (2, 4)

Step 5: Area of the Triangle

Base of the triangle = distance between (−1, 0) and (5, 0) = 6 units

Height of the triangle = y-coordinate of the vertex (2, 4) = 4 units

\[ \text{Area} = \frac{1}{2} \times 6 \times 4 = 12 \]

Answer

The graphical solution of the given system of equations is:

\[ (x, y) = (2, 4) \]

Area bounded by the given lines and the x-axis = 12 square units.

Conclusion

The triangle formed by the two given lines and the x-axis is shaded and its area is 12 square units.

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