Question:

The area of the triangle, formed by the straight lines \( y = 0 \), \( 12x - 5y = 0 \), and \( 3x + 4y = 7 \), is ________

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When one of the lines is an axis (like y=0 or x=0), calculating the area is often easier. You can use that line as the base of the triangle, and the height is simply the corresponding coordinate of the third vertex.
Updated On: Oct 14, 2025
  • \( \frac{28}{9} \)
  • \( \frac{14}{9} \)
  • \( \frac{35}{27} \)
  • \( \frac{35}{54} \)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
To find the area of a triangle formed by three lines, we first need to find the coordinates of its vertices. The vertices are the points of intersection of the lines, taken two at a time.
Step 2: Key Formula or Approach:
1. Find the three vertices by solving the systems of linear equations for each pair of lines. 2. Use the determinant formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \): \[ \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \] Step 3: Detailed Explanation:
Let the three lines be: L1: \( y = 0 \) L2: \( 12x - 5y = 0 \) L3: \( 3x + 4y = 7 \) Finding the Vertices:

Vertex A (Intersection of L1 and L2): Substitute \(y=0\) into L2: \( 12x - 5(0) = 0 \implies 12x = 0 \implies x = 0 \). So, Vertex A is (0, 0).
Vertex B (Intersection of L1 and L3): Substitute \(y=0\) into L3: \( 3x + 4(0) = 7 \implies 3x = 7 \implies x = \frac{7}{3} \). So, Vertex B is \( (\frac{7}{3}, 0) \).
Vertex C (Intersection of L2 and L3): From L2, we have \( 12x = 5y \implies y = \frac{12}{5}x \). Substitute this into L3: \( 3x + 4\left(\frac{12}{5}x\right) = 7 \). \[ 3x + \frac{48}{5}x = 7 \] \[ \frac{15x + 48x}{5} = 7 \] \[ \frac{63x}{5} = 7 \implies 9x = 5 \implies x = \frac{5}{9} \] Now find y: \( y = \frac{12}{5}x = \frac{12}{5} \left(\frac{5}{9}\right) = \frac{12}{9} = \frac{4}{3} \). So, Vertex C is \( (\frac{5}{9}, \frac{4}{3}) \).
Calculating the Area: The vertices are A(0,0), B\((\frac{7}{3}, 0)\), and C\((\frac{5}{9}, \frac{4}{3})\). Since one vertex is at the origin, the area formula simplifies to \( \frac{1}{2} |x_B y_C - x_C y_B| \). \[ \text{Area} = \frac{1}{2} \left| \left(\frac{7}{3}\right)\left(\frac{4}{3}\right) - \left(\frac{5}{9}\right)(0) \right| \] \[ \text{Area} = \frac{1}{2} \left| \frac{28}{9} - 0 \right| = \frac{1}{2} \times \frac{28}{9} = \frac{14}{9} \] Alternatively, we can consider the base of the triangle to be the segment AB, which lies on the x-axis (\(y=0\)). Base length = \( |\frac{7}{3} - 0| = \frac{7}{3} \). The height of the triangle is the perpendicular distance from vertex C to the base (the x-axis), which is simply the y-coordinate of C. Height = \( \frac{4}{3} \). \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{7}{3} \times \frac{4}{3} = \frac{28}{18} = \frac{14}{9} \] Step 4: Final Answer:
The area of the triangle is \( \frac{14}{9} \) square units.
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