Question

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

• A. \( \frac{28}{9} \)
• B. \( \frac{14}{9} \)
• C. \( \frac{35}{27} \)
• D. \( \frac{35}{54} \)

Hint:

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.

Answer 1

The Correct Option is B

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.