Question

Determine graphically, the coordinates of vertices of a triangle whose equations are \(2x - 3y + 6 = 0\); \(2x + 3y - 18 = 0\) and \(x = 0\). Also, find the area of this triangle.

Hint:

When the base of a triangle lies on a coordinate axis, the height is simply the perpendicular distance (absolute coordinate value) of the opposite vertex from that axis.

Answer 1

Step 1: Understanding the Concept:
To solve graphically, we find at least two points for each line, plot them, and identify the intersection points. The line \(x=0\) is the y-axis.
Step 2: Key Formula or Approach:
Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\).
Step 3: Detailed Explanation:
1. Line 1 (\(2x - 3y = -6\)): If \(x=0, y=2\). If \(y=0, x=-3\). Points: (0, 2), (-3, 0). 2. Line 2 (\(2x + 3y = 18\)): If \(x=0, y=6\). If \(y=0, x=9\). Points: (0, 6), (9, 0). 3. Vertices: - Intersection of Line 1 and y-axis (\(x=0\)): (0, 2). - Intersection of Line 2 and y-axis (\(x=0\)): (0, 6). - Intersection of Line 1 and 2: Adding equations \(4x = 12 \implies x=3\). Then \(2(3) + 3y = 18 \implies 3y = 12 \implies y=4\). Point: (3, 4). 4. Area: Base is on y-axis from \(y=2\) to \(y=6\), so base = 4 units. Height is the x-coordinate of the third vertex, so height = 3 units. \[ \text{Area} = \frac{1}{2} \times 4 \times 3 = 6 \text{ sq. units} \]
Step 4: Final Answer:
Vertices: (0, 2), (0, 6), (3, 4). Area: 6 sq. units.