Draw the graphs of the equations \(x - y + 1 = 0\) and \(3x + 2y - 12 = 0\). Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Consider the two lines:
\(L_1: x - y + 1 = 0\) and \(L_2: 3x + 2y - 12 = 0\), together with the x-axis \(y = 0\).
1. Intersection of \(L_1\) with the x-axis
For the x-axis, \(y = 0\). Substitute into \(x - y + 1 = 0\):
\[ x - 0 + 1 = 0 \Rightarrow x + 1 = 0 \Rightarrow x = -1. \]
So \(L_1\) meets the x-axis at the point \((-1, 0)\).
2. Intersection of \(L_2\) with the x-axis
Put \(y = 0\) in \(3x + 2y - 12 = 0\):
\[ 3x + 0 - 12 = 0 \Rightarrow 3x = 12 \Rightarrow x = 4. \]
So \(L_2\) meets the x-axis at \((4, 0)\).
3. Intersection of \(L_1\) and \(L_2\)
Solve the system
\[ x - y + 1 = 0 \quad \text{and} \quad 3x + 2y - 12 = 0. \]
From the first equation, express \(y\) in terms of \(x\):
\[ x - y + 1 = 0 \Rightarrow y = x + 1. \]
Substitute in the second equation:
\[ 3x + 2(x + 1) - 12 = 0 \Rightarrow 3x + 2x + 2 - 12 = 0 \Rightarrow 5x - 10 = 0 \Rightarrow 5x = 10 \Rightarrow x = 2. \]
Then
\[ y = x + 1 = 2 + 1 = 3. \]
Hence \(L_1\) and \(L_2\) intersect at the point \((2, 3)\).
4. Vertices of the triangle
The triangle is bounded by the two lines and the x-axis. Its three vertices are therefore:
Thus, the vertices of the triangle are \((-1, 0)\), \((4, 0)\) and \((2, 3)\). When these points are plotted and joined, the enclosed region is the required shaded triangular region.
Find where each line meets the x-axis by setting \(y = 0\), then solve the pair simultaneously for their intersection. Those three intersection points give the triangle’s vertices.