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. 30
Step-by-Step Solution
Key Concept: Find the intercepts of each line with the x‑axis, find the point of intersection of the two lines, and use these three points as vertices of the triangle.
1. Write the equations in slope‑intercept form
$$\begin{aligned}
x-y+1&=0 \\Rightarrow y = x+1 \\[4pt]
3x+2y-12&=0 \\Rightarrow 2y = 12-3x \\Rightarrow y = 6-\frac{3}{2}x
\end{aligned}$$
2. Find the points where each line meets the x‑axis (y=0).
- For $y=0$ in $x-y+1=0$: $x+1=0 \Rightarrow x=-1$. Hence $A(-1,0)$.
- For $y=0$ in $3x+2y-12=0$: $3x-12=0 \Rightarrow x=4$. Hence $B(4,0)$.
3. Find the intersection of the two lines.
Set the right‑hand sides equal:
$$x+1 = 6-\frac{3}{2}x$$
$$\frac{5}{2}x =5 \Rightarrow x =2$$
Substituting in $y = x+1$ gives $y =3$. Hence $C(2,3)$.
4. Plot the three points $A(-1,0), B(4,0), C(2,3)$ on the coordinate plane and join them with straight lines. The lines $AB$, $AC$ and $BC$ are respectively the x‑axis, the line $x-y+1=0$ and the line $3x+2y-12=0$.
5. Shade the triangular region bounded by the two given lines and the x‑axis. This is the triangle $\triangle ABC$.
Thus the vertices of the required triangle are $(-1,0), (4,0), (2,3)$.
Correct Answer: Vertices: $(-1,0), (4,0), (2,3)$. The triangle is bounded by the x‑axis, the line $x-y+1=0$ and the line $3x+2y-12=0$; shade the region inside this triangle.