Question

Draw the graph of the pair of linear equations $x - y + 2 = 0$ and $4x - y - 4 = 0$. Calculate the area of the triangle formed by the lines so drawn and the $x$-axis.

Hint:

Always check your intersection point by plugging it into both equations. For $(2, 4)$:
Line 1: $2 - 4 + 2 = 0$ (Correct).
Line 2: $4(2) - 4 - 4 = 8 - 8 = 0$ (Correct).

Answer 1

Step 1: Understand the Problem Conceptually 
We are asked to find the area of a triangle formed by two lines and the x-axis. 
To do this systematically: 
1️⃣ Find points where each line intersects the x-axis (\(y=0\)) — these give the base points of the triangle. 
2️⃣ Find the intersection point of the two lines — this gives the vertex opposite to the base. 
3️⃣ Use the formula for area of a triangle: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] 
Step 2: Find x- and y-Intercepts of the Lines 
Line 1: \(x - y + 2 = 0\) - For x-intercept (\(y=0\)): \[ x + 2 = 0 \implies x = -2 \] Point \(A(-2,0)\) - For y-intercept (\(x=0\)): \[ - y + 2 = 0 \implies y = 2 \] Point \(B(0,2)\) 
Line 2: \(4x - y - 4 = 0\) - For x-intercept (\(y=0\)): \[ 4x - 4 = 0 \implies x = 1 \] Point \(C(1,0)\) - For y-intercept (\(x=0\)): \[ - y - 4 = 0 \implies y = -4 \] Point \(D(0,-4)\) 
Step 3: Find the Intersection Point of the Two Lines 
Solve: \[ x - y + 2 = 0 \quad \text{(Line 1)} \] \[ 4x - y - 4 = 0 \quad \text{(Line 2)} \] Subtract Line 1 from Line 2: \[ (4x - y - 4) - (x - y + 2) = 0 \] \[ 4x - y - 4 - x + y - 2 = 0 \] \[ 3x - 6 = 0 \] \[ x = 2 \] Substitute \(x = 2\) in Line 1: \[ 2 - y + 2 = 0 \implies y = 4 \] Intersection point \(P(2,4)\) 
Step 4: Identify the Triangle and Base 
- Base is on the x-axis between \(A(-2,0)\) and \(C(1,0)\) - Length of base: \[ |1 - (-2)| = 3 \text{ units} \] - Height = y-coordinate of intersection \(P(2,4)\) \[ \text{Height} = 4 \text{ units} \] 
Step 5: Calculate the Area 
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \text{ square units} \] 
Step 6: Final Answer 
\[ \boxed{\text{Area of the triangle} = 6 \text{ sq units}} \] 
Step 7: Verification 
- Base points on x-axis: correct (\(-2,0\) and \(1,0\)) - Intersection point above x-axis: correct (\(2,4\)) - Area formula applied correctly. ✅