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: Understanding the Concept:
To draw the graph, find at least two solutions for each equation. The area of the triangle formed by two lines and the $x$-axis is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Step 2: Key Formula or Approach:
1. Find intercepts on $x$-axis for both lines ($y=0$).
2. Find intersection point of the two lines ($x, y$).
3. Area $= \frac{1}{2} \times |x_1 - x_2| \times |y_{\text{intersection}}|$.
Step 3: Detailed Explanation:
Line 1: $x - y + 2 = 0$
If $y = 0$, $x = -2$. Point $A(-2, 0)$.
If $x = 0$, $y = 2$. Point $B(0, 2)$.
Line 2: $4x - y - 4 = 0$
If $y = 0$, $4x = 4 \implies x = 1$. Point $C(1, 0)$.
If $x = 0$, $y = -4$. Point $D(0, -4)$.
Intersection Point:
Subtract Line 1 from Line 2:
$(4x - y - 4) - (x - y + 2) = 0$
$3x - 6 = 0 \implies x = 2$.
Substitute $x=2$ in Line 1: $2 - y + 2 = 0 \implies y = 4$.
Intersection Point $P(2, 4)$.
Area of Triangle formed by $A, C,$ and $P$:
Base is on the $x$-axis between $A(-2, 0)$ and $C(1, 0)$.
Length of base $= |1 - (-2)| = 3$ units.
Height is the $y$-coordinate of the intersection point $P(2, 4)$.
Height $= 4$ units.
Area $= \frac{1}{2} \times 3 \times 4 = 6$ square units.
Step 4: Final Answer:
The area of the triangle formed is 6 sq units.