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:

To find the height of a triangle formed with the x-axis, simply take the absolute value of the y-coordinate of the intersection point of the two lines!
This avoids extra vertical measurement steps on your graph.

Answer 1

Step 1: Understanding the Question:
We are given two linear equations in two variables:
\[ x - y + 2 = 0 \]
\[ 4x - y - 4 = 0 \]
We need to plot both equations on a graph and find the area of the triangle bounded by the two lines and the horizontal x-axis.

Step 2: Key Formula or Approach:
1. Find at least two solution points for each line to plot them.
2. Identify the coordinates of the vertices of the triangle formed by these lines and the x-axis.
3. The formula for the area of a triangle is:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]

Step 3: Detailed Explanation:

• Find points for the first line $x - y + 2 = 0 \implies y = x + 2$:
- If $x = 0$, then $y = 2$. Point is $(0, 2)$.
- If $y = 0$, then $x = -2$. Point is $(-2, 0)$.
- If $x = 2$, then $y = 4$. Point is $(2, 4)$.

• Find points for the second line $4x - y - 4 = 0 \implies y = 4x - 4$:
- If $x = 1$, then $y = 0$. Point is $(1, 0)$.
- If $x = 0$, then $y = -4$. Point is $(0, -4)$.
- If $x = 2$, then $y = 4$. Point is $(2, 4)$.

• Identify the vertices of the triangle formed by the lines and the x-axis:
- The first line intersects the x-axis ($y = 0$) at $A(-2, 0)$.
- The second line intersects the x-axis ($y = 0$) at $B(1, 0)$.
- The two lines intersect each other at $C(2, 4)$.
These three points $A$, $B$, and $C$ form the vertices of the triangle.

• Calculate the dimensions of $\Delta\text{ABC}$:
- Base ($AB$): The distance between $(-2, 0)$ and $(1, 0)$ on the x-axis is:
\[ \text{Base} = 1 - (-2) = 3 \text{ units} \]
- Height ($h$): The vertical distance from the x-axis to the intersection vertex $C(2, 4)$ is the y-coordinate of $C$:
\[ \text{Height} = 4 \text{ units} \]

• Calculate the area of the triangle:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
\[ \text{Area} = \frac{1}{2} \times 3 \times 4 = 6 \text{ square units} \]


Step 4: Final Answer:
The vertices of the triangle are $(-2,0)$, $(1,0)$, and $(2,4)$, and the area of the triangle is 6 square units.