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: Find where each line meets the x-axis algebraically.
Setting $y=0$: for $x-y+2=0$ we get $x=-2$, giving $A(-2,0)$; for $4x-y-4=0$ we get $x=1$, giving $B(1,0)$.
Step 2: Solve the two equations together to find where the lines cross each other.
From the first equation, $y=x+2$. Substituting into the second: $4x-(x+2)-4=0 \implies 3x-6=0 \implies x=2,\ y=4$. So the lines meet at $C(2,4)$.
Step 3: Use the coordinate (shoelace) formula instead of base times height.
\[ \text{Area}=\tfrac12\big|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\big| = \tfrac12\big|(-2)(0-4)+1(4-0)+2(0-0)\big| = \tfrac12|8+4+0| \]
\[ \boxed{6 \text{ square units}} \]