Question

Solve the following given system of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – 2y + 2 = 0, 2x + y – 6 = 0

Answer 1

1. Find points for equation 1

Rearrange x – 2y + 2 = 0 to express x in terms of y:

x = 2y – 2

Find at least two coordinate points to plot this line:

If y = 0: x = 2(0) – 2 = –2 → (–2, 0)

If y = 1: x = 2(1) – 2 = 0 → (0, 1)

2. Find points for equation 2

Rearrange 2x + y – 6 = 0 to express y in terms of x:

y = 6 – 2x

Find at least two coordinate points to plot this line:

If x = 0: y = 6 – 2(0) = 6 → (0, 6)

If x = 3: y = 6 – 2(3) = 0 → (3, 0)

3. Determine Intersection and Vertices

When plotted on a graph, the two lines intersect at a single point where both equations share the same values. Mathematically, substituting y = 6 – 2x into the first equation:

x – 2(6 – 2x) + 2 = 0

5x – 10 = 0 ⇒ x = 2  

Substituting x = 2 back gives y = 2. Thus, the intersection vertex is (2, 2). The base vertices lie along the x-axis (y = 0), which correspond to the x-intercepts found in steps 1 and 2: (–2, 0) and (3, 0).

4. Calculate area of the triangle

The base of the triangle lies entirely on the x-axis between x = –2 and x = 3:

Base = 3 – (– 2) = 5 units

The perpendicular height corresponds to the y-coordinate of the topmost intersection vertex:

Height = 2 units

Apply the standard triangle area formula:

Area = `1/2 xx "base" xx "height"`

Area = `1/2 xx 5 xx 2`

Area = 5 square units