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:

4x – 3y + 4 = 0, 4x + 3y – 20 = 0

Answer 1

1. Find points for line 1

For the first equation 4x – 3y + 4 = 0:

Substituting y = 0:

4x – 3(0) + 4 = 0

⇒ 4x = –4

⇒ x = –1

This gives the coordinate (–1, 0).

Substituting x = 2:

4(2) – 3y + 4 = 0

⇒ 8 + 4 = 3y

⇒ 3y = 12

⇒ y = 4

This gives the coordinate (2, 4).

2. Find points for line 2

For the second equation 4x + 3y – 20 = 0:

Substituting y = 0:

4x + 3(0) – 20 = 0

⇒ 4x = 20

⇒ x = 5

This gives the coordinate (5, 0).

Substituting x = 2:

4(2) + 3y – 20 = 0

⇒ 8 + 3y = 20

⇒ 3y = 12

⇒ y = 4

This gives the coordinate (2, 4).

3. Determine the intersection point

Plotting these coordinates reveals that both lines intersect at the point A(2, 4). Thus, the solution to the system of linear equations is x = 2 and y = 4.

4. Calculate the triangle area

The triangle is formed by the vertices A(2, 4), B(–1, 0) and C(5, 0):

Base (BC): Distance between (–1, 0) and (5, 0) = 5 – (–1) = 6 units.

Height (h): The perpendicular distance from vertex A to the x-axis, which corresponds to the y-coordinate of A = 4 units.

Using the area formula:

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

Area = `1/2 xx 6 xx 4`

Area = 12 square units