Question

Draw the graphs of the equations x – 5y + 14 = 0, 2x – y + 1 = 0 and x – 2y + 8 = 0. Write down the co-ordinates of the vertices of triangle formed.

Answer 1

1. Solve for intersection of lines 1 and 2

To find the first vertex, solve the system of equations for x – 5y + 14 = 0 and 2x – y + 1 = 0:

From the second equation: y = 2x + 1.

Substitute y into the first equation: x – 5(2x + 1) + 14 = 0.

Simplify: x – 10x – 5 + 14 = 0

⇒ –9x + 9 = 0

⇒ x = 1

Find y: y = 2(1) + 1 = 3.

Vertex A: (1, 3).

2. Solve for intersection of lines 2 and 3

Solve the system for 2x – y + 1 = 0 and x – 2y + 8 = 0:

From the second equation: x = 2y – 8.

Substitute x into the first equation: 2(2y – 8) – y + 1 = 0.

Simplify: 4y – 16 – y + 1 = 0

⇒ 3y – 15 = 0 

⇒ y = 5

Find x: x = 2(5) – 8 = 2.

Vertex B: (2, 5).

3. Solve for intersection of lines 3 and 1

Solve the system for x – 2y + 8 = 0 and x – 5y + 14 = 0:

Subtract the first equation from the second: (x – 5y + 14) – (x – 2y + 8) = 0.

Simplify: –3y + 6 = 0

⇒ 3y = 6

⇒ y = 2

Substitute y = 2 into x – 2y + 8 = 0:

x – 2(2) + 8 = 0

⇒ x – 4 + 8 = 0

⇒ x = –4

Vertex C: (–4, 2).

The coordinates of the vertices of the triangle formed are (1, 3), (2, 5) and (–4, 2).