Answer:
The three points for the line 2y = 5x + 11 are
point A( x₁ , y₁) ≡ ( 1 , 8)
point B( x₂ , y₂) ≡ (3 , 13)
point C(x₃ , y₃ ) ≡ (-1 , 3)
The Graph is attached below.
Step-by-step explanation:
Given:
........... equation of a line
Let the points be point A, B and point C
To Find:
point A( x₁ , y₁) ≡ ?
point B( x₂ , y₂) ≡ ?
point C(x₃ , y₃ ) ≡ ?
Solution:
For Drawing a graph we require minimum two points but we will have here three points.
For point A( x₁ , y₁)
Put x = 1 in the given equation we get
2y = 5 × 1 + 11
2y = 16
∴
∴ point A( x₁ , y₁) ≡ ( 1 , 8)
For point B( x₂ , y₂)
Put x= 3 in the given equation we get
2y = 5 × 3 + 11
2y = 26
∴
∴ point B( x₂ , y₂) ≡ (3 , 13)
For point C(x₃ , y₃ )
Put y = 3 in the given equation we get
2 × 3 = 5x+ 11
5x = -5
∴ point C(x₃ , y₃ )≡ (-1 , 3)
Therefore,
The three points for the line 2y = 5x + 11 are
point A( x₁ , y₁) ≡ ( 1 , 8) (blue color point on the graph)
point B( x₂ , y₂) ≡ (3 , 13) (green color point on the graph)
point C(x₃ , y₃ )≡ (-1 , 3) (purple color point on the graph)
The Graph is attached below..