Set of points that could be on the line that Sara graphs are:
Option B). (6,8) , (0,0) , (18,24)
Further explanation
Solving linear equation mean calculating the unknown variable from the equation.
Let the linear equation : y = mx + c
If we draw the above equation on Cartesian Coordinates , it will be a straight line with :
m → gradient of the line
( 0 , c ) → y - intercept
Gradient of the line could also be calculated from two arbitrary points on line ( x₁ , y₁ ) and ( x₂ , y₂ ) with the formula :
[tex]\large {\boxed{m = \frac{y_2 - y_1}{x_2 - x_1}}}[/tex]
If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :
[tex]\large {\boxed{y - y_1 = m ( x - x_1 )}}[/tex]
Let us tackle the problem.
[tex]\texttt{ }[/tex]
This problem is about Directly Proportional.
If (x₁ , y₁ ) and (x₂ , y₂) are on the line that represent a proportional relationship, then :
[tex]\large {\boxed { y_1 : x_1 = y_2 : x_2 } }[/tex]
Option A
Let:
(2,4) ⇒ (x₁ , y₁)
(3,9) ⇒ (x₂ , y₂)
[tex]4 : 2 \ne 9 : 3[/tex]
[tex]2 : 1 \ne 3 : 1[/tex]
[tex]2 \ne 3[/tex] → not proportional
[tex]\texttt{ }[/tex]
Option B
Let:
(6,8) ⇒ (x₁ , y₁)
(18,24) ⇒ (x₂ , y₂)
[tex]8 : 6 = 24 : 18[/tex]
[tex]4 : 3 = 4 : 3[/tex] → proportional
[tex]\texttt{ }[/tex]
Option C
Let:
(3,6) ⇒ (x₁ , y₁)
(9,4) ⇒ (x₂ , y₂)
[tex]6 : 3 \ne 4 : 9[/tex]
[tex]3 : 1 \ne 4 : 9[/tex] → not proportional
[tex]\texttt{ }[/tex]
Option D
Let:
(1,1) ⇒ (x₁ , y₁)
(2,1) ⇒ (x₂ , y₂)
[tex]1 : 1 \ne 1 : 2[/tex] → not proportional
[tex]\texttt{ }[/tex]
Learn more
- Infinite Number of Solutions : https://brainly.com/question/5450548
- System of Equations : https://brainly.com/question/1995493
- System of Linear equations : https://brainly.com/question/3291576
Answer details
Grade: High School
Subject: Mathematics
Chapter: Linear Equations
Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point
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