Answer:
A and B has the same constant of proportionality
Step-by-step explanation:
[tex]y \propto x[/tex]
[tex]y = kx ----1[/tex]
Where k is the constant of proportionality
We are supposed to find Which relationships have the same constant of proportionality between y and x as in the equation [tex]y=\frac{1}{2}x[/tex]
On comparing with 1
[tex]k = \frac{1}{2}[/tex]
A)6y = 3x
[tex]y = \frac{3}{6}x\\y = \frac{1}{2}x[/tex]
So, this equation has the same constant of proportionality
B)[tex](x_1,y_1)=(2,1)\\(x_2,y_2)=(4,2)[/tex]
To find the equation :
Formula : [tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]
So, [tex]y - 1=\frac{2-1}{4-2}(x-2)\\y-1=\frac{1}{2}(x-2)\\y-1=\frac{1}{2}x-1\\y=\frac{1}{2}x[/tex]
So, this equation has the same constant of proportionality
C)
[tex](x_1,y_1)=(1,2)\\(x_2,y_2)=(2,4)[/tex]
To find the equation :
Formula : [tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]
So, [tex]y - 2=\frac{4-2}{2-1}(x-1)\\y - 2=2(x-1)\\y - 2=2x-2\\y=2x[/tex]
So, this equation do not has the same constant of proportionality
D)
[tex](x_1,y_1)=(2,1)\\(x_2,y_2)=(3,2.5)[/tex]
To find the equation :
Formula :[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]
So, [tex]y - 1=\frac{2.5-1}{3-2}(x-2)[/tex]
[tex]y-1=1.5(x-2)\\y-1=1.5x-3\\y=1.5x-2[/tex]
So, this equation do not has the same constant of proportionality
Hence A and B has the same constant of proportionality
