The general equation of a line is given as
[tex]\begin{gathered} y=mx+c \\ \text{where} \\ m=Gradient \\ c=\text{intercept} \end{gathered}[/tex]
To calculate the slope/gradient of two points (x1,y1) and (x2,y2) is gotten as
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
From the table in the question, we will pick any two points
[tex]\begin{gathered} (x_1,y_1)=(1,5) \\ (x_2,y_2)=(5,10) \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} m=\frac{10-5}{5-1} \\ m=\frac{5}{4} \end{gathered}[/tex]
To calculate the equation of a line when the gradient/slope (m) is given and a point also is given, we will use the formula
[tex]m=\frac{y-y_1}{x-x_1}[/tex]
Let's take the points (x1,y1) to be ( 1,5) we will have the equation of the line to be
[tex]\begin{gathered} \frac{5}{4}=\frac{y-5}{x-1} \\ by\text{ cross multiplying, we will have} \\ y-5=\frac{5}{4}(x-1) \end{gathered}[/tex]
Therefore,
The relationship is linear ; y-5=5/4(x-1)
The correct answer is OPTION B
