The Solution:
We want to find the equation of a line S, which passes through point (6,2).
Let the required equation of line S be
[tex]y-y_1=m(x-x_1)[/tex]
Where m = the slope of line S, and
[tex]\begin{gathered} x_1=6 \\ y_1=2 \end{gathered}[/tex]
We are told that line S is parallel to line r, which is, y = 56x + 1. This implies that both lines have the same slope (m).
So, to find the slope (m) of line S, we shall compare y=56x+1 to the general form of equation of a line as given below:
[tex]y=mx+c[/tex]
Comparing both equations below:
[tex]\begin{gathered} y=mx+c \\ y=56x+1 \\ \text{ We have that,} \\ m=56 \end{gathered}[/tex]
Substituting 56 for m, and the given point (6,2) in the formula above, we get
[tex]y-2=56(x-6)[/tex]
Clearing the bracket, we get
[tex]\begin{gathered} y-2=56x-336 \\ y=56x-336+2 \\ y=56x-334 \end{gathered}[/tex]
Thus, the equation of line r is: y = 56x - 334
