For each equation, substitute the coordinates of the points one after another and check if both give an identity. In any other case, the equation does not describe that line.
A)
[tex]y-12=\frac{1}{4}(x-6)[/tex]Substitute y=4 and x=4:
[tex]\begin{gathered} 4-12=\frac{1}{4}(4-6) \\ \Rightarrow-8=\frac{1}{4}(-2) \\ \Rightarrow-8=-\frac{1}{2}! \end{gathered}[/tex]Since -8 is not equal to -1/2, then the point (4,4) does not belong to that line.
B)
[tex]y-6=4(x-12)[/tex]Substitute y=6 and x=12:
[tex]\begin{gathered} 6-6=4(12-12) \\ \Rightarrow0=4(0) \\ \Rightarrow0=0 \end{gathered}[/tex]Therefore, the point (12,6) belongs to this line. Now, check for the point (4,4).
Substitute y=4 and x=4
[tex]\begin{gathered} 4-6=4(4-12) \\ \Rightarrow-2=4(-8) \\ \Rightarrow-2=-32! \end{gathered}[/tex]Therefore, the point (4,4) does not belong to the line.
Therefore, the option B does not describe the desired line.
E)
[tex]y-4=\frac{1}{4}(x-4)[/tex]For the point (4,4), substitute x=4 and y=4:
[tex]\begin{gathered} 4-4=\frac{1}{4}(4-4) \\ \Rightarrow0=\frac{1}{4}(0) \\ \Rightarrow0=0 \end{gathered}[/tex]Therefore, the point (4,4) belongs to this line.
Now, for the point (12,6), substitute x=12 and y=6:
[tex]\begin{gathered} 6-4=\frac{1}{4}(12-4) \\ \Rightarrow2=\frac{1}{4}(8) \\ \Rightarrow2=2 \end{gathered}[/tex]Since both (4,4) and (12,6) belong to this line, the option E describes the desired line correctly.
Hint: there are only two options that apply.
