First, we fill the table. Given a value of $x,$ we can find the corresponding $y$ by substituting $x$ into the formula $y=-x+1.$ Therefore,

• if $x=0$ then $y=-(0)+1=1,$

• if $x=1$ then $y=-(1)+1=0,$

• if $x=2$ then $y=-(2)+1=-1,$ and

• if $x=3$ then $y=-(3)+1=-2.$

Finally, connecting the points gives us a straight line. This is the graph of $y=-x+1.$

First, we fill the table. Given a value of $x,$ we can find the corresponding $y$ by substituting $x$ into the formula $y=x-2.$ Therefore,

• if $x=0$ then $y=(0)-2=-2,$

• if $x=1$ then $y=(1)-2=-1,$

• if $x=2$ then $y=(2)-2=0,$ and

• if $x=3$ then $y=(3)-2=1.$

Finally, connecting the points gives us a straight line. This is the graph of $y=x-2.$

We need to find two distinct points that lie on our line. Any two points will do!

• For the first point, we will let the $x$-coordinate be $x=0.$ Then the $y$-coordinate is $$\begin{align*} y &= \dfrac{1}{2}x-2 \\ &= \dfrac{1}{2}(0)-2 \\ &= 0-2 \\ &= -2. \end{align*}$$ So, the first point is $(0,-2).$

• For the second point, we will let the $x$-coordinate be $x=6.$ Then the $y$-coordinate is $$\begin{align*} y &= \dfrac{1}{2}x-2 \\ &= \dfrac{1}{2}(6)-2 \\ &= 3-2 \\ &= 1. \end{align*}$$ So, the second point is $(6,1).$

Finally, we plot those two points on the $xy$-plane and draw a line through them. This is the graph of $y=\dfrac{1}{2}x-2.$

We need to find two distinct points that lie on our line. Any two points will do!

• For the first point, we will let the $x$-coordinate be $x=0.$ Then the $y$-coordinate is $$\begin{align*} y &= 2x-4 \\ &= 2(0)-4 \\ &= 0-4 \\ &= -4. \end{align*}$$ So, the first point is $(0,-4).$

• For the second point, we will let the $x$-coordinate be $x=2.$ Then the $y$-coordinate is $$\begin{align*} y &= 2x-4 \\ &= 2(2)-4 \\ &= 4-4 \\ &= 0. \end{align*}$$ So, the second point is $(2,0).$

Finally, we plot those two points on the $xy$-plane and draw a line through them. This is the graph of $y=2x-4.$

Let's pick two distinct points on the line. For example, we can choose $A$ and $B$ with coordinates $(0,2)$ and $(4,4)$, respectively.

Now, we substitute the coordinates of both points in our equations. If both points satisfy an equation, then we can conclude that the equation represents the line in the graph. Otherwise, if any point does not satisfy the equation, then we can conclude that the equation does not represent the line in the graph.

• Equation I: $y=x+\dfrac{1}{2}.$ Substituting the coordinates of $A(0,2)$, we obtain $$\begin{align*} (2)&=(0)+\dfrac{1}{2} \\ 2&=\dfrac{1}{2} \: {\color{red}\times} \end{align*}$$ Therefore, $y=x+\dfrac{1}{2}$ is not an equation of the line in the graph.

• Equation II: $y=\dfrac{1}{2}x+2.$ Substituting the coordinates of $A(0,2)$ and $B(4,4)$, we obtain $$\begin{align*} (2)&=\dfrac{1}{2}(0) + 2 \\ 2&=2 \: {\color{green}\checkmark} \\[5pt] (4)&=\dfrac{1}{2}(4) + 2 \\ 4&=4 \: {\color{green}\checkmark} \end{align*}$$ Therefore, $y=\dfrac{1}{2}x+2$ is an equation of the line in the graph.

• Equation III: $y=\dfrac{1}{2}x+\dfrac{1}{2}.$ Substituting the coordinates of $A(0,2)$, we obtain $$\begin{align*} (2)&=\dfrac{1}{2}(0)+\dfrac{1}{2} \\ 2&=\dfrac{1}{2} \: {\color{red}\times} \end{align*}$$ Therefore, $y=\dfrac{1}{2}x+\dfrac{1}{2}$ is not an equation of the line in the graph.

Of the given equations, the only equation that represented the line in the graph was $y=\dfrac{1}{2}x+2,$ corresponding to answer choice II. Therefore, the correct answer is "II only."

Let's pick two distinct points on the line. For example, we can choose $A$ and $B$ with coordinates $(0,4)$ and $(2,2)$, respectively.

Now, we substitute the coordinates of both points in our equations. If both points satisfy an equation, then we can conclude that the equation represents the line in the graph. Otherwise, if any point does not satisfy the equation, then we can conclude that the equation does not represent the line in the graph.

• Equation I: $x+2y=4.$ Substituting the coordinates of $A(0,4)$, we obtain $$\begin{align*} (0) + 2(4) &= 4 \\ 8 &= 4 \: {\color{red}\times} \end{align*}$$ Therefore, $x+2y=4$ is not an equation of the line in the graph.

• Equation II: $y=2-x.$ Substituting the coordinates of $A(0,4)$, we obtain $$\begin{align*} 4 &= 2 - (0) \\ 4 &= 2 \: {\color{red}\times} \end{align*}$$ Therefore, $y=2-x$ is not an equation of the line in the graph.

• Equation III: $y=4-x.$ Substituting the coordinates of $A(0,4)$ and $B(2,2)$, we obtain $$\begin{align*} (4)&=4-(0) \\ 4&=4 \: {\color{green}\checkmark} \\[5pt] (2)&=4-(2) \\ 2&=2 \: {\color{green}\checkmark} \end{align*}$$ Therefore, $y=4-x$ is an equation of the line in the graph.

Of the given equations, the only equation that represented the line in the graph was $y=4-x,$ corresponding to answer choice III. Therefore, the correct answer is "III only."

The plotted points correspond to the $xy$-pairs $A(2,7), B(4,5)$ and $C(7,2).$

To check whether a point lies on the line, we substitute the coordinates of the point into the equation and determine whether the resulting statement is true.

• Substituting the point $A(2,7)$ results in a false statement: $$\begin{align*} y &= x+1 \\ (7) &= (2)+1 \\ 7 &= 3 \; {\color{red}\times} \\ \end{align*}$$

  Therefore, the point $A$ does not lie on the line.

• Substituting the point $B(4,5)$ results in a true statement: $$\begin{align*} y &= x+1 \\ (5) &= (4)+1 \\ 5 &= 5 \; {\color{green}\checkmark} \end{align*}$$

  Therefore, the point $B$ does lie on the line.

• Substituting the point $C(7,2)$ results in a false statement: $$\begin{align*} y &= x+1 \\ (2) &= (7)+1 \\ 2 &= 8 \; {\color{red}\times} \end{align*}$$

  Therefore, the point $C$ does not lie on the line.

The plotted points correspond to the $xy$-pairs $A(-6,5)$, $B(1,5)$ and $C(6,5).$

To check whether a point lies on the line, we substitute the coordinates of the point into the equation and determine whether the resulting statement is true.

• Substituting the point $A(-6,5)$ results in a false statement: $$\begin{align*} y &= 8x-3 \\ (5) &= 8(-6)-3 \\ 5 &= -51 \; {\color{red}\times} \\ \end{align*}$$

  Therefore, the point $A$ does not lie on the line.

• Substituting the point $B(1,5)$ results in a true statement: $$\begin{align*} y &= 8x-3 \\ (5) &= 8(1)-3 \\ 5 &= 5 \; {\color{green}\checkmark} \end{align*}$$

  Therefore, the point $B$ does lie on the line.

• Substituting the point $C(6,5)$ results in a false statement: $$\begin{align*} y &= 8x-3 \\ (5) &= 8(6)-3 \\ 5 &= 45 \; {\color{red}\times} \end{align*}$$

  Therefore, the point $C$ does not lie on the line.