Solution:
To graph the line that passes through two points, we first need to determine the equation of the line.
The equation of the line that passes through two points A and B is expressed as
[tex]\begin{gathered} y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)\text{ ------ equation 1} \\ where \\ (x_{1,}y_1)\text{ and \lparen x}_{2,}y_2)\text{ are the coordinates of the points A and B respectively,} \\ through\text{ which the line passes.} \end{gathered}[/tex]Given that the line passes through the points (1, -4) and (-2, 2), this implies
[tex]\begin{gathered} x_1=1 \\ y_1=-4 \\ x_2=-2 \\ y_2=2 \end{gathered}[/tex]Substituting these values into equation 1, we have
[tex]\begin{gathered} y-(-4)=(\frac{2-(-4)}{-2-1})(x-1) \\ \Rightarrow y+4=(\frac{2+4}{-2-1})(x-1) \\ y+4=\frac{6}{-3}(x-1) \\ \Rightarrow y+4=-2(x-1) \\ subtract\text{ 4 from both sides of the equation} \\ y+4-4=-2(x-1)-4 \\ \Rightarrow y=-2(x-1)-4 \\ open\text{ parentheses} \\ y=-2x+2-4 \\ \Rightarrow y=-2x-2 \end{gathered}[/tex]Thus, the equation of the line that passes through the points (1, -4) and (-2, 2) is
[tex]y=-2x-2[/tex]To graph the line,
[tex]\begin{gathered} when\text{ x=0, we have} \\ y=-2(0)-2=0-2 \\ \Rightarrow y=-2 \\ when\text{ y=0, we have} \\ 0=-2x-2 \\ add\text{ 2 to both sides of the equation} \\ 0+2=-2x-2+2 \\ \Rightarrow2=-2x \\ divide\text{ both sides by the coefficient of x, which is -2} \\ \frac{2}{-2}=\frac{-2x}{-2} \\ \Rightarrow x=-1 \end{gathered}[/tex]Given the points (1, -4), (-2, 2), (0,-2) and (-1,0), the graph of the line is as shown below:
