Answer :
First, we obtain the gradient (slope) of the first parallel line
[tex]\text{gradient, m}_1\text{ = }\frac{y_2-y_1}{x_2-x_1}\text{ = }\frac{3-(-3)}{2-(-4)}=\frac{6}{6}\text{ = 1}[/tex]Recall that since both lines are parallel, we have that,
[tex]m_1=m_2[/tex]Thus
[tex]m_2\text{ = 1}[/tex]Hence, we can find the equation of the parallel line given that it passes through the points (-4, -3)
Using
[tex]\begin{gathered} y\text{ = mx + c} \\ \text{where m = m}_2\text{ = 1} \\ \text{and x = -4 , y = -3, we have} \\ -3\text{ = 1(-4) + c} \\ -3\text{ = -4 + c} \\ 4\text{ - 3 = c} \\ c\text{ = 1} \\ \text{Thus, the equation of the line is y = (1)x + 1} \\ y\text{ = x + 1} \end{gathered}[/tex]