The given line passes through the points (0, -3) and (2, 3).What is the equation, in point-slope form of the line that isparallel to the given line and passes through the point(-1, -1)?y+1 = - 3(x + 1)+1 = -(x + 1)v+1 = $(x + 1)5-3 -2 1/134y + 1 = 3(x + 1)Save and ExitMark this and return

The equation, in point-slope form of the line that is  parallel to the given line and passes through the point  (-1, -1) is y + 1 = 3(x + 1).Solution:Given that, a line passes through (0, -3) and (2, 3). We have to find the line equation which is parallel to above line and  passes through (-1, -1). Now, let us find the slope of the given line. [tex]\text { slope } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}, \text { where }\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right) \text { are points on that line }[/tex][tex]\text { Then, } m=\frac{3-(-3)}{2-0}=\frac{3+3}{2}=\frac{6}{2}=3[/tex]So, slope of given line is 3,  Then, slope of required line is also 3, as slopes of parallel lines are equal. Then, required line equation in point – slope form is given as:[tex]\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right)} \\\\ {y-(-1)=3(x-(-1))} \\\\ {\rightarrow y+1=3(x+1)}\end{array}[/tex]Hence, the line equation is y + 1 = 3(x + 1).