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).
