Answer :
The genral equation of line can be written as y=mx+c, where m is the slope of the line.
Comparing y=7/5x+4 with y= mx+c, we get slope m=7/5.
The slope of two parallel lines are equal. Hence, the slope of a parallel line to y=7/5x+4 is 7/5.
Now, the point slope form of a line which passes through a point (x1, y1) and habing slope m is,
[tex]\frac{y-y_1}{x-x_1}=m[/tex]Given (x1,y1)=(-7,-5). Therefore,
[tex]\begin{gathered} \frac{y-(-5)}{x-(-7)}=\frac{7}{5} \\ \frac{y+5}{x+7}=\frac{7}{5} \\ 5y+25=7x+49 \\ 5y=7x+49-25 \\ 5y=7x+24 \\ y=\frac{7}{5}x+\frac{24}{5} \end{gathered}[/tex]Therefore, the equation of line parallel to y=7/5x+4 and passing through point(-7,-5) is y=(7/5)x+(24/5).
(2)The slope of a line perpendicular to y=mx+c is -1/m.
So, slope of a line perpendicular to y=7/5x+4 is
[tex]Slope,m_1=\frac{-1}{m}=\frac{-1}{\frac{7}{5}}=\frac{-5}{7}[/tex]If (x1,y1)=(-7-5), then using point slope form,
[tex]\begin{gathered} \frac{y-y_1}{x-x_2}=m_1 \\ \frac{y-(-5)}{x-(-7)}=\frac{-5}{7} \\ \frac{y+5}{x+7}=\frac{-5}{7} \\ 7y+35=-5x-35 \\ 7y=-5x-70 \\ y=\frac{-5}{7}x-10 \end{gathered}[/tex]Therefore, the equation of line perpendicular to y=7/5x+4 and passing through point(-7,-5) is y=-(5/7)x-10.
