[tex]1.) \\ (4,3), \ \ \ (-4,-4)\\\\First \ find \ the \ slope \ of \ the \ line \ thru \ the \ points \: \\ \\ m= \frac{y_{2}-y_{1}}{x_{2}-x_{1} } \\ \\m=\frac{-4-3}{-4-4} = \frac{-7}{-8}=\frac{ 7}{ 8} \\ \\ Use \ point \ form \ of \ a \ line\ with \ one \ point:[/tex]
[tex]y-y_{1} =m(x-x _{1}) \\ \\y-3=\frac{7}{8} (x-4)\\\\y=\frac{7}{8}x-\frac{7}{2}+3\\\\y=\frac{7}{8}x-3.5+3\\\\y=\frac{7}{8}x-0.5[/tex]
[tex]2.)\\\\ ( \frac{3}{4} , 3.2)=( \frac{3}{4} , \frac{32}{10})=( \frac{3}{4} , \frac{16}{5}) , \\ (-\frac{4}{3} , \frac{7}{4} )\\\\First \ find \ the \ slope \ of \ the \ line \ thru \ the \ points \: \\ \\ m= \frac{y_{2}-y_{1}}{x_{2}-x_{1} }[/tex]
[tex]m=\frac{ \frac{7}{4}-\frac{16}{5}}{-\frac{4}{3}-\frac{3}{4} } = \frac{ \frac{35}{20}-\frac{64}{20}}{-\frac{16}{12}-\frac{9}{12} } =\frac{-\frac{29}{20}}{-\frac{25}{12}} =(-\frac{29}{20}):(-\frac{25}{12} )=(-\frac{29}{20})*(-\frac{12} {25} )= \frac{87}{125} \\ \\ Use \ point \ form \ of \ a \ line\ with \ one \ point: \\\\ y-y_{1} =m(x-x _{1}) \\ \\y- \frac{16}{5}=\frac{87}{125} (x-\frac{3}{4} )[/tex]
[tex]y- \frac{16}{5}=\frac{87}{125} x-\frac{87}{125} \cdot \frac{3}{4} \\\\ y =\frac{87}{125} x-\frac{261}{500} +\frac{16}{5}\\\\y=\frac{87}{125} x-\frac{261}{500} +\frac{1600}{500}\\\\y=\frac{87}{125} x +\frac{1339}{500}[/tex]
