Given:
There is a line given as
[tex]y=\frac{3}{7}x-4[/tex]
Required:
a)
Find equation of line that parallel which is passes through ( 9 , - 2 )
b)
Find equation of line that perpendicular which is passes through ( 9 , - 2 )
Explanation:
Here take
[tex](x_1,y_1)=(9,-2)[/tex]
a)
The slope of parallel line is same
so the slope m is 3/7
now to find the line which is passes through ( 9 , - 2 )
[tex]\begin{gathered} y-y_1=\frac{3}{7}(x-x_1) \\ \\ y+2=\frac{3}{7}(x-9) \\ \\ y=\frac{3}{7}x-\frac{27}{7}-2 \\ \\ y=\frac{3}{7}x-\frac{27+14}{7} \\ \\ y=\frac{3}{7}x-\frac{41}{7} \end{gathered}[/tex]
b)
Slope of perpendicular line is
[tex]\begin{gathered} m^{\prime}=-\frac{1}{m} \\ \\ m^{\prime}=-\frac{7}{3} \end{gathered}[/tex]
now to find the line which is passes through ( 9 , - 2 )
[tex]\begin{gathered} y-y_1=m^{\prime}(x-x_1) \\ y+2=-\frac{7}{3}(x-9) \\ \\ y=-\frac{7}{3}x+\frac{7}{3}*9-2 \\ \\ y=-\frac{7}{3}x+21-2 \\ \\ y=-\frac{7}{3}x+19 \end{gathered}[/tex]
Final answer:
a)
[tex]y=\frac{3}{7}x-\frac{41}{7}[/tex]
b)
[tex]y=-\frac{7}{3}x+19[/tex]
