Answer :
[tex](3,1) , \ \ y= 3x -9 \\ \\ The \ slope \ is :m _{1} =3 \\ \\ If \ m_{1} \ and \ m _{2} \ are \ the \ gradients \ of \ two \ perpendicular \\ \\ lines \ we \ have \ m _{1} \cdot m _{2} = -1 \\ \\3\cdot m_{2}=-1 \ \ /:3\\ \\ m_{2}=-\frac{1}{3}[/tex]
[tex]\Now \ your \ equation \ of \ line \ passing \ through \ (3,1) would \ be: \\ \\ y=m_{2}x+b \\ \\1=-\frac{1}{ 3} \cdot3 + b[/tex]
[tex]1=-1+b\\ \\ b=1+1\\ \\b=2 \\ \\ y =- \frac{1}{3}x+2[/tex]
[tex]\Now \ your \ equation \ of \ line \ passing \ through \ (3,1) would \ be: \\ \\ y=m_{2}x+b \\ \\1=-\frac{1}{ 3} \cdot3 + b[/tex]
[tex]1=-1+b\\ \\ b=1+1\\ \\b=2 \\ \\ y =- \frac{1}{3}x+2[/tex]
