Answer :
A perpendicular line has a slope that is the negative reciprocal of the line it's crossing. The negative reciprocal of 3 is -(1/3).
y=mx+b
5=-(1/3)(1)+b
5=(1/3)+b
4(2/3)=b
your final answer is:
y=(1/3)x+4(2/3)
y=mx+b
5=-(1/3)(1)+b
5=(1/3)+b
4(2/3)=b
your final answer is:
y=(1/3)x+4(2/3)
[tex]y=mx+b \perp y=3x+2 \\
\Downarrow \hbox{the product of the slopes is -1} \\
m \times 3=-1 \\
m=-\frac{1}{3} \\ y=-\frac{1}{3}x+b \\ \\
(-1,5) \\
x=-1 \\
y=5 \\
\Downarrow \\
5=-\frac{1}{3} \times (-1) + b \\
5=\frac{1}{3}+b \\
5-\frac{1}{3}=b \\
\frac{15}{3}-\frac{1}{3}=b \\
b=\frac{14}{3} \\
y=-\frac{1}{3}x+\frac{14}{3} \\ \\
y=-\frac{1}{3}x+\frac{14}{3} \\
\frac{1}{3}x+y=\frac{14}{3} \ \ \ |\times 3 \\
\boxed{x+3y=14}[/tex]
