Question

How do you find the equation of the line in standard form that is perpendicular to the line y=3x+2 and passes through (-1, 5)

please help!!!

Answer 1

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)

Answer 2

`y=mx+b \perp y=3x+2 \\ \Downarrow \hbox{the product of the slopes is -1} \\ m * 3=-1 \\ m=-(1)/(3) \\ y=-(1)/(3)x+b \\ \\ (-1,5) \\ x=-1 \\ y=5 \\ \Downarrow \\ 5=-(1)/(3) * (-1) + b \\ 5=(1)/(3)+b \\ 5-(1)/(3)=b \\ (15)/(3)-(1)/(3)=b \\ b=(14)/(3) \\ y=-(1)/(3)x+(14)/(3) \\ \\ y=-(1)/(3)x+(14)/(3) \\ (1)/(3)x+y=(14)/(3) \ \ \ |* 3 \\ \boxed{x+3y=14}`