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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!!!

Sagot :

jpmg2k
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)
naǫ
[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]
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