Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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]