Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
We have the following:
First we calculate the slope of the line where we are given two points (6,8) and (10,0)
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]repplacing:
[tex]m=\frac{0-8}{10-6}=\frac{-8}{4}=-2[/tex]now, when two lines are perpendicular:
[tex]\begin{gathered} m_1=-\frac{1}{m_2} \\ -2=-\frac{1}{m_2} \\ 2=\frac{1}{m_2} \\ m_2=\frac{1}{2} \end{gathered}[/tex]now,
[tex]y=mx+b[/tex]with the point (4,5), replacing:
[tex]\begin{gathered} 5=\frac{1}{2}\cdot4+b \\ 5=2+b \\ b=5-2 \\ b=3 \end{gathered}[/tex]Therefore, the equation is:
[tex]\begin{gathered} y=\frac{1}{2}x+3 \\ y=\frac{x}{2}+3 \end{gathered}[/tex]check:
[tex]\begin{gathered} y=\frac{4}{2}+3 \\ y=2+3 \\ y=5 \end{gathered}[/tex]Therefore, the answer is y = x/2 + 3
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
