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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
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