2. What is the equation of the line that passes through (5, 2) and isperpendicular to y =10x + 7?AC.y10x +yX+1021y = 10x - 4810B.-+52+52D.
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Step 1
Given; What is the equation of the line that passes through (5, 2) and is
perpendicular to y =
10x + 7?
Step 2
The slope of the given line is;
[tex]\begin{gathered} m=10 \\ since,\text{ when we compare y=mx+b} \\ m=10 \end{gathered}[/tex]Slope of perpendicular lines have the following relationship;
[tex]\begin{gathered} m_1=-\frac{1}{m_2} \\ 10=-\frac{1}{m_2} \\ m_2=-\frac{1}{10} \end{gathered}[/tex]Therefore the required equation will be in the form of;
[tex]\begin{gathered} y=-\frac{1}{10}x+b \\ y=2 \\ x=5 \end{gathered}[/tex]Find b, the y-intercept
[tex]\begin{gathered} 2=-\frac{1}{10}(5)+b \\ 2=-\frac{1}{2}+b \\ 4=-1+2b \\ 2b=5 \\ b=\frac{5}{2} \end{gathered}[/tex]Thus the answer will be; Option B
[tex]\begin{gathered} y=-\frac{1}{10}x+\frac{5}{2} \\ \\ \\ \\ \end{gathered}[/tex]