Given the points:
(5, 12), (11, 9)
Let's find the slope of a line perpendicular to the line that passes through the given points.
The slope of a perpendicular line is the negative reciprocal of the slope of the original line:
[tex]m_1m_2=-1[/tex]
Where:
m1 is the slope of the origginal line
m2 is the slope of the perpendicular line.
To find the slope of the original line, apply the slope formula:
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
Thus, we have:
(x1, y1) ==> (5, 12)
(x2, y2) ==> 11, 9
[tex]\begin{gathered} m=\frac{9-12}{11-5} \\ \\ m=\frac{-3}{6} \\ \\ m=-\frac{1}{2} \end{gathered}[/tex]
The slope of the original line is -1/2.
To fine the slope of the perpendicular line substitute -1/2 for m1 in the equation (m1m2 = -1).
Thus, we have:
[tex]\begin{gathered} m_1m_2=-1 \\ \\ -\frac{1}{2}m_2=-1 \\ \\ Multiply\text{ both sides by 2:} \\ -\frac{1}{2}m_2\times2=-1\times2 \\ \\ -1m_2=-2 \end{gathered}[/tex]
Divide both sides by -1:
[tex]\begin{gathered} -\frac{1m_2}{-1}=\frac{-2}{-1} \\ \\ m=2 \end{gathered}[/tex]
Therefore, the slope of the perpendicular line is = 2
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