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Sagot :
Given that,
The equation of line is y=7/5x+ 6 and that passes through the point (2,-6).
To find,
The equation of line that is perpendicular to the given line.
Solution,
The given line is :
y=7/5x+ 6
The slope of this line = 7/5
For two perpendicular lines, the product of slopes of two lines is :
[tex]m_1m_2=-1\\\\m_2=\dfrac{-1}{7/5}\\\\=\dfrac{-5}{7}[/tex]
Equation will be :
y=-5x/7+ b
Now finding the value of b. As it passes through (2,-6). The equation of line will be :
[tex]-6=\dfrac{-5}{7}(2)+b\\\\ -6=\dfrac{-10}{7}+b\\\\b=-6+\dfrac{10}{7}\\\\b=\dfrac{-32}{7}[/tex]
So, the required equation of line is :
y=-5x/7+ (-32/7)
[tex]y=\dfrac{-5x}{7}-\dfrac{-32}{7}[/tex]
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