Answered

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Write the equation of a line that is perpendicular to y=7/5x+ 6 and that passes through the point (2,-6).

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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