# How do you write an equation of the line containing the given point and perpendicular to the given line: (6, -9); 6x+5y=7?

Jul 7, 2016

$\textcolor{b l u e}{y = \frac{5}{6} x - 14}$

#### Explanation:

Given:$\text{ "6x+5y=7 larr" point } {P}_{1} \to \left(x , y\right) = \left(6 , - 9\right)$

Change the equation into the standard form of $y = m x + c$

$\implies y = - \frac{6}{5} x + \frac{7}{5}$............. Equation for line (1)
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From this the gradient of the given line is:$\text{ } - \frac{6}{5}$

So the gradient of a line perpendicular (normal) to this is:

$\left(- 1\right) \times \frac{1}{m} \to \left(- 1\right) \times - \frac{5}{6} = + \frac{5}{6}$

$\implies y = + \frac{5}{6} x + c$.....................Equation for line (2)
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But the second line passes through ${P}_{1}$

$\implies {P}_{1} \to \left(x , y\right) = \left(6 , - 9\right) \to - 9 = \frac{5}{\cancel{6}} \left(\cancel{6}\right) + c$

$\implies c = - 9 - 5 = - 14$ giving:

$\textcolor{b l u e}{y = \frac{5}{6} x - 14}$

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