# What is the equation of the line perpendicular to y=-5/8x  that passes through  (-6,3) ?

Jan 16, 2016

$y = \frac{8}{5} x + \frac{126}{10}$

#### Explanation:

Consider the standard equation form of a strait line graph:

$y = m x + c$ where m is the gradient.

A straight line that is perpendicular to this will have the gradient: $- \frac{1}{m}$

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$\textcolor{b l u e}{\text{Find the generic equation of the line perpendicular to the original}}$

Given equation: ${y}_{1} = - \frac{5}{8} x$...............................(1)

The equation perpendicular to this will be

$\textcolor{w h i t e}{\times \times \times \times} \textcolor{b l u e}{{y}_{2} = + \frac{8}{5} x + c}$......................................(2)

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$\textcolor{b l u e}{\text{To find the value of the constant}}$

We know it passes through the point $\left(x , y\right) \to \left(- 6 , 3\right)$

Substitute this point into equation (2) giving:

${y}_{2} = 3 = \frac{8}{5} \left(- 6\right) + c$

${y}_{2} = 3 = - \frac{48}{5} + c$

$c = 3 + \frac{48}{5} = \frac{15 + 48}{5}$

$c = 12.6$

So equation (2) becomes:

$y = \frac{8}{5} x + \frac{126}{10}$

I opted for fractional form for consistency of format. This is because the 5 in $\frac{8}{5}$ is prime. Thus division (convert to decimal) would introduce an error.

Jan 16, 2016

$y = - \frac{5}{8} x$

If $y = m x + c$ then $m$ is called the slope of the line.

Here $y = - \frac{5}{8} x + 0$

Therefore slope of the given line is $- \frac{5}{8} = {m}_{1} \left(S a y\right)$.

If two lines are perpendicular then the product of their slopes is $- 1$.

Let the slope of the line perpendicular to the given line be ${m}_{2}$.

Then by definition ${m}_{1} \cdot {m}_{2} = - 1$.

$\implies {m}_{2} = - \frac{1}{m} _ 1 = - \frac{1}{- \frac{5}{8}} = \frac{8}{5} \implies {m}_{2} = \frac{8}{5}$

This is the slope of required line and the line required line also passes through $\left(- 6 , 3\right)$.

Using point slope form

$y - {y}_{1} = {m}_{2} \left(x - {x}_{1}\right)$

$\implies y - 3 = \frac{8}{5} \left(x - \left(- 6\right)\right)$

$\implies y - 3 = \frac{8}{5} \left(x + 6\right)$

$\implies 5 y - 15 = 8 x + 48$

$\implies 8 x - 5 y + 63 = 0$

This is the required line.