# What is the equation of the line perpendicular to y=-7/8x  that passes through  (-5,1) ?

Nov 28, 2015

$y = \frac{8}{7} x + 6 \frac{5}{7}$
Looks a lot in the explanation. This is because I have explained in a lot of detail what is happening. Standard calculations would not do that!

#### Explanation:

The standared equation of a straight line graph is:

$\textcolor{b r o w n}{{y}_{1} = m {x}_{1} + c}$

Where $m$ is the gradient (slope) Let this first gradient be ${m}_{1}$

Any slope that is perpendicular to this line has the gradient of:

$\textcolor{b l u e}{- 1 \times \frac{1}{m} _ 1}$

~~~~~~~~~~~~~~ Comment ~~~~~~~~~~~~~~~~~~~~~~~
I have done it this way to help with signs. Suppose that $m$ is negative. Then the perpendicular would have the gradient of :

$\left(- 1 \times \frac{1}{- {m}_{1}}\right)$ This would give you: $+ \frac{1}{m} _ 1$
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$\textcolor{b r o w n}{\text{To find the gradient of the perpendicular}}$
Given: ${y}_{1} = - \frac{7}{8} {x}_{1} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

Let the gradient of the line perpendicular be ${m}_{2}$

$\textcolor{g r e e n}{{m}_{2}} = \textcolor{b l u e}{- 1 \times \frac{1}{m} _ 1} = - 1 \times \left(- \frac{8}{7}\right) = \textcolor{g r e e n}{+ \frac{8}{7}}$

So the equation of the perpendicular line is:

$\textcolor{b l u e}{{y}_{2} = \textcolor{g r e e n}{\frac{8}{7}} {x}_{2} + c} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$
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$\textcolor{b r o w n}{\text{To find the value of c}}$

This new line passes through $\left({x}_{2} , {y}_{2}\right) \to \left(- 5 , 1\right)$

So
${y}_{2} = 1$
${x}_{2} = \left(- 5\right)$

Substitute these into (2) giving:

$1 = \left(\frac{8}{7}\right) \left(- 5\right) + c$

$\textcolor{b r o w n}{1 = - \frac{40}{7} + c}$ .......Watch those signs!

$\textcolor{w h i t e}{. . \times x .}$ .......................................................
$\textcolor{w h i t e}{. . \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times .}$

Add $\textcolor{b l u e}{\frac{40}{7}}$ to both sides to 'get rid of it' on the right

$\textcolor{b r o w n}{1 \textcolor{b l u e}{+ \frac{40}{7}} = \left(- \frac{40}{7} \textcolor{b l u e}{+ \frac{40}{7}}\right) + c}$

But $1 + \frac{40}{4} = \frac{47}{7} \mathmr{and} + \frac{40}{7} - \frac{40}{7} = 0$ giving:

$\frac{47}{7} = 0 + c$

So$\textcolor{w h i t e}{\ldots} \textcolor{g r e e n}{c} = \frac{47}{7} = \textcolor{g r e e n}{6 \frac{5}{7}}$
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So
$\textcolor{b l u e}{{y}_{2} = \frac{8}{7} {x}_{2} + c}$

Becomes:
$\textcolor{b l u e}{{y}_{2} = \frac{8}{7} {x}_{2} + \textcolor{g r e e n}{6 \frac{5}{7}}}$