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

The standared equation of a straight line graph is:

`color(brown)(y_1=mx_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:

`color(blue)(-1xx1/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 :

`(-1xx1/(-m_1))` This would give you: `+1/m_1`
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`color(brown)("To find the gradient of the perpendicular")`
Given: `y_1=-7/8 x_1 .........................................(1)`

Let the gradient of the line perpendicular be `m_2`

`color(green)(m_2)=color(blue)(-1xx1/m_1)=-1xx(-8/7) = color(green)( +8/7)`

So the equation of the perpendicular line is:

`color(blue)(y_2=color(green)(8/7)x_2+c)............................(2)`
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`color(brown)("To find the value of c")`

This new line passes through `(x_2,y_2) -> (-5,1)`

So
`y_2=1`
`x_2=(-5)`

Substitute these into (2) giving:

`1=(8/7)(-5)+c`

`color(brown)(1=-40/7+c)` .......Watch those signs!

`color(white)(..xxx.)` .......................................................
`color(white)(..xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx.)`

Add `color(blue)(40/7)` to both sides to 'get rid of it' on the right

`color(brown)(1 color(blue)(+40/7)=(-40/7color(blue)(+40/7))+c)`

But `1+40/4 =47/7 and +40/7-40/7=0` giving:

`47/7=0+c`

So`color(white)(...) color(green)(c)=47/7 = color(green)(6 5/7)`
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So
`color(blue)(y_2=8/7x_2+c)`

Becomes:
`color(blue)(y_2=8/7x_2+color(green)(6 5/7))`