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))#