[tex]\bf \stackrel{line~J}{y+8=-(x-2)}\implies y+8=-x+2
\\\\\\
y=-x-6\implies y=\stackrel{slope}{-1}x-6[/tex]
now we know the slope of line J is -1.
a line perpendicular to J, will have a negative reciprocal slope, therefore,
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -1\implies \cfrac{-1}{\underline{1}}\\\\
negative\implies +\cfrac{1}{{{ \underline{1}}}}\qquad reciprocal\implies + \cfrac{{{ \underline{1}}}}{1}\implies 1[/tex]
therefore, line K has a slope of 1, and passes through 3,4,
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1\\
% (a,b)
&&(~{{ 3}} &,&{{ 4}}~)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}\implies 1
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-4=1(x-3)\implies y-4=x-3
\\\\\\
y=x+1[/tex]
