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Perpendicular to the line given by 10 y + 3x= -2
For this part we desire the line to be perpendicular to 10 y + 3x= -2 & so we know we can determine the new slope as follows,
m 2 =-=1/(-3/10)=10/3
Then, just as we did in the earlier part we can employ the point-slope form of the line to obtain the equation of the new line. Here it is,
y=2+(10/3)(x-8)
=2+(10/3)x-(80/3)
y =(10/3)x-74/3
(4 sqrt3+5 sqrt2)/(sqrt48+ srt18)
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