SOLUTION: Find the equation of the line perpendicular to y=7x+1 and passing at a distance of √2 from (4,-2). Then graph
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-> SOLUTION: Find the equation of the line perpendicular to y=7x+1 and passing at a distance of √2 from (4,-2). Then graph
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Question 1027800: Find the equation of the line perpendicular to y=7x+1 and passing at a distance of √2 from (4,-2). Then graph Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! The desired line(s) should have the form 7y + x +c = 0, for some undetermined constant(s) c.
The formula fro the distance between a point and the line is given by .
==> .
==> c - 10 = 10, or c - 10 = -10
==> c = 20, or c = 0.
Therefore there are two lines satisfying the given conditions, namely
7y + x + 20 = 0, or 7y + x = 0,
on both sides of the point (4,-2) and perpendicular to y=7x+1.
