Question 1027803
Find the equation of the line perpendicular to y=7x+1 and passing at a distance of sqrt(2) from (4,-2). Then graph (i need it ASAP i beg of you)
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Another method:
All points a distance of sqrt(2) from (4,-2) is a circle,
{{{(x-4)^2 + (y+2)^2 = 2}}}
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Find the 2 points on the circle with a slope of the tangent of -1/7.
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m = -(x-h)/(y-k) for a circle with its center at (h,k)
(x-4)/(y+2) = 1/7
7x-28 = y+2
y = 7x-30
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Sub for y in {{{(x-4)^2 + (y+2)^2 = 2}}} and solve for x, then y.
Then you have 2 points and a slope of -1/7