Math
posted by Bob .
Find the point on the line 1x+7y4=0 which is closest to the point (4,6)

distance will be along the perpendicular. The line perpendicular will have a slope= 7 (negative reciprocal of slope of the curve)
y=mx+b=7x+b but the point 4,6) is on the line, so
6=28+b, or b=22 check that
finally, the curve and the line intersect, so
y=x/7+4/7
y=7x+22
solve for x, y.
Now the distance formula
distance=sqrt[(x+4)^2+(y+6)^2 using x,y above. 
isn't x,y above just 4,6 ?
what do i do with the distance formula?
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