what a stupid note. Of course it will be a point with (x,y) coordinates! This is a calculus class - you know all that already.
Given a point (x,y) on the line, y=1-2x
The distance from (x,1-2x) to (3,1) is
d^2 = (x-3)^2 + (1-(1-2x))^2 = 5x^2-6x+9
so, we want minimum d, when dd/dx = 0:
2d dd/dx = 10x-6
dd/dx = 10x-6/2sqrt(blah blah)
dd/dx=0 when x = 3/5.
So, minimum d^2 is
5(3/5)^2 - 6(3/5) + 9 = 36/5
minimum d is 6/√5
Or, as we all know, the distance from a point (x,y) to a line ax+by+c=0 is
|ax+by+c|/√(a^2+b^2) = (6(3)+3(1)-3)/√(36+9) = 18/√45 = 6/√5
Find the point on the line 6x + 3y-3 =0 which is closest to the point (3,1).
Note: Your answer should be a point in the xy-plane, and as such will be of the form (x-coordinate,y-coordinate)
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