Find the point on the line -5x+5y-1=0 which is closest to the point (2,-4) .

Answer 1

Let d be the distance

y =x +1/5

y' = 1

d² = (x-2)²+(y+4)²

Take the derivative dd/dx, set to zero and solve for x

2dd' = 2(x-2)+2(y+4)y'

d' = (x-2 + y+4)/d

x-2+x+1/5+4=0

x=-11/10

y = -1.1 + .2 = -.9

Answer 2

Alternatively, we can use a precalc approach, recalling that the shortest distance from a point to a line is the perpendicular distance.

Thus, the slope of the given line is 1, so the perpendicular slope is - 1.

An eqn of the line thru (2 , -4) with slope - 1 is y + 4 = -1(x - 2) or y = - x - 2.

The point of intersection between this line and the given line is as follows:

-5x + 5(-x - 2) - 1 = 0

-10x = 11 and x = - 11/10 and y = 11/10 - 2 = - 9/10

(-1.1 , -.9) , gladly, the same solution as above.