4x-4y-4=0

4x+4y=4

y+x=1

y=-x+1

Let x,y be a point on the line.

Distance-squared from (x,y) to (4,3) is;

D = (x-4)^2 + (y-3)^2

D = (x-4)^2 +(-x+1-3)^2

D = (x-4)^2 + (-x-2)^2

dD/dx = 2(x-4) + 2(-x-2)(-1) = 0

2x-8 +2x+4 = 0

4x = 4

x=1

y= -x+1 = -1+1 = 0

(1,0) is the point on the line

d^2D/dx^2 = 2 +2 = 4 > 0 so D is minumum at x=1

Let's rewrite the line as x + y = 1.

Any line perpendicular to it will be of the form y = x + b.

The through (4,3) and perpendicular to 4x+4y=4 will be y = x - 1.

The intersection of y = x - 1 with x + y = 1 will be (1,0).

The answer is (1,0).

4x+4y–4=0 is just x + y – 1 = 0, so d = 1*4-1*3 = 1, and the near point is:

( (1*1–1*-1)/(1²+1²), (-1*1–1*-1)/(1²+1²) ) = (2/2, 0/2) = (1,0) ← ANSWER

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The nearest point to (x₀,y₀) on the line ax+by+c=0 is ( (bd–ac)/(a²+b²), (-ad–bc)/(a²+b²) ) where d = bx₀–ay₀.

The perpendicular line through (4, 3) can be written as

.. (x -4) -(y -3) = 0

It intersects the given line at

.. (x, y) = (1, 0) . . . . . . . . . the point you seek