prove shortest distance from point(x1,y1) to the function f(x) is perpendicular to the tangent line.?

Answer 1

This is a situation where implicit differentiation is very handy: Minimizing the distance from P=(x1, y1) to (x, f(x)) is the same as => minimizing the square of the distance <=

Thus, set the derivative of (x-x1)^2 + (f(x)-y1)^2 to 0:
2(x-x1) + 2(f(x)-y1) f'(x) = 0
f'(x) = -(x-x1) / (f(x) - y1)

Now the slope of the segment from P to (x, f(x)) is
(f(x) - y1) / (x - x1). The slope of any line perpendicular to this is the negative reciprocal, which matches f'(x).
QED

Answer 2

It's a little too late, after a little too much wine, for me to do a detailed answer. Start by picking an arbitrary (x, y) on the graph of the function, setting D = √[(x - x1)² + (y - y1)²], minimizing D by computing D', then showing f'(x) at (x,y) is perpendicular (slopes are negative reciprocals) to the line through (x,y) and (x1,y1). Not a nice easy thing to do at midnight after a bottle of chardonnay. Much ugly algebra. Better if f(x) is a known function, not general.

If you picture f(x) as some squiggle of a function, and a T connecting with (x1,y1) so that the foot of the T sits on the point, the crossbar of the T is a line, and stretch the T so that the line is tangent to f(x), you know theorem is true. Just a whole lotta work to prove it. So much like life, eh?

Answer 3

Let the point on the function closest to (x1,y1) be (x, f(x))
The line between these points has slope [f(x) - y1] / [x - x1].

The distance between them is

D = √[(x-x1)^2 + (f(x)-y1)^2]
= √(x^2 - 2xx1 + (x1)^2 + f^2(x) - 2f(x)y1 + (y1)^2 )

dD/dx =
(1/2)[(x-x1)^2 + (f(x)-y1)^2]^(-1/2) * [2x - 2x1 + 2f(x)f '(x) - 2f '(x)y1]

Setting this equal to zero to find the minimum,
2x - 2x1 + 2f(x)f '(x) - 2f '(x)y1 = 0
2x - 2x1 + f '(x) *[2f(x) - 2y1] = 0
f '(x) = (x-x1) / (y1 - f(x)) , the slope of the tangent line at (x,f(x)).

This is the negative reciprocal of the line between the points. Therefore, that line and the tangent line are perpendicular.