Given $p_1=(1,1), p_2=(6,3), p_3=(4,5)$ and the straight $y = 1.5-(x-4)$ what is the point $p=(x,y)$ pertaining to the straight, minimizing $norm(p_1-p) + norm(p_2-p)+norm(p_3-p)$ ?

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Answer 1 · 2016-09-06T10:30:26

$p = (0.696452, 6.19645)$

The line $y = 1.5-(x-4)$ can be represented in parametric form

$L->p=p_4+lambda vec v$ . The sought distance is given by

$delta = norm(p_4+lambda vec v-p_1)+norm(p_4+lambda vec v-p_2)+norm(p_4+lambda vec v-p_3)$ . Here

$norm(p_4+lambda vec v-p_k) = sqrt(norm(p_4-p_k)^2-2lambda << p_4-p_k, vec v >> + lambda^2 norm(vec v)^2)$

The minimum distance obeys the condition

$(d delta)/(d lambda) = 0$ and

$d/(d lambda)norm(p_4+lambda vec v-p_k) = (lambda norm(vec v)^2 - << p_4-p_k, vec v >> )/norm(p_4+lambda vec v-p_k)$ for $k=1,2,3$

Solving for $lambda$ we get

$lambda = lambda_0=0.696452$ and the point is

$p_0=p_4+lambda_0vec v = (0.696452, 6.19645)$

Given p_1=(1,1), p_2=(6,3), p_3=(4,5)p_1=(1,1), p_2=(6,3), p_3=(4,5) and the straight y = 1.5-(x-4)y = 1.5-(x-4) what is the point p=(x,y) p=(x,y) pertaining to the straight, minimizing norm(p_1-p) + norm(p_2-p)+norm(p_3-p)norm(p_1-p) + norm(p_2-p)+norm(p_3-p)?