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)#?

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)#?

1 Answer
Sep 6, 2016

Answer:

#p = (0.696452, 6.19645)#

Explanation:

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)#