3412) Let
A=z_1 (0,0); B=z_2 (1,0); C=z_3 (0,1); D= z_4 (3,4)
Then
|z-z_1|+|z-z_4|+|z-z_2|+|z-z_3| \ge |z_1-z_4| + |z_2-z_3| = 5+\sqrt2
with minimum attained for z corresponding to the intersection of AD and BC
$(1) Find \underline{\underline{Least value}} in each cases:\\\\ (i) $3\sqrt{x^2+y^2}+4\sqrt{(x-3)^2+(y-4)^2}$\\\\ (ii) $\sqrt{x^2+y^2}+\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}$\\\\ (iii) $\sqrt{(x-1)^2+(y-1)^2}+\sqrt{(x-2)^2+(y-2)^2}+\sqrt{(x+1)^2+(y+1)^2}$\\\\ (iv) $\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}$\\\\\\ (2)Find Max. value of \\\\ $\sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}-\sqrt{x^2+y^2}-\sqrt{(x-1)^2+y^2}$
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9 Answers
62First one is about finding the minimum value of 3a+4b where a and b are distances from 2 points which are at a distance of 5 from each other
so we have to minimize 3x+4(5-x) x being at least zero and at most 5
so the answer is 15
Use this logic to solve the other questions :)
3411) Let z1(0,0) and zz (3,4) be two points in the complex plane
3|z-z_1|+4|z-z_2| \ge 3|z_1-z_2|+|z-z_2| \ge 3|z_1-z_2|=15 with both the inequalities being satisified for z=z2
341341z_1 (1,1); z_2 (2,2);z_3 (-1,-1)
Then |z-z_1|+|z-z_2|+|z-z_3| \ge |z-z_1| + |z_2-z_3| \ge |z_2-z_3| with both inequalities being satisfied by z=z1.
Hence the minimum is 3√2
341(2) Let
A=z_1 (0,1); B=z_2(3,4); C=z_3(0,0); D=z_4(1,0)
the point lying on the intersection of AC and BD.
For any point O, we have by triangle inequality, we have OA-OC ≥AC ≥ 1 and OB-OD ≥ BD = 2√5
We notice that the minimum of OA+OB-OC-OD = (OA-OC)+(OB-OD) is attained for the point lying on the intersection of AC and BD and therefor the minimum is 2√5+1
341For Qn1, part 4, we need to find the Fermat point of (0,1), (1,0) and (3,4). (Please see http://en.wikipedia.org/wiki/Fermat_point)
71@Prophet Sir , @ Nishant Sir,
May you please devote or create a post with detailed idea of how does this concept build up with examples. Wherefrom to get this and how to approach in typical situation. Much like an article.
I'll be thankful to you.. Please
