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a man starts from the point p(-3,4) and will reach the point q(0,1) touching the line2x+y=7 at R. The coordinates R on the line so that he will travel in the shortest distance is
10 months ago

P( — 3, 4) and Q(0,1) Let the pd nt R be (x ,y).
Now PR + RQ is minimum.
By distance Formula
√(x-(-3))²+ (y —4)² + √(x —0)²+ (y —1)² is minimum.
The point is lying on 2x+y=7 , so y=7-2x
Replacing y by x we get , = √(x + 3)²+ (7-2x — 4)²+ √x² + (7-2x —1)²
= {√(x + 3)²+ ( -2 + 3)²} + {√x² + ( -2 + 6)²}
= √x²+9+6x+4x²+9-12x   +   √x²+4x²-24x+36
= L-- √5x²+18 —6x + √5x²-24x +36
Now Differentiating it w.r.t to x ,we get L' — 10x-6 / (2√5x²-6x +18)   +   10x-24 / (2√5x²-24x +36)  = 0 (equating to0 to find the critical point.)
10x-6 / (2√5x²-6x +18)  =  - 10x-24 / (2√5x²-24x +36)
5x-3 / (2√5x²-6x +18)   =   12-5x / (2√5x²-24x +36)
Squaring both side and after solving, we will get
25x²-192x + 252= 0
x =6and 84/ 50
So the value is minimum for x=84/50 and r 91/25

Regards
Arun (askIITians forum expert)

10 months ago
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