Question

A line segment $AB$ of length $16$ is divided into two parts $AP$ and $PB$ by a point $P$. If $A{P}^2+P{B}^2$ is minimum, then

• A. $P$ trisects seg $AB$
• B. $AP:PB=3:1$
• C. $P$ bisects seg $AB$
• D. $noneofthese$

Answer 1

The correct option is C $P$ bisects seg $AB$

Solution -

REF. Image

$f\left(x\right)=x^2+(16-x{)}^2$

$f^{\prime }\left(x\right)=2x+2(16-x)\times -1=0$.

$2x=2(16-x)$

$x=16-x$

$x=8$ AP = 8 PB = 8

P bisects AB

C is correct.