Question

Let $S$ be the set of points whose abscissas and ordinates are natural numbers. Let $P\in S$ such that the sum of the distance of $P$ from $(8,0)$ and $(0,12)$ is minimum among all elements in $S$. Then the number of such points $P$ in $S$ is

• A. $1$
• B. $3$
• C. $5$
• D. $11$

Answer 1

The correct option is B $3$

The sum of the distances of $P$ from $A(8,0),B(0,12)$ is minimum if $P$ lies on line joining $AB$ and between them

$AP+PB=AB$

$A{P}^{\prime }+P^{\prime }B>AB$ by triangle property

So let $P(x,y)$

The line $AB$ is $\frac{x}{8}+\frac{y}{12}=1$

ie $3x+2y=24$

$\Rightarrow 3x=2(12-y)$

So $y$ can be $3,6,9$

$y=3,x=6\to 1$ point

$y=6,x=4\to 1$ point

$y=9,x=2\to 1$ point

So in total three points.