Question

$P(-2,3),Q(3,7)$ . The point A on the x-axis for which $PA+AQ$ is least is

• A. $(\frac{-1}{2},0)$
• B. $(\frac{1}{2},0)$
• C. $(0,\frac{-1}{2})$
• D. $(0,\frac{1}{2})$

Answer 1

The correct option is A $(\frac{-1}{2},0)$

Let $A=(x-0)$

$PA+AQ=\sqrt{(x+2{)}^2+9}+\sqrt{(x-3{)}^2+49}$

$\frac{d\left(d\right)}{dx}=\frac{(x+2)}{\sqrt{(x+2{)}^2+9}}+\frac{(x-3)}{\sqrt{(x-3{)}^2+49}}$

$0=(x+2)\sqrt{(x-3{)}^2+49}+(x-3)\sqrt{(x+2{)}^2+9}$

$(x+2{)}^2((x-3{)}^2+49)=(x-3{)}^2((x+2{)}^2+9)$

$49(x+2{)}^2=9(x-3{)}^2$

$7(x+2)=\pm 3(x-3)$

$10x=-5$

$x=\frac{-1}{2}$