Question

A straight line L with negative slope passes through the point $(8,2)$ and cuts the positive coordinates axes at points P and Q. Find the absolute minimum value of $OP + OQ$, as L varies, where O is the origin

Answer 1

Let the line be
$y -2=- k ( x -8) k >0 $
$\text { Point } P\left(\frac{8 k +2}{ k }, 0\right) ; \text { Point } Q (0,8 k +2) $
$\therefore OP + OQ =\frac{8 k +2}{ k }+8 k +2=10+\frac{2}{ k }+8 k =10+2\left(4 k +\frac{1}{ k }\right)$
$= 10+2\left[\left(2 \sqrt{ k }-\frac{1}{\sqrt{ k }}\right)^2+4\right]$
$\therefore $ Minimum value of $OP + OQ$ when
$ 2 \sqrt{ k }-\frac{1}{\sqrt{ k }}=0 \Rightarrow 2 k -1=0 \Rightarrow k =\frac{1}{2} $
$\therefore$ Minimum value $= 10 + 8 = 18$
$\therefore y _{\min }=| a + b |$
$\text { equation } \frac{ x }{ a }+\frac{ y }{ b }=1 a >0, b >0$

$\frac{8}{a}+\frac{2}{b}=1 $
$\frac{2}{b}=\left(1-\frac{8}{a}\right)=\frac{a-8}{a}$
$b=\frac{2 a}{a-8}$
$z=a+\frac{2 a}{a-8}=\left|\frac{a^2-8 a+2 a}{a-8}\right|=\left|\frac{a^2-6 a}{a-8}\right| $
$(a-8)(4 a+6)-\left(a^2-6 a\right)=0 $
$2 a^2-44 a+48-a^2+6 a$
$a=|2| \text { or } a=y$
$a^2-16 a+48 $
$(a-12)(a x-y)=0$