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The line $AB$ joins the points $A(a, 0 )$, $B(0, b)$ on the $x$ and $y$ axes respectively and passes through the points $(8, 27)$. Find the positions of $A$ and $B$ which minimizes the length $AB$.

Anyone have any idea, I drew it out and for equations for $a, b$ and tried creating a general line with gradients in terms of $a$ and $b$ but this didn't really help me. Ive been trying this question for some time now and want to know if anyone knows how to go about solving it.

Note: I recognize that if you make $a$ and $b = 0$ then the line that goes through the origin and the point $(8, 27)$ would mean the distance is $0$.

JKnecht
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user2250537
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1 Answers1

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Hint : consider a line passing through $(8,27)$, whose slope is $m<0$ : $y=m(x-8)+27$ (I assume that $m<0$ because I think you should have $a,b>0$, otherwise you could have $a=0=b$ which is uninteresting). It intersects the $x$-axis at $x(m)=-27/m+8$, and the $y$-axis at $y(m)=-8m+27$. Then you just have to find $m$ in order to minimize $g(m) = x(m)^2 + y(m)^2$.

The blue point indicates $(8,27) :$ enter image description here

The idea is that when $m_0$ is the minimum of $g$, then $a=x(m_0),b=y(m_0)$ satisfy the desired conditions.

You will find $g'(m) = \frac{d}{dm}((-8 m+27)^2+(-27/m+8)^2) = (2 (8 m-27) (8 m^3+27))/m^3$. Then you can prove that $m=-3/2<0$ satisfies $g'(m)=0$ is actually a minimum of $g$.

Watson
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