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The equation of a line through the point (1,2) Whose distance from the point (3,1) has the greatest value is

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Solution: Any line through $(1,2)$

$y-2=m(x-1)$ $..........(1)$

$Rightarrow$ $mx-y+2-m=0$

Length of perp. from $(3,1)$ to Line $(1)$

$h=frac2m+1sqrtm^2+1$ $............(2)$

$Rightarrow$ $fracmathrmd hmathrmd m=frac2-m(m^2-11)^frac32$ $........(3)$

$Rightarrow$ $fracmathrmd hmathrmd m=0$ gives $m=2$

Now , $fracmathrmd hmathrmd m> 0$ for $m< 2$ and $fracmathrmd hmathrmd m< 0$ for $m> 2$

$Rightarrow$ h is maximum for $m=2$

Therefore , $(1)$ becomes for $y=2x$ .

Posted by

Deependra Verma

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