Question

A straight line through the point (h,k) where h > 0 and k > 0, makes positive intercepts on the coordinate axes. Then the minimum length of the line intercepted between the coordinate axes is

• A. $(h^{\frac{2}{3}}+k^{\frac{2}{3}}{)}^{\frac{3}{2}}$
• B. $(h^{\frac{3}{2}}+k^{\frac{3}{2}}{)}^{\frac{2}{3}}$
• C. $(h^{\frac{2}{3}}-k^{\frac{2}{3}}{)}^{\frac{3}{2}}$
• D. $(h^{\frac{3}{2}}-k^{\frac{3}{2}}{)}^{\frac{2}{3}}$

Answer 1

The correct option is A $(h^{\frac{2}{3}}+k^{\frac{2}{3}}{)}^{\frac{3}{2}}$

Equation of line having slope m and passing through (h,k) is given by
$y-k=m(x-h)$
Let the line intersects x-axis at A i.e.$y=0$
$\Rightarrow x=h-\frac{k}{m}$
Coordinates of A are $(h-\frac{k}{m},0)$
Let the line intersects y-axis at B i.e.$x=0$
$\Rightarrow y=k-mh$
Coordinates of B are $(0,k-mh)$
Length of line intercepted between coordinate axis is
${AB}^2=(h-\frac{k}{m}{)}^2+(k-mh{)}^2$
$f\left(m\right)=(h-\frac{k}{m}{)}^2+(k-mh{)}^2$
$f^{\prime }\left(m\right)=\frac{2k}{m^2}(h-\frac{k}{m})+2m(mh-k)$
For maxima or minma,
$f^{\prime }\left(m\right)=0$
$\Rightarrow \frac{2k}{m^2}(h-\frac{k}{m})+2m(mh-k)=0$
$\Rightarrow 2(mh-k)(\frac{k}{m^2}+m)=0$
$\Rightarrow m=\frac{k}{h},m=(-\frac{k}{h}{)}^{\frac{1}{3}}$
$f^{\prime }\left(m\right)>0$ at $m=(-\frac{k}{h}{)}^{\frac{1}{3}}$
Hence, f(m) has a minimum at $m=(-\frac{k}{h}{)}^{\frac{1}{3}}$
Minimum length $=(h^{\frac{2}{3}}+k^{\frac{2}{3}}{)}^{3/2}$