# A straight line is drawn through a given point

Question:

A straight line is drawn through a given point $P(1,4)$. Determine the least value of the sum of the intercepts on the coordinate axes.

Solution:

The equation of line passing through $(1,4)$ with slope $m$ is given by

$y-4=m(x-1)$          ...(1)

Substituting $y=0$, we get

$0-4=m(x-1)$

$\Rightarrow \frac{-4}{m}=x-1$

$\Rightarrow x=\frac{m-4}{m}$

Substituting $x=0$, we get

So, the intercepts on coordinate axes are $\frac{m-4}{m}$ and $-(m-4)$.

Let $S$ be the sum of the intercepts. Then,

$\mathrm{S}=\frac{m-4}{m}-(m-4)$

$\Rightarrow \frac{d S}{d m}=\frac{4}{m^{2}}-1$

For maximum or minimum values of $S$, we must have

$\frac{d S}{d m}=0$

$\Rightarrow \frac{4}{m^{2}}-1=0$

$\Rightarrow \frac{4}{m^{2}}=1$

$\Rightarrow m^{2}=4$

$\Rightarrow m=\pm 2$

Now,

$\frac{d^{2} S}{d m^{2}}=\frac{-8}{m^{3}}$

$\left(\frac{d^{2} S}{d m^{2}}\right)_{m=2}=\frac{-8}{2^{3}}=-1<0$

So, the sum is minimum at $m=2$.

$\left(\frac{d^{2} S}{d m^{2}}\right)_{m=-2}=\frac{-8}{(-2)^{3}}=1>0$

So, the sum is maximum at $m=-2$.

Thus, the minimum value is given by

$S=\frac{-2-4}{-2}-(-2-4)=3+6=9$