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.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{passing}\mathrm{through}\left(1,4\right)\mathrm{with}\mathrm{slope}m\mathrm{is}\mathrm{given}\mathrm{by} \\ y-4=m\left(x-1\right)...\left(1\right) \\ \mathrm{Substituting}y=0,\mathrm{we}\mathrm{get} \\ 0-4=m\left(x-1\right) \\ \Rightarrow \frac{-4}{m}=x-1 \\ \Rightarrow x=\frac{m-4}{m} \\ \mathrm{Substituting}x=0,\mathrm{we}\mathrm{get} \\ y-4=m\left(0-1\right) \\ \Rightarrow y=-m+4 \\ \Rightarrow x=-\left(m-4\right) \\ \mathrm{So},\mathrm{the}\mathrm{intercepts}\mathrm{on}\mathrm{coordinate}\mathrm{axes}\mathrm{are}\frac{m-4}{m}\mathrm{and}-\left(m-4\right). \\ \mathrm{Let}S\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{intercepts}.\mathrm{Then}, \\ \mathrm{S}=\frac{m-4}{m}-\left(m-4\right) \\ \Rightarrow \frac{dS}{dm}=\frac{4}{m^2}-1 \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}S,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dS}{dm}=0 \\ \Rightarrow \frac{4}{m^2}-1=0 \\ \Rightarrow \frac{4}{m^2}=1 \\ \Rightarrow m^2=4 \\ \Rightarrow m=\pm 2 \\ \mathrm{Now}, \\ \frac{d^{2}S}{d{m}^2}=\frac{-8}{m^3} \\ {\left(\frac{d^{2}S}{d{m}^2}\right)}_{m=2}=\frac{-8}{2^3}=-1<0 \\ \mathrm{So},\mathrm{the}\mathrm{sum}\mathrm{is}\mathrm{minimum}\mathrm{at}m=2. \\ {\left(\frac{d^{2}S}{d{m}^2}\right)}_{m=-2}=\frac{-8}{{\left(-2\right)}^3}=1>0 \\ \mathrm{So},\mathrm{the}\mathrm{sum}\mathrm{is}\mathrm{maximum}\mathrm{at}m=-2. \\ \mathrm{Thus},\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ S=\frac{-2-4}{-2}-\left(-2-4\right)=3+6=9 \end{gathered}$$