Question

A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinate axes at the points P and Q respectively.

When O is the origin, find the minimum value of OP + OQ, as L varies ?

Answer 1

$$\begin{gathered} GiventhatastraightlineLwithnegativeslopepassesthroughthepoint\left(8,2\right) \\ andcutsthepositivecoordinateaxesatthepointsPandQrespectively. \\ So,letslopeofthelineL=-m\left(wherem>0\right) \\ Thus,equationoflineLis, \\ y-2=-m\left(x-8\right) \\ \Rightarrow y-2=-mx+8m \\ \Rightarrow mx+y=8m+2 \\ \Rightarrow \frac{mx}{8m+2}+\frac{y}{8m+2}=1 \\ \Rightarrow \frac{x}{\left({\displaystyle \frac{8m+2}{m}}\right)}+\frac{y}{\left(8m+2\right)}=1 \\ So, \\ OP=\frac{8m+2}{m}And,OQ=8m+2 \\ LetS=OP+OQ \\ =\frac{8m+2}{m}+8m+2 \\ =8+\frac{2}{m}+8m+2 \\ =10+\frac{2}{m}+8m \\ =10+{\left(\sqrt{\frac{2}{m}}\right)}^2+{\left(2\sqrt{2m}\right)}^2+8-8 \\ =\left[{\left(\sqrt{\frac{2}{m}}\right)}^2+{\left(2\sqrt{2m}\right)}^2-8\right]+18 \\ ={\left(2\sqrt{2m}-\sqrt{\frac{2}{m}}\right)}^2+18 \\ Thus,\mathit{m}\mathit{i}\mathit{n}\mathit{i}\mathit{m}\mathit{u}\mathit{m}\mathbf{}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{S}\mathbf{=}\mathbf{18},when\left(2\sqrt{2m}-\sqrt{\frac{2}{m}}\right)=0 \\ \Rightarrow 2\sqrt{2m}=\sqrt{\frac{2}{m}} \\ \Rightarrow 8m=\frac{2}{m} \\ \Rightarrow m=\pm \frac{1}{2} \end{gathered}$$