PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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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 ? - PUC Karnataka Science Class 12 - Mathematics

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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

\[\text { The equation of line passing through }\left( 1, 4 \right) \text { with slope m is given by } \]

\[y - 4 = m\left( x - 1 \right) ................. \left( 1 \right)\]

\[\text { Substituting y = 0, we get }\]

\[0 - 4 = m\left( x - 1 \right)\]

\[ \Rightarrow \frac{- 4}{m} = x - 1\]

\[ \Rightarrow x = \frac{m - 4}{m}\]

\[\text{ Substituting x = 0, we get } \]

\[y - 4 = m\left( 0 - 1 \right)\]

\[ \Rightarrow y = - m + 4\]

\[ \Rightarrow x = - \left( m - 4 \right)\]

\[\text { So, the intercepts on coordinate axes are } \frac{m - 4}{m} \text { and }- \left( m - 4 \right) . \]

\[\text { Let S be the sum of the intercepts . Then }, \]

\[S = \frac{m - 4}{m} - \left( m - 4 \right)\]

\[ \Rightarrow \frac{dS}{dm} = \frac{4}{m^2} - 1\]

\[\text { For maximum or minimum values of S, we must have }\]

\[ \frac{dS}{dm} = 0\]

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

\[ \Rightarrow \frac{4}{m^2} = 1\]

\[ \Rightarrow m^2 = 4\]

\[ \Rightarrow m = \pm 2\]

\[\text {Now}, \]

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

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

\[\text { So, the sum is minimum at m = 2} . \]

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

\[S\text { o, the sum is maximum at m = - 2 } . \]

\[\text { Thus, the minimum value is given by}\]

\[S = \frac{- 2 - 4}{- 2} - \left( - 2 - 4 \right) = 3 + 6 = 9\]

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Solution 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 ? Concept: Graph of Maxima and Minima.
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