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

#### 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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