A cat is located at the point $(- 6, - 4)$ in xy plane. A bottle of milk is kept at $(5, 11)$ . The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take its milk.

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Solution

Equation of the line joining the point is
$\begin{matrix} \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{x - x_{1}}{x_{2} - x_{1}} \frac{y + 4}{11 + 4} = \frac{x + 6}{5 + 6} \frac{y + 4}{15} = \frac{x + 6}{11} 15 (x + 6) = 11 (y + 4) 15 x + 90 = 11 y + 44 15 x - 11 y + 90 - 44 = 0 15 x - 11 y + 46 = 0 \end{matrix}$
The equation of the path is $15 x - 11 y + 46 = 0$

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A cat is located at the point (−6,−4) in xy plane. A bottle of milk is kept at (5,11). The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take its milk.

Equation of the line joining the point is y−y1y2−y1=x−x1x2−x1y+411+4=x+65+6y+415=x+61115(x+6)=11(y+4)15x+90=11y+4415x−11y+90−44=015x−11y+46=0 The equation of the path is 15x−11y+46=0

A cat is located at the point $(- 6, - 4)$ in xy plane. A bottle of milk is kept at $(5, 11)$ . The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take its milk.