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. $[3]$

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Solution

Given that,
Lets take the position of the cat A(-6, -4)
Position of the milk B(5,11)
We know that the shortest path between two points is equation of line passing through those two points
$[1]$
Slope of Points A and B (m) = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{11 - (- 4)}{5 - (- 6)} = \frac{15}{11} (y - y_{1}) = m (x - x_{1}) \Rightarrow y - (- 4) = \frac{15}{11} (x - (- 6)) \Rightarrow (y + 4) \times 11 = 15 \times (x - (- 6)) \Rightarrow 11 y + 44 = 15 x + 90 \Rightarrow 15 x - 11 y + 46 = 0$ $[2]$

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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. [3]

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. $[3]$