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Q. A point moves such that the area of the triangle formed by it with the points (1, 5) and $ (3,-7) $ is 21 sq unit. Then locus of the point is:

KEAMKEAM 2002

Solution:

Let $ (x,\text{ }y) $ be the required point $ \therefore $ $ \frac{1}{2}\left| \begin{matrix} x & y & 1 \\ 1 & 5 & 1 \\ 3 & -7 & 1 \\ \end{matrix} \right|=21 $ $ \Rightarrow $ $ x(5+7)-y(1-3)+1(-7-15)=42 $ $ \Rightarrow $ $ 12x+2y+22=42 $ $ \Rightarrow $ $ 12x+2y-64=0 $ $ \Rightarrow $ $ 6x+y-32=0 $