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Point \mathrm{A}(2,1), \mathrm{B}(3,-7), \mathrm{C} is any point on the line 3 \mathrm{x}-2 \mathrm{y}=1, then locus of point \mathrm{D} such that \mathrm{ABCD} is a parallelogram

Option: 1

\mathrm{3 x-2 y=20}


Option: 2

\mathrm{3 x-y=20}


Option: 3

\mathrm{2 x+3 y=20}


Option: 4

\mathrm{3 x-2 y+18=0}


Answers (1)

best_answer

 Point \mathrm{A}(2,1), \mathrm{B}(3,-7), \mathrm{C} \ldots \ldots \ldots \ldots

Let the coordinates of \mathrm{C} be \left(\alpha, \frac{3 \alpha-1}{2}\right) and that of \mathrm{D} be \mathrm{(\mathrm{h}, \mathrm{k})},then

\mathrm{2+\alpha=3+\mathrm{h}\: and\: \frac{3 \alpha-1}{2}+1=\mathrm{k}-7}
\mathrm{\therefore 3+3 \mathrm{~h}=2 \mathrm{k}-15}

\mathrm{\therefore\: locus \: is \: 3 \mathrm{x}-2 \mathrm{y}+18=0}

Posted by

Devendra Khairwa

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