document.write( "Question 281405: Points P(1,2), Q(-1,-1), R(1,-3) are given: determine the coordinates or point S for PQRSto be a parallelogram \n" ); document.write( "

Algebra.Com's Answer #204416 by subudear(62) $\"\"$ $\"About$

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For a parallelogram PQRS you can consider either PQ and RS lines or PS and QR\r
\n" ); document.write( "\n" ); document.write( "Lets take PQ ans RS line. Let coordinate for S be (x,y)\r
\n" ); document.write( "\n" ); document.write( "Then slope for PQ and RS should be same.\r
\n" ); document.write( "\n" ); document.write( "Slope of PQ line = (-1 - 2)/(-1 - 1) = 3/2
\n" ); document.write( "Slope of RS line = (y - -3)/(x - 1) = (y+3)/(x - 1)\r
\n" ); document.write( "\n" ); document.write( "equating both of them
\n" ); document.write( "(y+3)/(x - 1) = 3/2
\n" ); document.write( "2(y+3) = 3(x-1)\r
\n" ); document.write( "\n" ); document.write( "2y -3x = -9\r
\n" ); document.write( "\n" ); document.write( "Now Lets take PS and QR line. Let coordinate for S be (x,y)\r
\n" ); document.write( "\n" ); document.write( "Then slope for PS and QR should be same.\r
\n" ); document.write( "\n" ); document.write( "Slope of PS line = (y - 2)/(x - 1)
\n" ); document.write( "Slope of QR line = (-3 - -1)/(1 - -1) = -1\r
\n" ); document.write( "\n" ); document.write( "equating both of them
\n" ); document.write( "(y-2)/(x - 1) = -1
\n" ); document.write( "(y-2) = -(x-1)\r
\n" ); document.write( "\n" ); document.write( "y +x = 3\r
\n" ); document.write( "\n" ); document.write( "solving the above 2 equations we get \r
\n" ); document.write( "\n" ); document.write( "x = 3, y = 0 \n" ); document.write( "

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