document.write( "Question 30627: (1)Find the Equation of a line through (0,10) which is perpendicular to the line
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document.write( "-x-15y=10.\r
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document.write( "(2)The line segments with endpoints (0,5) and (5,0) is the diameter of a circle. write the eguation of this circle in standard form. \n" );
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Algebra.Com's Answer #17363 by venugopalramana(3286)![]() ![]() You can put this solution on YOUR website! (1)Find the Equation of a line through (0,10) which is perpendicular to the line \n" ); document.write( "-x-15y=10. \n" ); document.write( "EQN. OF A LINE PERPENDICULAR TO THIS \n" ); document.write( "15X-Y=K.......WHERE K IS A CONSTANT TO BE FOUND..THIS GOES THROUGH (0,10)..SO \n" ); document.write( "15*0-10=K \n" ); document.write( "K=-10 \n" ); document.write( "SO EQN. OF REQD.LINE IS \n" ); document.write( "15X-Y=-10 \n" ); document.write( "15X-Y+10=0 \n" ); document.write( "(2)The line segments with endpoints (0,5) and (5,0) is the diameter of a circle. write the eguation of this circle in standard form. \n" ); document.write( "TAKE P(X,Y) ANY POINT ON CIRCLE.THE 2LINES JOINING P TO ENDS OF DIAMETER ARE AT RIGHT ANGLES.HENCE PRODUCT OF THEIR SLOPES =-1....SO..... \n" ); document.write( "EQN OF CIRCLE IS GIVEN BY \n" ); document.write( "{(Y-5)/(X-0)}*{(Y-0)/X-5)}=-1 \n" ); document.write( "(Y-5)*Y/(X*(X-5))=-1 \n" ); document.write( "Y^2-5Y=-X^2+5X \n" ); document.write( "X^2+Y^2-5X-5Y=0\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |