document.write( "Question 30627: (1)Find the Equation of a line through (0,10) which is perpendicular to the line
\n" ); document.write( "-x-15y=10.\r
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\n" ); document.write( "\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.
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Algebra.Com's Answer #17363 by venugopalramana(3286)\"\" \"About 
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(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
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