Question 30627
(1)Find the Equation of a line through (0,10) which is perpendicular to the line 
-x-15y=10. 
EQN. OF A LINE PERPENDICULAR TO THIS 
15X-Y=K.......WHERE K IS A CONSTANT TO BE FOUND..THIS GOES THROUGH (0,10)..SO
15*0-10=K
K=-10
SO EQN. OF REQD.LINE IS 
15X-Y=-10
15X-Y+10=0
(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.
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.....
EQN OF CIRCLE IS GIVEN BY 
{(Y-5)/(X-0)}*{(Y-0)/X-5)}=-1
(Y-5)*Y/(X*(X-5))=-1
Y^2-5Y=-X^2+5X
X^2+Y^2-5X-5Y=0