SOLUTION: Find the equation of the line through the point (-2,5) and the perpendendular to the line 3x+y=4. How do I do this problem?

Question 31108: Find the equation of the line through the point (-2,5) and the perpendendular to the line 3x+y=4. How do I do this problem?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING AND TRY

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Graphs/30627: (1)Find the Equation of a line through (0,10) which is perpendicular to the line
-x-15y=10.
(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.
1 solutions
Answer 17363 by venugopalramana(1167) About Me on 2006-03-19 10:20:13 (Show Source):
(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
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