Question 35060: find the equation of a straight line perpendicular to a line (5;-2) and (0;5) and through the point (3;-2)
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! find the equation of a straight line perpendicular to a line (5;-2) and (0;5) and through the point (3;-2)
SLOPE OF LINE JOINING (X1,Y1)=(5,-2) AND (X2,Y2)=(0,5) IS M=(Y2-Y1)/(X2-X1)=(5+2)/(0-5)=-7/5...HENCE SLOPE OF LINE PERPENDICULAR TO THIS IS M'=5/7..IT IS PASSING THROUGH (X3,Y3)=(3,-2)...IS
Y-Y3=M'(X-X3)
Y+2=(5/7)(X-3)
7Y+14=5X-15
7Y=5X-29...OR....Y=5X/7 - 29/7 IN SLOPE INTERCEPT FORM
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