Question 64698
FIRST WE FIND THE SLOPE OF THE LINE THROUGH THE 2 POINTS THUS
(Y2-Y1)/(X2-X1)
(-4+5)/(-2-7)
1/-9 IS THE SLOPE OF THE LINE THROUGH ((7,-5)(-2,-4)
NOW TO IDENTIFY A LINE PERPENDICULAR TO THIS LINE IT HAS TO HAVE A SLOPE OF A NEGATIVE RECIPRICAL OF THIS SLOPE. THUS THE SL;OPE OF A LINE PERPENDICULAR TO THIS LINE THROUGH (3,-5) HAS A SLOPE OF 9. NOW WE SUBSTITUTE THESE X&Y POINTS IN THE FORMULA USING THE NEW SLOPE WE GET
-5=9*3+b
-5=27+b
b=-5-27
b=-32 THUS THUIS EQUATION IS
Y=9X-32