SOLUTION: the straight line y= 2x + 3 meets the curve xy+20=5y at the points A and B.Find the equation of the perpendicular bisector of AB. i cant just cant seem to work this out

Question 96947: the straight line y= 2x + 3 meets the curve xy+20=5y at the points A and B.Find the equation of the perpendicular bisector of AB. i cant just cant seem to work this out
Answer by mathslover(157) (Show Source): You can put this solution on YOUR website!
First we need to find the intersection points of the curve xy + 20 =5y and the straight line y= 2x + 3 to get the points A and B
Substituting y= 2x+3 in the equation of the curve we have,
x(2x + 3) + 20 = 5(2x + 3)
2x^2 + 3x + 20 = 10x +15
2x^2 -7x + 5 =0
Using the quadratic formula to get the values of x

x = 7 +- (sqrt(49 -40))/(4)
x=5/2 , 1
putting these values of x in the equation y = 2x + 3
y = 8 and y = 5
Therefore the points of intersection of the straight line and the curve is
(5/2,8) and (1,5)
The perpendicular bisector of AB divides the line segment AB in equal halves and therefore if D is the point of bisection the co-ordinates of D are
((5/2 +1)/2 , (8 + 5)/2 ) Using the mid point formula
((x1 + x2)/2, (y1+ y2)/2 )
= (7/4, 13/2)
we are left with finding the equation of the line through (7/4, 13/2) and perpendicular to the line y =2x +3
slope of the line y=2x +3 is 2 ( y=mx +c where m is the slope )
if p is the slope of the perpendicular than p* 2 = -1 (since product of the slope of a line and its perpendicular is -1)
therefore p =-1/2
so if we represent y=mx + c as the equation of the perpendicular
we have
13/2= -1/2 * 7/4 + c ( Substituting values of (x,y) and m )
C= 13/2 + 7/8
c = 59/8
Hence equation of the line is y = -1/2x + 59/8
Multiplying by 8 on both sides
8y = -4x + 59

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