Question 1177539


Show that the equation of the perpendicular bisector of the points (t,t+1) and (3t,t+3) is y+tx= 2rsquared + t + t . If this perpendicular bisector passes through the point (5,2), Calculate the values of t.

(t,t+1) and (3t,t+3) 

slope of line = [(t+3)-(t+1)]/(3t-t)=1/t
slope of perpendicular bisector = -t ( negative reciprocal)

mid point =x~ (t+3t)/2 =4t/2 = 2t
y~ (t+1+t+3)/2 = (2t+4)/2 = (t+2)

Point  2t,(t+2) and slope = -t  

y=mx+c

t+2=-t(2t)+c

t+2 =-2t^2 +c

c= 2t^2 +t+2

y= -t(x) +2t^2+t+2

y+tx = 2t^2+t+2
(5,2) the point on perpendicular bisector

plug 5 &2

2+5t = 2t^2+t+2

2t^2-4t=0
2t(t-2)=0
t =0 OR t=2