Question 1177698: Show that the equation of the perpendicular bisector of the points (t,t+1) and (3t,t+3) is y+tx = 2tsquared + t + 2. If this perpendicular bisector passes through the point (5,2),calculate the values of t.
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! 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
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