Coordinate Geometry

Question

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

Answer 1

We have two points (t, t+1) and (3t, t+3)

 

To find the perpendicular bisector we want the equation of the line perpendicular to

the line through these points that crosses this line at it's midpoint....

 

First:  find the slope of the original line

m = (y₂-y₁)/(x₂-x₁) = (t+3 - (t+1))/(3t-t)

 

= (t+3-t-1)/2t = 2/2t = 1/t    the slope of the original line is 1/t
The slope of the perpendicular bisector is  m_⊥ = -1/m = -1/(1/t) = -t

 

We have he slope for our new line.  Now we need a point through which it runs so we can use

the point slope form  y-y₁ = m(x - x₁) to find the equation of the new line

 

We know the new line bisects the original line segment so it passes through the point halfway

between  (t, t+1) and (3t, t+3) so we need to find the midpoint of these...

 

the midpoint of (x₁, y₁) , (x₂, y₂) is  ((x₁+x₂)/2 , (y₁+y₂)/2)  

 

The midpoint here is...  ((t+3t)/2 , (t+1)+(t+3))/2)

                                  = (4t/2 , (2t+4)/2)

                                  = (2t, t+2)

 

We now have the perpendicular slope m⊥ = -t   and a point on the line (2t, t+2)

 

so use the point slope formula gives:

 

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

y -t -2 = -tx + 2t²

Simply rearranging this appropriately we have

 

y + tx = 2t² + t + 2

 

**********************************

 

Now we want to find the value of t if this line passes through (5,2)

that means we substitute in 5 for x and 2 for y so...

 

2 + 5t = 2t² + t + 2

2t² -5t + t +2 - 2 = 0

2t² -4t = 0

2t(t-2) = 0

 

t = 0  or  t = 2