+0  
 
0
23
1
avatar

The equation of a straight line AC is 2y=-3x+6. Given the point A(-2,6) and AB:BC = 1:5. Find the equation of straight line that passes through point B and perpendicular to line AC.

 Feb 15, 2021
 #1
avatar+116126 
+1

Unless  we know  some  specific  value  for C, there  could  be  many answers  to  this

 

Here' s one thing  that would work

 

Construct  two   circles  with  radii  of 1   and  6   that  are centered at  ( -2,6)

 

This will guarantee  that  if  B lies  the  first circle and the  line, and  C lies  on the  second  circle  and the line, then  AB : BC  =1:5

 

Rearrangng the equation of the line we  haxe  that   y = (-3/2)x  + 3     (1)

 

The  equation of the  first circle  is ( x +2)^2  + ( y - 6)    = 1     (2)

 

Putting (1)  into (2)  and solving   for   either  x    we have  that

 

(x + 2)^2  +  (-3x/2  + 3  - 6)^2   = 1

 

x  = 2/sqrt (13)   - 2  =       [ 2 - 2sqrt (13)] /sqrt (13)

 

And  y  =  (-3/2) [ 2/sqrt (13)  - 2  ]  +  3  =  6 - 3/sqrt (13) =   [6 sqrt (13) - 3 ]  /sqrt (13)

 

A perpendicular  line at B will have a slope of  2/3

 

And the equation of this line  is

 

y =  (2/3) ( x -[ 2/sqrt/13 -2]) + (6 - 3/sqrt(13) )

 

See  here :  https://www.desmos.com/calculator/zu8eebvbso

 

 

cool cool cool

 Feb 15, 2021

34 Online Users

avatar
avatar