STRAIGHT LINES
Asked by devendra.mittal
| 5th Jan, 2010,
09:57: PM
Circum centre is the point of intersection of the perpendicular bisectors of the triangle so coordinates of the circumcentre is the point of intersection of lines
x+y=3 and x-y=1 which is (2,1) lets call it C
orthocentre is the point of intersection of altitudes of the triangle
Altitude from A to BC will be a line through the point (0,0) and with slope = slope of the perpendicular bisector =1/2
Equation of Altitude through B is
(x - 0) = 1/2(y-0)
2x = y......(i)
Also CA=CB and slope of AB will be negative reciprocal of slope of x+y=3
so slope of AB =1
let (h,k) be coordinates of the vertex B so slope of AB = k/h=1 or h=k
using CA2=CB 2
we get 5 =(h-2)2+(k-1)2 =(h-2)2+(h-1)2 =2h2-6h+5 which gives h =3 (neglect 0)
so coordinates of B are (3,3)
So equation of alitude from B to AC will be x-3=-1(y-3)
x+y-6=0....(ii)
Sove (i) and (ii) to get coordinates of orthocentre as (4,2)
Answered by
| 8th Jan, 2010,
01:07: PM
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