STRAIGHT LINES

Asked by devendra.mittal | 5th Jan, 2010, 09:57: PM

Expert Answer:

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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