Circumcentre of a triangle, which is formed by the lines y=x, y=2x and y= 3x+4 is
Hence we have x = 3x + 4 ; -2x = 4 ; x = -2
Since y = x ; hence y = -2
Hence the points will intersect at B(-2,-2)
y = 2x and y = 3x + 4
2x=3x+4
x = 4
y = -8
(-4,-8)
Vertices of triangle are A(0,0), B(-2,-2), C(-4,-8)
Circumcentre P(x,y) should be equidistant from the vertices of the triangle
Let(x,y) be the co-ordinates of the circum centre.
AP=CP and AP=BP
x2+y2=(x+4)2+(y+8)2
⇒ 8x+16y+80=0
Or 4x+8y+40=0 .....(1)
AP=BP
x2+y2=(x+2)2+(y+2)2
⇒ 4x+4y+8=0 .....(2)
By solving (1) and (2), we get x=6 and y=-8
Coordinates of circumcentre P=(6,-8)