Let points A,B and C lie on lines y-x=0, 2x-y=0 and y-3x=0, respectively. Also, AB passes through fixed point P(1,0) and BC passes through fixed point Q(0,-1). Then prove that AC also passes through a fixed point and find that point.

Explanation

Let the coordiantes of points A,B and C be `(alpha, alpha), (beta,2beta) " and " (gamma, 3gamma)`, respectively.
Points A,B, P are collinear.
`therefore |{:(1,0,1),(alpha,alpha, 1),(beta, 2beta,1):}| = 0`
`rArr alpha-2beta+alphabeta=0 " " (1)`
Also, points B,C,Q are collinear.
`therefore |{:(0, -1,1),(beta,2beta,1),(gamma, 3gamma,1):}| = 0`
`rArr beta-gamma +beta gamma = 0`
`rArr beta = (gamma)/(1+gamma) " " (2)`
Putting value of `beta` in equation (1), we get `alpha +2alpha gamma = 2gamma.`
Let AC pass through fixed point R(h, k).
Since C, A and R are collinear,
`|{:(alpha,alpha,1),(gamma,3 gamma, 1),(h, k,1):}| = 0`
`rArr h(alpha-3gamma) - k(alpha-gamma) +2alphagamma = 0`
`rArr h(alpha-3gamma) - k(alpha-gamma) +2gamma-alpha = 0`
`rArr alpha(h-k-1) + gamma(-3h+k+2)=0 " for all "alpha,gamma`
`therefore h-k-1=0 " and "-3h+k+2 =0`
`therefore h =(1)/(2), k =-(1)/(2)`
`"Thus, AC passes through the point " ((1)/(2), -(1)/(2)).`