### Suppose that $a, b, c$ are distinct numbers such that $( c - b )^2 - 4( c - a )( a - b ) = 0,$ find the value of $\dfrac{ c - a }{ a - b }.$

$1$
1. Given, $(c - b)^2 - 4(c - a)(a - b) = 0,$ we need to find the value of $\dfrac{ c-a } { a-b }.$
2. \begin{align} & (c - b)^2 -4(c - a)(a - b) = 0 \\ \implies & (c - a + a - b)^2 -4(c - a)(a - b) = 0 \\ \implies & \Big( (c - a) + (a - b) \Big)^2 -4(c - a)(a - b) = 0 \\ \implies & (c - a)^2 + (a - b)^2 + 2(c - a)(a - b) -4(c - a)(a - b) = 0 \\ \implies & (c - a)^2 + (a - b)^2 - 2(c - a)(a - b) = 0 \\ \implies & \Big( (c - a) - (a - b) \Big)^2 = 0 \\ \implies & (c - a) - (a - b) = 0 \\ \implies & (c - a) = (a - b) \\ \implies & \dfrac{ c - a } { a - b } = 1 \end{align}