For the point of intersection of two straight lines $\(y=a x+b \& y=c x+d\)$, we need to solve them for $\(x\)$ and $\(y\)$.

Subtracting the equations we get

$$
\(\begin{aligned}
& y=a x+b \\
& \underline{y}=c x+d \\
& 0=(a-c) x+(b-d) \Rightarrow x=\frac{(b-d)}{(c-a)}
\end{aligned}\)
$$


Now, substituting the value of $x$ in any one of the two equations we get the value of $y$ as

$$
\(y=a \times \frac{(b-d)}{(c-a)}+b=\frac{a b-a d+b c-a b}{c-a}=\frac{b c-a d}{c-a}\)
$$


Hence the answer is (C).