7. Write the equation of the line passing through the points:
1) \(A(3 ; 0)\) and \(B(3 ; 5)\);
3) \(E(0 ; 4)\) and \(K(2 ; 0)\)
2) \(C(2 ; 2)\) and \(D(-1 ;-1)\);
4) \(M(3 ; 2)\) and \(P(6 ; 3)\).
7. 1) Equation of the line: \(x=3\). 2) \(y=x\). 3) S o l u t i o n. Equation of the line: \(y=k x+b\). We need to find \(k\) and \(b\). When \(x=0\) and \(y=4\), we get: \(4=0 \cdot x+b\), hence, \(b=4\). When \(x=2\) and \(y=0\), we have: \(0=k \cdot 2+4\), hence, \(k=-2\). Equation of the line: \(y=-2 x+4\).

4) S o l u t i o n. Let the equation of the desired line be: \(y=k x+b\). We need to find \(k\) and \(b\). Since the line passes through the point \(M(3 ; 2)\), then \(2=k \cdot 3+b\), and since, in addition, this line passes through the point \(P(6 ; 3)\), then \(3=k \cdot 6+b\). Therefore, solving the system of equations

\[
\left\{\begin{array}{l}
3 k+b=2 \\
6 k+b=3
\end{array}\right.
\]

we find \(k=\frac{1}{3}\) and \(b=1\). Equation of the line: \(y=\frac{1}{3} x+1\).