Find an equation of the line $L$ that passes through the point $(-6,4)$ and satisfies the condition. (Let $x$ be the independent variable and $y$ be the dependent variable.) $L$ passes through the point $\left(2, \frac{7}{2}\right)$.

Step 1 ：First, we can find the slope of the line through these two points. The slope will be equal to: \(\frac{y_2-y_1}{x_2-x_1}=\frac{\frac{7}{2}-4}{2-(-6)}=\frac{\frac{1}{2}}{8}=\frac{1}{16}\)

Step 2 ：Therefore, the line will have the form \(y=\frac{1}{16}x+b\). To solve for \(b\), substitute in one of the given points for \(x\) and \(y\) as shown: \begin{align*} 4&=\frac{1}{16}(-6)+b\\ \Rightarrow\qquad 4&=-\frac{3}{8}+b\\ \Rightarrow\qquad \frac{35}{8}&=b \end{align*}

Step 3 ：Since \(b\) in a line of the form \(y=mx+b\) is the \(y\)-intercept, and this is what we are trying to find, the answer is \(\boxed{\frac{35}{8}}\).

Step 4 ：Finally, the equation of the line \(L\) that passes through the point \((-6,4)\) and \(\left(2, \frac{7}{2}\right)\) is \(\boxed{y=\frac{1}{16}x+\frac{35}{8}}\).