help quadratic

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in the Cartesian plane is given by the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$

In this case, we are given the points $(x_1, y_1) = (5, 6)$ and $(x_2, y_2) = (3x - 1, ax + 5)$, and we know that the distance $d$ is $4$:

$$4 = \sqrt{(3x - 1 - 5)^2 + (ax + 5 - 6)^2}.$$

Simplify inside the square root:

$$16 = (3x - 6)^2 + (ax - 1)^2.$$

Expand:

$$16 = 9x^2 - 36x + 36 + a^2x^2 - 2ax + 1.$$

Combine like terms:

$$16 = (9 + a^2)x^2 - (36 + 2a)x + 37.$$

Now, we want to find the value of $a$ for which there is exactly one value of $x$ that satisfies this equation. For there to be exactly one solution, the quadratic equation must have a discriminant of $0$:

$$(36 + 2a)^2 - 4(9 + a^2)(37) = 0.$$

Simplify and solve for $a$:

$$1296 + 288a + 4a^2 - 1332 - 4a^2 = 0.$$

Simplify:

$$288a - 36 = 0.$$

Solve for $a$:

$$a = \frac{36}{288} = \frac{1}{8}.$$

Therefore, the value of $a$ is $\boxed{\frac{1}{8}}$.