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There is exactly one value of $x$ for which the distance from $(5,6)$ to $(3x-1,ax+5)$ is $4$. If $a \neq 0,$ what is $a$?

Aug 26, 2023

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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 $$\boxed{\frac{1}{8}}$$.

Aug 26, 2023