Find all of the points of the form $(x, −1)$ which are $4$ units from the point $(3, 2)$

Question

Find all points of the form $(x, −1)$, which are $4$ units from the point $(3, 2)$.

I understand the distance formula, I think. Also, I don't know how to format for math on here yet so I apologize for that.

\begin{align} &\text{dist} = 4 = \sqrt{(x-3)^2 + (2-(-1))^2)} \\ &\implies 16 = (x-3)^2 + (2 + 1)^2 \tag{square both sides} \\ &\implies 16 = x^2 -6x + 18 \tag{expand} \\ &\implies 0 = x^2 -6x +2 \tag{zero on the left} \end{align}

Next should be finding the factors which would give me the answers but I have no idea what would work for this.

$$(x + \text{something})(x - \text{something}).$$

I must be doing something wrong or missing something. Please correct me.

Answer 1

Your computations are correct up to the factoring $$ (x+\text{something})(x-\text{somethign}).$$ In general, there is no reason to expect that your polynomial will factor like a difference of squares. In fact, the presence of a linear term (the $-6x$) indicates that this won't work. Instead, you need to either factor the polynomial somehow, or solve the quadratic using some other technique. In this case, completing the square seems reasonable.

Every quadratic polynomial can be written in the form $$ A(x-h)^2 + k, $$ where $A$, $h$, and $k$ are real numbers. Hence we have $$x^2 - 6x + 2 =A(x-h)^2 + k = Ax^2 - 2Ahx + Ah^2 + k. \tag{1}$$ Since two polynomials are equal to each other if and only if they have the same coefficients, we can equate the coefficients on the left with the coefficients on the right in order to obtain $$ \begin{cases} 1 = A, \\ -6 = -2Ah, \\ 2 = Ah^2 + k. \end{cases} $$ Solving this system, we immediately have that $A = 1$. Substituting this into the second equation, we get $-6 = -2h$, which implies that $h = 3$. Finally, plugging these into the last equation, we get $2 = 3^2 + k$, which implies that $k = -7$. Putting these back into (1), we get $$ x^2 - 6x + 2 = (x-3)^2 - 7.$$ Since we are looking for the roots of this polynomial, we set it equal to zero, then solve by extracting roots. This gives us \begin{align} 0 = (x-3)^2 - 7 &\implies (x-3)^2 = 7 \\ &\implies x-3 = \pm \sqrt{7} \tag{extract the roots} \\ &\implies x = 3\pm \sqrt{7}. \end{align} Hence the two points on the line $x=-1$ that are a distance of 4 from the point $(3,2)$ are the points $$\left( 3+\sqrt{7}, -1 \right) \qquad\text{and}\qquad \left( 3-\sqrt{7}, -1 \right).$$

Answer 2

$$x^2-6x+9-7=0$$ or $$(x-3)^2-(\sqrt7)^2=0$$ or $$(x-3-\sqrt7)(x-3+\sqrt7)=0$$ and it's just $x=3+\sqrt7$ or $x=3-\sqrt7$ and you get two points: $$(3+\sqrt7,-1)$$ and $$(3-\sqrt7,-1)$$

Answer 3

Suppose you have two points in a 2-D coordinate system such as $(x_{1}, y_{1}) = (3, 2)$ and $(x_{2}, y_{2}) = (x, -1)$ and distance between them is 4 units. Hence you can write as

\begin{align} & d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} \\ \implies & 4 = \sqrt{(3-x)^{2} + (2-(-1))^{2}} \\ \implies & 16 = 9 + x^{2} - 6x +9 \tag{squaring both sides}\\ \implies & x^{2} - 6x + 2 = 0 \end{align}

This is a quadratic equation. The solution to a quadratic equation of the form $ax^{2} + bx + c$ can be found as $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$. Hence comparing the standard form and the equation above, we have $a = 1, b = -6$ and $c = 2$. Hence the roots will be

$$x = \dfrac{6 \pm \sqrt{36 - 8}}{2} = \dfrac{6 \pm 2\sqrt{7}}{2} = 3 \pm \sqrt{7}$$

Answer 4

You can work this out geometrically. Plot the point $(3,2)$ and the horizontal line $y=-1$. You want to find the two points on the the line which are $4$ units away from $(3,2)$. Just mark about where they are. Then there are two right triangles with vertices $(3,2)$, $(3,-1)$ and either of the points you marked. The hypotenuse of each triangle is $4$. The vertical leg is $3$ so the length of the other leg is $\sqrt{4^2-3^2} = \sqrt{7}.$ If you move $\sqrt{7}$ units left and right of $(3,-1)$, you get your two points $(3 \pm \sqrt{7},-1).$