Triangle Vertices: Finding Possible Values Of X

When dealing with geometric problems, especially those involving triangles and their properties, a clear understanding of coordinate geometry is essential. In this article, we'll dive into a specific problem: finding the possible values of x given the coordinates of three vertices, F(5, 1), G(x, 7), and H(8, 2), and a crucial relationship between two sides – the length of side FG is twice the length of side FH. This type of problem requires us to leverage the distance formula and algebraic manipulation to arrive at the solution. We'll break down each step, ensuring that by the end, you'll have a solid grasp of how to approach similar challenges in coordinate geometry. The journey begins with recalling the distance formula, a cornerstone of calculating lengths between two points in a Cartesian plane. Remember, the distance d between two points (x1, y1) and (x2, y2) is given by $d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$. This formula will be our primary tool as we set up equations based on the given information about the triangle's sides. The problem statement provides us with the coordinates of three vertices: F at (5, 1), G at (x, 7), and H at (8, 2). We are also told that the length of the side FG is exactly twice the length of the side FH. This relationship is key to forming an equation that will allow us to solve for the unknown coordinate x. So, let's start by calculating the square of the lengths of FG and FH. This often simplifies calculations by avoiding the square root until the very end. The square of the distance between F(5, 1) and G(x, 7), denoted as $FG^2$, is $(x - 5)^2 + (7 - 1)^2$. Simplifying this, we get $(x - 5)^2 + 6^2 = (x - 5)^2 + 36$. Next, let's find the square of the distance between F(5, 1) and H(8, 2), denoted as $FH^2$. Using the distance formula again, we have $(8 - 5)^2 + (2 - 1)^2$. This simplifies to $3^2 + 1^2 = 9 + 1 = 10$. Now we have the expressions for the squared lengths of our two sides. The problem states that FG is twice the length of FH, which means $FG = 2 imes FH$. Squaring both sides of this equation gives us $FG^2 = (2 imes FH)^2$, which simplifies to $FG^2 = 4 imes FH^2$. This is the core equation we will use to solve for x. Substituting our calculated expressions for $FG^2$ and $FH^2$ into this equation, we get: $(x - 5)^2 + 36 = 4 imes 10$. The equation now becomes $(x - 5)^2 + 36 = 40$. Our next step is to isolate the term containing x. Subtracting 36 from both sides of the equation, we get $(x - 5)^2 = 40 - 36$, which simplifies to $(x - 5)^2 = 4$. Now, to solve for (x - 5), we need to take the square root of both sides. Remember that when taking the square root of a number, there are always two possible results: a positive and a negative value. Therefore, $x - 5 = \sqrt{4}$ or $x - 5 = -\sqrt{4}$. This means $x - 5 = 2$ or $x - 5 = -2$. We now have two separate linear equations to solve for x. In the first case, $x - 5 = 2$. Adding 5 to both sides gives us $x = 2 + 5$, so $x = 7$. In the second case, $x - 5 = -2$. Adding 5 to both sides gives us $x = -2 + 5$, so $x = 3$. Thus, the possible values of x are 7 and 3. This thorough breakdown illustrates how the distance formula and careful algebraic manipulation can solve problems involving geometric relationships between points. It's a fundamental skill that opens the door to more complex geometry problems.

Understanding the Distance Formula and Its Application

The distance formula is a fundamental concept in coordinate geometry, derived directly from the Pythagorean theorem. It allows us to calculate the straight-line distance between two points in a two-dimensional plane. Given two points, $(x_1, y_1)$ and $(x_2, y_2)$, the distance $d$ between them is expressed as: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. This formula essentially treats the distance between the two points as the hypotenuse of a right-angled triangle, where the legs are the horizontal and vertical differences between the coordinates. The horizontal leg has a length of $|x_2 - x_1|$ and the vertical leg has a length of $|y_2 - y_1|$. Applying the Pythagorean theorem ($a^2 + b^2 = c^2$), we get $(x_2 - x_1)^2 + (y_2 - y_1)^2 = d^2$, which leads directly to the distance formula by taking the square root. In our specific problem, we used this formula to find the lengths of sides FG and FH. For FG, with F(5, 1) and G(x, 7), the distance is $FG = \sqrt{(x - 5)^2 + (7 - 1)^2} = \sqrt{(x - 5)^2 + 6^2} = \sqrt{(x - 5)^2 + 36}$. For FH, with F(5, 1) and H(8, 2), the distance is $FH = \sqrt{(8 - 5)^2 + (2 - 1)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$. The problem states that the length of side FG is twice the length of side FH, meaning $FG = 2 imes FH$. Substituting the expressions we found: $\sqrt{(x - 5)^2 + 36} = 2 \times \sqrt{10}$. To solve for x, it's often easier to work with the squares of the distances. Squaring both sides of the equation eliminates the square roots: $(\sqrt{(x - 5)^2 + 36})^2 = (2 \times \sqrt{10})^2$. This simplifies to $(x - 5)^2 + 36 = 4 imes 10$, which further simplifies to $(x - 5)^2 + 36 = 40$. This is the algebraic equation we need to solve. It's important to recognize that when we encounter an equation of the form $(x - a)^2 = b$, where $b > 0$, there will be two possible solutions for $(x - a)$, namely $\sqrt{b}$ and $-\sqrt{b}$. This is because both $(\sqrt{b})^2$ and $(-\sqrt{b})^2$ equal $b$. In our case, $(x - 5)^2 = 4$. Thus, $x - 5$ can be either $\sqrt{4}$ or $-\sqrt{4}$. This leads to $x - 5 = 2$ or $x - 5 = -2$. Solving these two linear equations for x gives us $x = 7$ and $x = 3$. The application of the distance formula, combined with understanding the properties of squares and square roots, is crucial for accurately solving such geometric problems. It highlights the power of translating geometric relationships into algebraic equations.

Algebraic Steps to Find the Value of x

Once we have established the core equation based on the given geometric relationship, the process of finding the possible values of x becomes a matter of careful algebraic manipulation. We derived the equation $(x - 5)^2 + 36 = 40$. The primary goal here is to isolate the variable x. The term $(x - 5)^2$ is the part directly involving x, so we'll work to get that term by itself on one side of the equation. The first step is to eliminate the constant term (+36) from the left side. We achieve this by subtracting 36 from both sides of the equation. So, we have: $(x - 5)^2 + 36 - 36 = 40 - 36$. This simplifies to $(x - 5)^2 = 4$. Now, the term $(x - 5)^2$ is isolated. To solve for $(x - 5)$, we need to undo the squaring operation. The inverse operation of squaring is taking the square root. When we take the square root of both sides of an equation like this, it's imperative to remember that a positive number has two square roots: a positive one and a negative one. For example, the square root of 9 is both 3 and -3, because $3^2 = 9$ and $(-3)^2 = 9$. Applying this principle to our equation, $(x - 5)^2 = 4$, we get: $x - 5 = \sqrt{4}$ or $x - 5 = -\sqrt{4}$. Evaluating the square root of 4, we know it is 2. So, our two possibilities are: $x - 5 = 2$ or $x - 5 = -2$. These are now two simple linear equations. For the first equation, $x - 5 = 2$, we isolate x by adding 5 to both sides: $x = 2 + 5$, which gives us $x = 7$. For the second equation, $x - 5 = -2$, we again add 5 to both sides: $x = -2 + 5$, which gives us $x = 3$. Therefore, the two possible values for x are 7 and 3. These are the values of x for the coordinate of point G such that the length of side FG is exactly twice the length of side FH. This step-by-step algebraic process ensures that all potential solutions are found, reflecting the dual nature of square roots. It's a critical part of solving equations that arise from geometric contexts.

Geometric Interpretation and Verification

It's always a good practice to mentally (or even visually, if possible) verify the solutions obtained. We found two possible values for x: 3 and 7. Let's see what these values mean for the position of point G. Remember, G has coordinates (x, 7).

Case 1: x = 3

If x = 3, then point G is at (3, 7). Let's recalculate the lengths of FG and FH.

• Length of FG: Using F(5, 1) and G(3, 7). $FG = \sqrt{(3 - 5)^2 + (7 - 1)^2} = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40}$.

• Length of FH: Using F(5, 1) and H(8, 2). $FH = \sqrt{(8 - 5)^2 + (2 - 1)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.

Now, let's check the condition: Is FG = 2 * FH? Is $\sqrt{40} = 2 \times \sqrt{10}$? We know that $\sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2 \times \sqrt{10}$. So, $2 \times \sqrt{10} = 2 \times \sqrt{10}$. This condition holds true. Thus, x = 3 is a valid solution.

Case 2: x = 7

If x = 7, then point G is at (7, 7). Let's recalculate the lengths of FG and FH.

• Length of FG: Using F(5, 1) and G(7, 7). $FG = \sqrt{(7 - 5)^2 + (7 - 1)^2} = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40}$.

• Length of FH: Using F(5, 1) and H(8, 2). $FH = \sqrt{(8 - 5)^2 + (2 - 1)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.

Again, let's check the condition: Is FG = 2 * FH? Is $\sqrt{40} = 2 \times \sqrt{10}$? As we saw in the previous case, $\sqrt{40}$ is indeed equal to $2 \times \sqrt{10}$. So, $2 \times \sqrt{10} = 2 \times \sqrt{10}$. This condition also holds true. Thus, x = 7 is also a valid solution.

The geometric interpretation shows that there are two possible locations for point G on the line y = 7 such that the distance from F to G is twice the distance from F to H. These locations are (3, 7) and (7, 7). This verification step reinforces the accuracy of our algebraic solution. It's satisfying to see how the coordinates and distances work together in a geometric system.

Conclusion: Mastering Coordinate Geometry Problems

Solving problems like finding the possible values of x for triangle vertices F(5, 1), G(x, 7), and H(8, 2), where FG is twice the length of FH, requires a systematic approach. We began by understanding the problem and identifying the key mathematical tools needed: the distance formula and basic algebraic manipulation. By applying the distance formula to segments FG and FH, we set up expressions for their lengths. The crucial condition, $FG = 2 imes FH$, was then translated into an equation involving the squared lengths to simplify calculations: $FG^2 = 4 imes FH^2$. Substituting the derived expressions led to the quadratic equation $(x - 5)^2 + 36 = 40$. Isolating the squared term, $(x - 5)^2 = 4$, and then taking the square root of both sides revealed the two possible values for $(x - 5)$, which are 2 and -2. Solving the resulting linear equations, $x - 5 = 2$ and $x - 5 = -2$, yielded the final possible values for x as 7 and 3. We concluded by verifying these solutions, confirming that both values of x satisfy the given condition. This problem serves as an excellent example of how coordinate geometry allows us to solve geometric puzzles using algebraic methods. Mastering these techniques is fundamental for tackling more complex challenges in mathematics and related fields.

For further exploration into coordinate geometry and related theorems, you might find the resources at MathWorld or Wikipedia's Coordinate Geometry page invaluable.