Question

Find all points having a \(y\) -coordinate of -6 whose distance from the point (1,2) is 17 (a) By using the Pythagorean Theorem. (b) By using the distance formula.

Short answer

The points are (16, -6) and (-14, -6).

Step-by-step solution

Set up the known values

We need to find all points \((x, -6)\) whose distance from the point (1, 2) is 17 units.

Apply the Pythagorean Theorem (a)

The Pythagorean Theorem states that for any right triangle with legs of lengths \(a\) and \(b\), and hypotenuse of length \(c\), we have: \[ a^2 + b^2 = c^2 \] Here, let \(a = x – 1\), \(b = -6 - 2 = -8\), and \(c = 17\). Substitute in and solve for \(x\): \[ (x – 1)^2 + (-8)^2 = 17^2 \] \[ (x – 1)^2 + 64 = 289 \] \[ (x – 1)^2 = 225 \] \[ x – 1 = \pm 15 \] So, \[ x = 16 \text{ or } -14 \] Therefore, the points are \((16, -6)\) and \((-14, -6)\).

Apply the distance formula (b)

The distance formula states: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Here, let \((x_1, y_1) = (1, 2)\) and \((x_2, y_2) = (x, -6)\). We know \frac{d = 17\}. Substitute in and solve for \((x, -6)\): \[ 17 = \sqrt{(x - 1)^2 + (-6 - 2)^2} \] \[ 17 = \sqrt{(x - 1)^2 + 64} \] Square both sides: \[ 289 = (x - 1)^2 + 64 \] \[ (x - 1)^2 = 225 \] \[ x - 1 = \pm 15 \] So, \[ x = 16 \text{ or } -14 \] Therefore, the points are \((16, -6)\) and \((-14, -6)\).

Key concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry. It deals with right triangles and reveals a relation involving the lengths of their sides. In any right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse, which can be expressed as: \[ a^2 + b^2 = c^2 \]

• Here, \(a\) and \(b\) represent the legs.
• \(c\) represents the hypotenuse.

For our problem, we need to find points \((x, -6)\) at a distance of 17 units from the point \((1, 2)\).Given this arrangement, we substitute \(a\), \(b\), and \(c\) as follows:

• \(a = x - 1\)
• \(b = -6 - 2 = -8\)
• \(c = 17\)

By substituting these values into the Pythagorean equation, we get:\[ (x - 1)^2 + (-8)^2 = 17^2 \]Simplify it step by step until we isolate and solve for \(x\).Finally, the solutions are: \(x = 16\) or \(x = -14\), giving us the points \((16, -6)\) and \((-14, -6)\).

Coordinate Geometry

Coordinate geometry, or analytic geometry, bridges algebra and geometry. It provides tools to determine the coordinates of points and find distances between points or equations of lines using a coordinate plane. In our context, we are looking for points with fixed \(y\)-coordinate (\(-6\)) that are a specified distance (17 units) from another point \((1, 2)\).We can visualize the problem:

• \( (x_1, y_1) = (1, 2)\)
• \(( x_2, y_2) = (x, -6)\).

By representing these points on a coordinate plane and embedding the distance formula into this setup, we solidify our understanding. This geometric approach helps easily solve for variables like \(x\).Understanding how changes in coordinates affect the distance between points is key to mastering coordinate geometry. For this problem, it means realizing that for the points we seek, the direct line connecting them will necessarily be the hypotenuse of our right triangle set up above.

Solving Equations

Solving equations is a universal mathematical method applicable across various topics including our current problem. In this exercise, we use it twice for validating our points.First, we solve the equation derived from the Pythagorean Theorem:

• \((x - 1)^2 + 64 = 289\)

We rearrange and solve step by step:

• Subtract 64 from both sides: \((x - 1)^2 = 225\)
• Take the square roots of both sides: \(x - 1 = \pm 15\)
• Solve each equation stemming from the \pm to get final values: \(x = 16\), \(x = -14\)

Similarly, by solving the distance formula expression derivatively, we reaffirm solutions to ensure accuracy.Overall, combining these concepts and systematically solving step-by-step elevates our grasp of handling such geometric problems with confidence and precision.