Problem 13 - Precalculus Essentials

Question

Find the distance between each pair of points. If necessary, express answers in simplified radical form and then round to two decimals places. $$(0,-\sqrt{3}) \text { and }(\sqrt{5}, 0)$$

Step-by-Step Solution

Answer

The distance between the points (0, -√3) and (√5, 0) is \(\sqrt{8}\) or 2.83 when rounded to two decimal places.

Step 1: Identify coordinates of the two points

The coordinates of the first point are (0, -√3) and the coordinates of the second point are (√5, 0).

Step 2: Substitute into the distance formula

Now, substitute the values into the distance formula which is: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). But here, \(x_1=0\), \(y_1=-\sqrt{3}\), \(x_2=\sqrt{5}\), and \(y_2=0\). After substituting these values into the distance formula it becomes: \(\sqrt{(\sqrt{5} - 0)^2 + (0 - (-\sqrt{3}))^2}\).

Step 3: Simplify the expression inside the square root

Next, simplify the expression inside the square root to get: \(\sqrt{(\sqrt{5})^2 + (\sqrt{3})^2}\). Which is then simplified further to: \(\sqrt{5 + 3}\).

Step 4: Calculate the final value

Finally, calculate the value inside the square root to find the distance: \(\sqrt{8}\) or rounded to two decimal places, which is 2.83.

Key Concepts

Simplified Radical Form

When you calculate distances in geometry, sometimes you'll end up with a square root that doesn't simplify neatly into an integer. This is where simplified radical form comes in handy. Simplified radical form involves reducing the square root to its simplest possible expression without decimals.
\[\text{For example, if you have the square root of 8:}\,<\[\sqrt{8}\,\],\]
you might start by factoring 8 to make it easier to simplify. Notice that 8 can be broken down as \(4 \times 2\), and since \(4\) is a perfect square (\(2^2\)), the square root becomes:
\[\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}.\]
Instead of using a calculator to get a decimal, we leave it in this form. This method is beneficial because it gives a more exact representation of the number rather than an approximate decimal. It is also essential for ensuring that you handle and work with radicals correctly in higher mathematics.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, uses algebraic methods to solve geometrical problems. It involves plotting points, lines, and shapes on a coordinate plane, a grid divided into four quadrants by the x-axis (horizontal) and y-axis (vertical). Each point on this grid is identified by an ordered pair \((x, y)\).

• In this exercise, we have two sets of coordinates: the first point is \((0, -\sqrt{3})\) and the second is \((\sqrt{5}, 0)\).
• These coordinates indicate that the first point lies on the y-axis (since its x-coordinate is 0), and the second point lies on the x-axis (since its y-coordinate is 0).

To find the distance between these two points, we use the distance formula:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\].
This formula helps measure the straight-line distance between any two given points in the coordinate plane by evaluating the change along each axis separately and then combining these changes. This approach bridges algebra and geometry, allowing precise calculations and insights into spatial relationships.

Rounding

Rounding numbers is an essential mathematical skill used to tidy up complex or long numbers by reducing them to simpler forms while maintaining their overall value. This becomes particularly useful when you want to express solutions in a more digestible form.
After calculating the distance between points using the formula and simplifying it to \(\sqrt{8}\), you may obtain a decimal value by further evaluating the expression with a calculator. This decimal is approximately \(2.828427\) - quite cumbersome for everyday use!
Therefore, to make it more manageable, we round this number to two decimal places. To round effectively, we look at the third decimal place to decide:

• If it's 5 or more, increase the second decimal place by one.
• If it's less than 5, leave the second decimal place as is.

Using this method, \(2.828427\) becomes \(2.83\). Rounding is crucial in giving a quick, easy to interpret answer that is still acceptably close to the exact value, especially in practical contexts where precision to several decimal places is unnecessary.