Problem 70 - Precalculus

Question

Find the distance between the point and the line. $$\begin{array}{cc}\text{Point} && \text{Line} \\ (1,-3) && 4 x-3 y=-7\end{array}$$

Step-by-Step Solution

Answer

The distance between the point and the line is 6/5 or 1.2.

Step 1: Substitute Values into the Formula

Now that we've identified our values, we can substitute them into the formula to solve for the distance. The formula is now: \(d = \frac{|4(1) - 3(-3) - 7|}{\sqrt{4^2 + (-3)^2}}\).

Step 2: Simplify the Numerator

Now we can simply use the order of operations to simplify the numerator. We get that the numerator is |4 + 9 - 7| = |6| = 6.

Step 3: Simplify the Denominator

Now we need to simplify the denominator. Following the order of operations, \(\sqrt{4^2 + (-3)^2} =\sqrt{16 + 9} = \sqrt{25} = 5\).

Step 4: Division

Finally, we divide the numerator by the denominator which gives us the required distance. Hence, \(d = \frac{6}{5}\).

Key Concepts

Distance formula

The distance formula is a vital tool for finding the shortest path between a point and a line. Imagine you're trying to find the closest distance from a spot on a map to a road. The same concept applies in geometry. To find the distance between a point \((x_1, y_1)\) and a line given by \(Ax + By + C = 0\), we use this special formula:

• Distance, \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)

The numerator, \(|Ax_1 + By_1 + C|\), represents a measure involving the point's coordinates and the line equation. It's essentially calculating how far off from zero this point is when plugged into the linear equation for the line.
The denominator, \(\sqrt{A^2 + B^2}\), normalizes this distance with the line's orientation or its 'steepness'. It essentially accounts for the 'tilt' of the line relative to vertical and horizontal axes.
Since we use absolute values in the numerator, the distance is always non-negative, just as distances in real life are measured as positive lengths.

Coordinate geometry

Coordinate geometry, also called analytic geometry, is about using algebra to analyze geometric shapes and figures. It combines the elements of geometry and algebra, allowing us to use coordinates to make sense of the spatial properties.
Here's how it plays a role in the given exercise:

• We represent points using coordinates. For the given point \((1, -3)\), 1 is the x-coordinate and -3 is the y-coordinate.
• Lines are expressed with equations like \(4x - 3y = -7\), which tells us how all points on that line relate algebraically.

This approach makes it easier to compute distances, find midpoints, or calculate the slopes of lines. It serves as a bridge between numerical computations and spatial reasoning.
This blend of geometry and algebra is what allows us to solve problems like finding the distance between a point and a line, transforming abstract geometric problems into solvable algebraic ones.

Linear equations

Linear equations form the backbone of coordinate geometry, simplifying the representation of lines. A line gets its shape and direction from a linear equation, typically written in the form

• \(Ax + By = C\)

Each term here impacts the line:

• \(A\) and \(B\) determine the slope and direction of the line.
• \(C\) influences the positioning or 'shift' from the origin.

For example, the line \(4x - 3y = -7\) has specific slope and orientation, thanks to its coefficients. In solving for the distance in our exercise, understanding the line's nature was key.
Setting it equal to zero emphasizes the line's consistency in solution contexts. Linear equations sum up the interactions of x and y, and any changes in their coefficients will affect the line's slope or tilt relative to the coordinate axes.
They are fundamental not just in geometry but in describing real-world situations where relationships between two variables need to be analyzed.