Problem 64 - Precalculus

Question

Find the distance between the point and the line. $$\begin{array}{cc}\text{Point} && \text{Line} \\ (4,-4) && y=-2 x-3\end{array}$$

Step-by-Step Solution

Answer

The distance from the given point to the line is \( 11 / \sqrt {5} \).

Step 1: Identify the coefficients

The equation of the given line is \(y=-2x-3\). To put it into the canonical form \(Ax + By + C = 0\), rewrite it as \(2x + y + 3 = 0\). Thus, \(A = 2\), \(B = 1\), and \(C = 3\). The given point is \((4, -4)\), so \(x = 4\) and \(y = -4\).

Step 2: Substitute the values into the formula

Substitute the above values into the distance formula. The distance, \(d\), is given by \( |2*4 + (-1)*(-4) + 3| / \sqrt {2^2 + 1^2} \) , which simplifies to \( |11| / \sqrt {5} \).

Step 3: Calculate the distance

Now, the final step is to calculate the distance which is \( 11 / \sqrt {5} \). Simplify this to get the final answer.

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This method enables problems involving geometry to be translated into algebraic equations, which can then be solved using mathematical techniques.

In the context of finding the distance from a point to a line, we use coordinate geometry to represent both the point and the line within the same coordinate system—usually the Cartesian plane, where points are defined by their x (horizontal) and y (vertical) coordinates.

For instance, the point (4, -4) is located 4 units right and 4 units down from the origin (0, 0), while the line given by the equation y = -2x - 3 can be graphed to show its slope and y-intercept. By using coordinates and equations, we can analyze and solve various geometric problems efficiently.

Distance Formula

The distance formula is a vital tool in coordinate geometry, allowing us to calculate the shortest distance between two objects in a coordinate system, typically points. It's derived from the Pythagorean theorem and represented as:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
When finding the distance between a point and a line, we modify this formula. For a point \( P(x_1, y_1) \) and a line \( Ax + By + C = 0 \) the formula becomes:
\[d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\]
This formula gives us the perpendicular distance from the point to the line, which is important in mathematical problems requiring precision measurements. The absolute value ensures the distance is non-negative, while the denominator normalizes the equation of the line.

Linear Equations

Linear equations form the foundation for lines in a coordinate system. An equation of the form \( y = mx + c \) describes a straight line with slope \( m \) and y-intercept \( c \). These lines are central to many aspects of coordinate geometry and can be manipulated to describe relationships between points and to find distances.

In our exercise, the linear equation is given as \( y = -2x - 3 \) which is a slope-intercept form. It means the line slopes downward as it moves from left to right (a negative slope) and crosses the y-axis at -3 (the y-intercept). To find the distance from a point to this line, we first convert the equation into the standard form \( Ax + By + C = 0 \) by rearranging the terms and matching coefficients A, B, and C. Understanding how to manipulate these linear equations is key to tackling geometry problems involving points and lines.