Problem 59 - Precalculus

Question

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

Step-by-Step Solution

Answer

The shortest distance between the point (1,1) and the line \(y=x+1\) is \(d = \frac{1}{\sqrt{2}}\).

Step 1: Standard Form of The Line

Rewrite the line equation into the standard form. The given equation \(y = x + 1\) can be rewritten as \(x - y + 1 = 0\), here \(A=1\), \(B=-1\), and \(C=1\).

Step 2: Substitute Point Values Into Distance Formula

Substitute the given point (1,1) and coefficients \(A=1\), \(B=-1\), and \(C=1\) into the distance formula. Hence, the distance \(d\) becomes \(d = \frac{|1\cdot1 +(-1)\cdot1+1|}{\sqrt{1^2 +(-1)^2}}\).

Step 3: Simplifying

Simplify the equation to get the final answer. The distance \(d\) simplifies to \(d = \frac{1}{\sqrt{2}}\).

Key Concepts

Standard Form of a Line

In geometry, understanding the standard form of a line is crucial for solving various problems, such as finding the distance between a point and a line. The standard form is represented by the equation:
\[ Ax + By + C = 0 \]
where A, B, and C are real numbers, and A and B are not both zero. This form is particularly useful because it clearly reflects the coefficients of the x and y terms, which can be used in other calculations like the distance formula.

When a line's equation is not in this form, it can be rearranged through basic algebraic manipulation. For instance, the equation \( y = x + 1 \) can be converted into the standard form by subtracting y from both sides and adding any constants to the other side, resulting in \( x - y + 1 = 0 \). This step is typically the starting point for finding the perpendicular distance from a point to a line, as it sets the stage for substitution into the distance formula.

Distance Formula

For those grappling with geometry, the distance formula is an indispensable tool, permitting one to ascertain the shortest distance between a point and a line. The formula is predicated on the coordinates of the point and the coefficients of the line's standard form equation. Expressed in terms of these parameters, the formula is:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(d\) denotes the distance, \(Ax_1 + By_1 + C\) is the absolute value of the expression obtained by plugging the x and y coordinates of the point into the line's equation, and \(\sqrt{A^2 + B^2}\) is the square root of the sum of the squares of the line's coefficients.

It is crucial to apply the absolute value because distance cannot be negative. The presence of the square root in the denominator reflects Pythagoras' theorem, tying this distance to the right triangle formed by the point, the closest point on the line, and the line's projection.

Line Equations

Equations of lines are fundamental in understanding geometric relations between different figures. There are various forms of line equations, each useful in different contexts. Besides the standard form, there are:

• Slope-intercept form: \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept, showing directly how steep the line is and where it intersects the y-axis.
• Point-slope form: \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope, helping when a point on the line and the slope are known.
• Two-point form: \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \) where \( (x_1, y_1) \) and \( (x_2, y_2) \) are two distinct points on the line, useful for finding an equation from two known points on the line.

While different forms are used for diverse purposes, converting between them allows one to analyze the geometry of a line in multiple contexts. For calculating the distance to a point, using the standard form is often the most straightforward approach.