Problem 49 - Precalculus

Question

(a) Sketch the line \(y=\frac{1}{2} x-5\) and the point \(P(1,3) .\) Follow parts (b)-(d) to calculate the perpendicular distance from point \(P(1,3)\) to the line. (b) Find an equation of the line that passes through \(P(1,3)\) and is perpendicular to the line \(y=\frac{1}{2} x-5\) (c) Find the coordinates of the point where these two lines intersect. Hint: From intermediate algebra, to find where two lines \(y=m x+b\) and \(y=M x+B\) intersect, set the expressions \(m x+b\) and \(M x+B\) equal to each other, and solve for \(x\) (d) Use the distance formula to find the perpendicular distance from \(P(1,3)\) to the line \(y=\frac{1}{2} x-5\)

Step-by-Step Solution

Answer

The perpendicular distance from point \(P(1,3)\) to the line is \(3\sqrt{5}\).

Step 1: Sketch the Line and Plot the Point

Plot the line given by the equation \( y = \frac{1}{2}x - 5 \). Begin by finding the y-intercept \((0, -5)\) and use the slope \(\frac{1}{2}\) to find another point, such as \((2, -4)\). Draw the line through these points. Now, plot the point \( P(1, 3) \) on the graph.

Step 2: Find Perpendicular Slope

The slope of the line \( y = \frac{1}{2}x - 5 \) is \( \frac{1}{2} \). Because the slope of the perpendicular line is the negative reciprocal, the perpendicular slope is \(-2\).

Step 3: Equation of Perpendicular Line

Using point-slope form, \( y - y_1 = m(x - x_1) \), with point \( P(1, 3) \) and slope \( -2 \), the equation becomes: \[ y - 3 = -2(x - 1) \] Simplify to get the equation of the perpendicular line: \[ y = -2x + 5 \]

Step 4: Find Intersection Point

Set the equations of the lines equal to find the intersection: \[ \frac{1}{2}x - 5 = -2x + 5 \] Solving, we get: \[ \frac{1}{2}x + 2x = 10 \] \[ \frac{5}{2}x = 10 \] \[ x = 4 \] Substitute \( x = 4 \) back into one of the line equations, \( y = -2(4) + 5 \), to find \( y = -3 \). The intersection point is \((4, -3)\).

Step 5: Calculate Perpendicular Distance

Use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] where \((x_1, y_1) = (1, 3)\) and \((x_2, y_2) = (4, -3)\). This becomes: \[ d = \sqrt{(4 - 1)^2 + (-3 - 3)^2} \] \[ d = \sqrt{3^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \]

Key Concepts

Slope of a Line

The slope of a line is a fundamental concept in algebra and geometry. It indicates the steepness and direction of the line. Mathematically, the slope is defined as the ratio of the rise over the run, which represents vertical change over horizontal change between two points on a line. In the equation of a line, such as \( y = \frac{1}{2}x - 5 \), the coefficient of \( x \) (here, \( \frac{1}{2} \)) is the slope.

When you see a positive slope like \( \frac{1}{2} \), it means the line is rising as it moves from left to right. A negative slope would indicate a line falling as it moves to the right. For example, a line with a slope of \( -2 \) would be steeper and decreasing. The slope tells you how many units you move up or down for every unit you move to the right along the line.

Understanding slopes is crucial when comparing the orientation of lines, determining parallelism, and finding perpendicular lines.

Point-Slope Form

The point-slope form is a useful method to write the equation of a line when you know a point on the line and its slope. The formula is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope of the line, and \((x_1, y_1)\) is a known point on the line.

For the given line in the exercise, the slope between the point \( P(1, 3) \) and the line equation \( y = \frac{1}{2}x - 5 \) was determined as \(-2\). This negative reciprocal indicates that the new line is perpendicular to the original line.

By inserting \( m = -2 \) and \((x_1, y_1) = (1, 3)\) into the point-slope form, the equation transforms into \( y - 3 = -2(x - 1) \). Simplifying gives us \( y = -2x + 5 \), representing the line passing through \( P(1, 3) \) and perpendicular to the original line.

Intersection of Lines

The point of intersection between two lines is where they cross each other on the graph. To find this point, we solve the two line equations simultaneously.

In this exercise, the equations \( y = \frac{1}{2}x - 5 \) and \( y = -2x + 5 \) were set equal because at the intersecting point, both y-values are the same. Solving \( \frac{1}{2}x - 5 = -2x + 5 \) allowed us to isolate \( x \).

By rearranging and simplifying, we found \( x = 4 \). Substituting \( x = 4 \) back into either original line equation yields \( y = -3 \). Hence, the intersection point is \((4, -3)\), which marks the spot where both lines meet.

Distance Formula

The distance formula is a vital tool in geometry for calculating the distance between two points in a Cartesian plane. The formula is derived from the Pythagorean theorem and is expressed as \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].

Applying this formula to the exercise, we need to find the distance from point \( P(1, 3) \) to the intersection point \((4, -3)\). Substituting these coordinates into the formula gives us:

• \( (x_1, y_1) = (1, 3) \)
• \( (x_2, y_2) = (4, -3) \)

Calculating step-by-step, the x-difference \( (4 - 1) \) is 3, and the y-difference \( (-3 - 3) \) is -6. So the distance becomes \( \sqrt{3^2 + (-6)^2} \), or \( \sqrt{9 + 36} \), which simplifies to \( \sqrt{45} = 3\sqrt{5} \). This result represents the shortest (perpendicular) distance from the point \( P \) to the line.