Problem 57 - Calculus Single Variable

Question

Find a point \((a, b)\) with distance 1 unit from the origin and a point \((c, d)\) with distance 5 units from the origin so that the line through \((a, b)\) and \((c, d)\) has slope 2 .

Step-by-Step Solution

Answer

(a, b) = (1, 0), (c, d) can be either (-1, -4) or (\frac{21}{5}, \frac{32}{5}).

Step 1: Understand the problem requirements

We need to find two points - \( (a, b) \) that is 1 unit away from the origin - \( (c, d) \) that is 5 units away from the origin.Furthermore, the slope of the line connecting these points must be 2. Recall that slope \( m \) can be calculated as \( m = \frac{d - b}{c - a} \).

Step 2: Solve for \\( (a, b) \\) with distance 1

From the formula for distance from the origin, we know \( \sqrt{a^2 + b^2} = 1 \). Squaring both sides gives \( a^2 + b^2 = 1 \). Any \( (a, b) \) that satisfies this equation is 1 unit away.

Step 3: Solve for \\( (c, d) \\) with distance 5

Similarly, for the point \( (c, d) \) 5 units away, \( \sqrt{c^2 + d^2} = 5 \), leading to \( c^2 + d^2 = 25 \). The point \( (c, d) \) that satisfies this is exactly 5 units away.

Step 4: Determine equation for slope 2

We want \( \frac{d - b}{c - a} = 2 \), which rearranges to \( d - b = 2(c - a) \). This equation must hold true for the line's slope to be 2.

Step 5: Solve the system of equations

To find specific values, let's use trial and error or systematic solving.Let \( (a, b) = (1, 0) \), which satisfies \( a^2 + b^2 = 1 \).Use the slope condition: \( d = 2(c - 1) \).Substitute \( c, d \) in the circle equation: \( c^2 + (2c - 2)^2 = 25 \).Solve: \[ c^2 + 4(c^2 - 2c + 1) = 25 \]\[ 5c^2 - 8c + 4 = 25 \]\[ 5c^2 - 8c - 21 = 0 \]By solving this quadratic equation using the quadratic formula, \( c = -1 \text{ or } c = \frac{21}{5} \).

Step 6: Calculate corresponding \\( d \\) values

Using \( c = -1 \): \( d = 2(-1 - 1) = -4 \).Using \( c = \frac{21}{5} \): \( d = 2( \frac{21}{5} - 1) = \frac{32}{5} \).Verify both fit \( c^2 + d^2 = 25 \)(they do).So solutions are \( (a, b) = (1,0) \), \( (c, d) = (-1, -4) \) and \((a, b) = (1,0) \), \( (c, d) = (\frac{21}{5}, \frac{32}{5}) \).

Key Concepts

Distance Formula

In analytic geometry, determining the distance between two points in the coordinate plane is essential. The distance formula helps find the straight-line distance between two points, \( (x_1, y_1) \) and \( (x_2, y_2) \).
The formula is derived from the Pythagorean theorem: \[ Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

• \( (a, b) \), a point with distance 1 from the origin, follows: \( \sqrt{a^2 + b^2} = 1 \).
• After squaring both sides, you get \( a^2 + b^2 = 1 \).
• Similarly, for point \( (c, d) \) with distance 5: \( c^2 + d^2 = 25 \).

These equations state that \( (a, b) \) lies on a circle of radius 1, and \( (c, d) \) lies on a circle of radius 5, both centered at the origin. This concept is crucial in finding locations based on given constraints.

Slope of a Line

The slope of a line represents its steepness and direction. Mathematically, the slope (\( m \)) is defined as the ratio of the change in the y-coordinates to the change in the x-coordinates between two points on the line.
The formula for the slope between points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

• In our exercise, the slope requirement is \( m = 2 \).
• Given points \( (a, b) \) and \( (c, d) \), the formula turns into \( \frac{d - b}{c - a} = 2 \).

Rearranging yields \( d - b = 2(c - a) \), an equation ensuring the line connecting these points has the desired slope. The slope is central for linear relationships and helps us understand how changes in one variable relate to changes in another.

System of Equations

When given multiple conditions to meet simultaneously, solving a system of equations becomes necessary. This involves finding solutions that satisfy each equation in the system.

• Our particular problem involves three conditions:
• Point \( (a, b) \) distance from origin: \( a^2 + b^2 = 1 \).
• Point \( (c, d) \) distance from origin: \( c^2 + d^2 = 25 \).
• Slope of the line: \( \frac{d - b}{c - a} = 2 \).

By solving these equations together, we ensure all conditions are met. This might involve:- Substitution: Replacing a variable with an expression from another equation.- Elimination: Adding or subtracting equations to remove a variable.Systematically working through the equations allows us to find valid points \( (a, b) \) and \( (c, d) \) that satisfy all constraints.

Quadratic Equation

Quadratic equations are a step into solving more complex problems within algebra. A typical quadratic equation in its standard form is represented as \( ax^2 + bx + c = 0 \).
It's solved using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• In our given solution, \( 5c^2 - 8c - 21 = 0 \) arises from manipulations involving expected conditions.
• Using the quadratic formula, solutions for \( c \) are found as \( c = -1 \) or \( c = \frac{21}{5} \).

Quadratic equations are powerful tools for modeling numerous scenarios, including those in physics and geometry. They provide two potential solutions leading to possible different scenarios. Incorporating quadratic equations into this problem allows for the accurate determination of point coordinates that meet all initial requirements.