Problem 53
Question
Use the results of Exercises \(49-52\) to find a set of parametric equations to represent the graph of the line or conic. Line: passes through (0,0) and (3,6)
Step-by-Step Solution
Verified Answer
The parametric equations for the line are \(x = t\) and \(y = 2t\).
1Step 1: Define the Slope of the Line
The slope of the line can be determined by using two given points (0,0) and (3,6). The formula to calculate the slope (m) is \(m = \frac{(y2 - y1)}{(x2 - x1)} \). Substituting the given point values into the formula, we get \(m = \frac{(6 - 0)}{(3 - 0)} = 2\).
2Step 2: Finding the Line Equation
A line equation is generally written as \(y = mx + c\). Since the line passes through the origin (0,0), the value of c (the y-intercept) is zero. Thus, the line equation becomes \(y = 2x\).
3Step 3: Converting into Parametric Form
To convert into parametric form, we can set \(x = t\). Consequently, \(y = 2t\). Thus, the parametric equations are \(x = t\) and \(y = 2t\), where \(t\) is the parameter.
Key Concepts
Slope of a LineLine EquationParametric FormPoint-Slope Formula
Slope of a Line
Understanding the slope of a line is crucial in algebra and geometry as it describes the steepness and direction of the line. When you have two points on a line, say point A with coordinates (0,0) and point B with coordinates (3,6), the slope (m) can be found using the formula
\(m = \frac{(y_2 - y_1)}{(x_2 - x_1)}\).
The slope indicates how much the y-coordinate changes for a one-unit increase in the x-coordinate. In our case, substituting the points into the formula gives us a slope of 2, meaning that for every step right along the x-axis, the line rises by 2 units.
\(m = \frac{(y_2 - y_1)}{(x_2 - x_1)}\).
The slope indicates how much the y-coordinate changes for a one-unit increase in the x-coordinate. In our case, substituting the points into the formula gives us a slope of 2, meaning that for every step right along the x-axis, the line rises by 2 units.
Line Equation
The equation of a line in the Cartesian plane is usually expressed in the form \(y = mx + c\), where \(m\) represents the slope and \(c\) represents the y-intercept, which is where the line crosses the y-axis. When given two points, you can identify the slope and then derive the equation. In the exercise, since our line goes through the origin (0,0), the y-intercept \(c\) is 0, simplifying our equation to \(y = 2x\). This form of the equation is clear and concise, illustrating a direct proportional relationship between x and y.
Parametric Form
Parametric equations offer a way to represent lines and curves using a third variable, often denoted \(t\), known as the parameter. Converting the line equation into parametric form results in separate equations for \(x\) and \(y\) in terms of \(t\). For the line in the given exercise, by letting \(x = t\), the parametric equation for \(y\) becomes \(y = 2t\). This parametric representation is extremely useful when analyzing motion, as \(t\) can represent time, allowing us to track the position of a point on the line at any given moment.
Point-Slope Formula
The point-slope formula is another tool to craft the equation of a line when you know a point on the line and its slope. The formula is written as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the known point and \(m\) is the slope. This is particularly useful if the line does not pass through the origin, as it provides a quick way to get to the line's equation. Since our line in this problem passes through the origin, this formula reinforces that the simple equation \(y = 2x\) is the result of zeroing out the values of \(x_1\) and \(y_1\) in the point-slope formula.
Other exercises in this chapter
Problem 53
A point in rectangular coordinates is given. Convert the point to polar coordinates. $$(-\sqrt{3},-\sqrt{3})$$
View solution Problem 53
Use a graphing utility to graph the polar equation. Describe your viewing window. $$r=8 \cos \theta$$
View solution Problem 53
Find the angle \(\theta\) (in radians and degrees) between the lines. $$\begin{aligned} &0.05 x-0.03 y=0.21\\\ &0.07 x+0.02 y=0.16 \end{aligned}$$
View solution Problem 53
Sketch (if possible) the graph of the degenerate conic. $$x^{2}+y^{2}+2 x-4 y+5=0$$
View solution