First, draw the vector  = (3, 4); this is represented in red below. Then, plot the point P₁ (1, 4), and draw a line  (represented in blue) through it that is parallel to the vector .

We must find the equation of line . For any point P₂ (x, y) on , . Since  is on line  and is parallel to ,  for some value of t. By substitution, we have . Therefore, the equation  is a vector equation describing all of the points (x, y) on line  parallel to  through P₁ (1, 4).

This is true. The independent variable  in this equation is called a parameter. 

A line through a point (x₁,y₁) that is parallel to the vector  = (a₁, a₂) has the following parametric equations, where t is any real number.

Using the given vector and point, we get the following:

x = 3t

y = 5 + 2t

Each value of t creates a distinct (x, y) ordered pair. You can think of these points as representing positions of an object, and of t as representing time in seconds. Evaluating the parametric equations for a value of t gives us the coordinates of the position of the object after t seconds have passed.

A line through a point (x₁,y₁) that is parallel to the vector  = (a₁, a₂) has the following parametric equations, where t is any real number.

Using the given vector and point, we get the following:

x = 4 - 7t

y = -3 + 3.5t

Each value of t creates a distinct (x, y) ordered pair. You can think of these points as representing positions of an object, and of t as representing time in seconds. Evaluating the parametric equations for a value of t gives us the coordinates of the position of the object after t seconds have passed.

In the equation y = -3x +1.5, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables. 

Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:

x = t

y = -3t +1.5

In the equation y = 5x - 3, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables. 

Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:

x = t

y = 5t - 3

Start by solving each parametric equation for t:

Next, write an equation involving the expressions for t; since both are equal to t, we can set them equal to one another:

Multiply both sides by the LCD, 4:

Get y by itself to represent this as an equation in slope-intercept (y = mx + b) form:

Start by solving each parametric equation for t:

Next, write an equation involving the expressions for t; since both are equal to t, we can set them equal to one another:

Multiply both sides by the LCD, 6:

Get y by itself to represent this as an equation in slope-intercept (y = mx + b) form:

To solve this problem, we need to know that the path of a projectile can be described with the following equations:

In these equations, t is time and g is the acceleration due to gravity.

First, you need to write the position of the ball as a pair of parametric equations that define the path of the ball at anytime, t, in seconds:

As you set up the equation for y, use the value g = -32.

Finally, find x and y when t = .05:

Use a calculator to solve, making sure you are in degree mode: 

This means that after 0.5 seconds, the ball has travelled 17.5 feet horizontally and 5.7 feet vertically. 

To solve this problem, we need to know that the path of a projectile can be described with the following equations:

In these equations, t is time and g is the acceleration due to gravity.

First, you need to write the position of the ball as a pair of parametric equations that define the path of the ball at anytime, t, in seconds:

As you set up the equation for y, use the value g = -32.

Finally, find x and y when t = .05:

To find the height that the ball will be at when it has travelled horizontally, plug 400 feet (horizontal distance) in for x.

This tells us that the ball has travelled 400 feet horizontally when 2.84 seconds have passed. Let's use the value of t = 2.84 and plug it into our equation for y to see how high the ball is at this time:

This means that when the ball has travelled 400 feet horizontally, 2.84 seconds have passed, and the ball is 16.65 feet above the ground.