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:

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.

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).

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.

To find the vector in component form given the initial and terminal points, simply subtract the initial point from the terminal point.

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

To figure out the horizontal component, set up an equation involving cosine, since that side of the implied triangle is adjacent to the 48-degree angle:

To solve for x, first find the cosine of 48, then multiply by 11:

To figure out the vertical component, set up an equation involving sine, since that side of the implied triangle is opposite the 48-degree angle:

to solve for y, just like x, first find the sine of 48, then multiply by 11:

Putting this in component form results in the vector

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.

In vector addition, you simply add each component of the vectors to each other.

x component: .

y component: .

z component: .

The new vector is

.

To find the magnitude we use the formula,

Thus its magnitude is 5.

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: