Algebra I

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Questions 1 - 10

Which of the following lines is parallel to

$y = -\frac{1}{2}x -4$

$y = \frac{1}{2}x - 5$

Explanation

When comparing two lines to see if they are parallel, they must have the same slope. To find the slope of a line, we write it in slope-intercept form

where m is the slope.

The original equation

will need to be written in slope-intercept form. To do that, we will divide each term by 4

$\frac{4y}{4} = \frac{-2x}{4} + \frac{8}{4}$

$y = \frac{-1x}{2} + 2$

$y = - \frac{1}{2}x + 2$

Therefore, the slope of the original line is $-\frac{1}{2}$ . A line that is parallel to this line will also have a slope of $-\frac{1}{2}$ .

Therefore, the line

$y = -\frac{1}{2}x -4$

is parallel to the original line.

What is the midpoint of a line segment with the end points and ?

Explanation

In order to solve for the midpoint of a line segment when the end points are given, the x and y values must be averaged.

A formula for this would be:
$midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

This probem may be quickly solved for by substituting in the given information.

$midpoint=(\frac{1+7}{2},\frac{2+8}{2})$

$midpoint=(\frac{8}{2},\frac{10}{2})$

$midpoint={\color{Blue} (4,5)}$

A straight line passes through the points and .

What is the -intercept of this line?

Explanation

First calculate the slope:

$(\text{change in Y})/(\text{change in x}) = (1-2)/(3-2) = -1$

The standard equation for a line is .

In this equation, is the slope of the line, and is the -intercept. All points on the line must fit this equation. Plug in either point (1,3) or (2,2).

Plugging in (1,3) we get .

Therefore, .

Our equation for the line is now:

To find the -intercept, we plug in :

Thus, the -intercept the point (4,0).

What is the equation of a circle that has its center at $(5, \frac{1}{4})$ and a radius length of $\frac{1}{5}$ ?

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{25}$

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{10}$

$(x-\frac{1}{4})^2+(y+5)^2=\frac{1}{25}$

$(x-5)^2-(y+\frac{1}{4})^2=\frac{1}{5}$

Explanation

Recall the standard form for the equation of a circle:

In this equation, represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{25}$

What line is perpendicular to $y=\frac{1}{2}x+3$ through ?

Explanation

Perdendicular lines have slopes that are opposite reciprocals. The slope of the old line is $\frac{1}{2}$ , so the new slope is .

Plug the new slope and the given point into the slope intercept equation to calculate the intercept:

or , so .

Thus , or .

Find the midpoint of a line segment with the following endpoints:

Explanation

Finding the midpoint of a line segment only requires that we find the midpoint, or average of both the x and y components. In order to do that, we use the following formula:

$(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

For the points (2,6) and (12,12) plug in the numbers and solve:

$(\frac{2+12}{2},\frac{6+12}{2})=(\frac{14}{2},\frac{18}{2})=(7,9)$

This gives a final answer of

What line is perpendicular to $y=\frac{1}{2}x+3$ through ?

Explanation

Perdendicular lines have slopes that are opposite reciprocals. The slope of the old line is $\frac{1}{2}$ , so the new slope is .

Plug the new slope and the given point into the slope intercept equation to calculate the intercept:

or , so .

Thus , or .

What is the y-intercept of the line with the equation ?

Explanation

You should recognize that the given equation is in the point-slope form.

In order to find the y-intercept, rearrange the equation into slope-intercept form, .

Since, , the y-intercept must be located at

Find the point that corresponds to the following ordered pair:

Explanation

In order to get to the point , start at the origin and move right unit and up units.

This is point .

The points and are plotted on a quadrant. Which graph depicts the correct points?

Explanation

A quadrant is set up in such a way that the positive numbers are to the right and top, while the negative numbers are to the left and bottom. The x-value (the first number in the ordered pair) is the distance left or right from the center. The y-value (the second number in the ordered pair) is the distance above or below the axis.

will be units to the right, and units up.