Coordinate Geometry - SSAT Upper Level Quantitative
Card 0 of 1512
What is the slope of the given linear equation?
2x + 4y = -7
What is the slope of the given linear equation?
2x + 4y = -7
We can convert the given equation into slope-intercept form, y=mx+b, where m is the slope. We get y = (-1/2)x + (-7/2)
We can convert the given equation into slope-intercept form, y=mx+b, where m is the slope. We get y = (-1/2)x + (-7/2)
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Find the slope of the line 6X – 2Y = 14
Find the slope of the line 6X – 2Y = 14
Put the equation in slope-intercept form:
y = mx + b
-2y = -6x +14
y = 3x – 7
The slope of the line is represented by M; therefore the slope of the line is 3.
Put the equation in slope-intercept form:
y = mx + b
-2y = -6x +14
y = 3x – 7
The slope of the line is represented by M; therefore the slope of the line is 3.
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What is the slope of the line:

What is the slope of the line:
First put the question in slope intercept form (y = mx + b):
–(1/6)y = –(14/3)x – 7 =>
y = 6(14/3)x – 7
y = 28x – 7.
The slope is 28.
First put the question in slope intercept form (y = mx + b):
–(1/6)y = –(14/3)x – 7 =>
y = 6(14/3)x – 7
y = 28x – 7.
The slope is 28.
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If 2x – 4y = 10, what is the slope of the line?
If 2x – 4y = 10, what is the slope of the line?
First put the equation into slope-intercept form, solving for y: 2x – 4y = 10 → –4y = –2x + 10 → y = 1/2*x – 5/2. So the slope is 1/2.
First put the equation into slope-intercept form, solving for y: 2x – 4y = 10 → –4y = –2x + 10 → y = 1/2*x – 5/2. So the slope is 1/2.
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What is the slope of the line with equation 4_x_ – 16_y_ = 24?
What is the slope of the line with equation 4_x_ – 16_y_ = 24?
The equation of a line is:
y = mx + b, where m is the slope
4_x_ – 16_y_ = 24
–16_y_ = –4_x_ + 24
y = (–4_x_)/(–16) + 24/(–16)
y = (1/4)x – 1.5
Slope = 1/4
The equation of a line is:
y = mx + b, where m is the slope
4_x_ – 16_y_ = 24
–16_y_ = –4_x_ + 24
y = (–4_x_)/(–16) + 24/(–16)
y = (1/4)x – 1.5
Slope = 1/4
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What is the slope of a line that is parallel to the line 11x + 4y - 2 = 9 – 4x ?
What is the slope of a line that is parallel to the line 11x + 4y - 2 = 9 – 4x ?
We rearrange the line to express it in slope intercept form.
Any line parallel to this original line will have the same slope.

We rearrange the line to express it in slope intercept form.
Any line parallel to this original line will have the same slope.
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Find the slope of the line that passes through the points 
Find the slope of the line that passes through the points
Use the following formula to find the slope:


Use the following formula to find the slope:
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Given the graph of the line below, find the equation of the line.

Given the graph of the line below, find the equation of the line.
To solve this question, you could use two points such as (1.2,0) and (0,-4) to calculate the slope which is 10/3 and then read the y-intercept off the graph, which is -4.
To solve this question, you could use two points such as (1.2,0) and (0,-4) to calculate the slope which is 10/3 and then read the y-intercept off the graph, which is -4.
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Define an operation
as follows:
For all complex numbers
,

Evaluate 
Define an operation as follows:
For all complex numbers ,
Evaluate


Multiply both numerator and denominator by the conjugate of the denominator,
, to rationalize the denominator:






Multiply both numerator and denominator by the conjugate of the denominator, , to rationalize the denominator:
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Raise
to the power of 4.
Raise to the power of 4.
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Give the equation of the line through
and
.
Give the equation of the line through and
.
First, find the slope:

Apply the point-slope formula:





Rewriting in standard form:


First, find the slope:
Apply the point-slope formula:
Rewriting in standard form:
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A line can be represented by
. What is the slope of the line that is perpendicular to it?
A line can be represented by . What is the slope of the line that is perpendicular to it?
You will first solve for Y, to get the equation in
form.
represents the slope of the line, which would be
.
A perpendicular line's slope would be the negative reciprocal of that value, which is
.
You will first solve for Y, to get the equation in form.
represents the slope of the line, which would be
.
A perpendicular line's slope would be the negative reciprocal of that value, which is .
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Examine the above diagram. What is
?
Examine the above diagram. What is ?
Use the properties of angle addition:





Use the properties of angle addition:
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Give the equation of a line that passes through the point
and has an undefined slope.
Give the equation of a line that passes through the point and has an undefined slope.
A line with an undefined slope has equation
for some number
; since this line passes through a point with
-coordinate 4, then this line must have equation 
A line with an undefined slope has equation for some number
; since this line passes through a point with
-coordinate 4, then this line must have equation
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Give the equation of a line that passes through the point
and has slope 1.
Give the equation of a line that passes through the point and has slope 1.
We can use the point slope form of a line, substituting
.





or

We can use the point slope form of a line, substituting .
or
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Find the equation the line goes through the points
and
.
Find the equation the line goes through the points and
.
First, find the slope of the line.

Now, because the problem tells us that the line goes through
, our y-intercept must be
.
Putting the pieces together, we get the following equation:

First, find the slope of the line.
Now, because the problem tells us that the line goes through , our y-intercept must be
.
Putting the pieces together, we get the following equation:
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A line passes through the points
and
. Find the equation of this line.
A line passes through the points and
. Find the equation of this line.
To find the equation of a line, we need to first find the slope.

Now, our equation for the line looks like the following:

To find the y-intercept, plug in one of the given points and solve for
. Using
, we get the following equation:

Solve for
.


Now, plug the value for
into the equation.

To find the equation of a line, we need to first find the slope.
Now, our equation for the line looks like the following:
To find the y-intercept, plug in one of the given points and solve for . Using
, we get the following equation:
Solve for .
Now, plug the value for into the equation.
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What is the equation of a line that passes through the points
and
?
What is the equation of a line that passes through the points and
?
First, we need to find the slope of the line.

Next, find the
-intercept. To find the
-intercept, plug in the values of one point into the equation
, where
is the slope that we just found and
is the
-intercept.

Solve for
.


Now, put the slope and
-intercept together to get 
First, we need to find the slope of the line.
Next, find the -intercept. To find the
-intercept, plug in the values of one point into the equation
, where
is the slope that we just found and
is the
-intercept.
Solve for .
Now, put the slope and -intercept together to get
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Are the following two equations parallel?


Are the following two equations parallel?
When two lines are parallal, they must have the same slope.
Look at the equations when they are in slope-intercept form,
where b represents the slope.
We must first reduce the second equation since all of the constants are divisible by
.
This leaves us with
. Since both equations have a slope of
, they are parallel.
When two lines are parallal, they must have the same slope.
Look at the equations when they are in slope-intercept form, where b represents the slope.
We must first reduce the second equation since all of the constants are divisible by .
This leaves us with . Since both equations have a slope of
, they are parallel.
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Reduce the following expression:

Reduce the following expression:
For this expression, you must take each variable and deal with them separately.
First divide you two constants
.
Then you move onto
and when you divide like exponents you must subtract the exponents leaving you with
.
is left by itself since it is already in a natural position.
Whenever you have a negative exponential term, you must it in the denominator.
This leaves the expression of
.
For this expression, you must take each variable and deal with them separately.
First divide you two constants .
Then you move onto and when you divide like exponents you must subtract the exponents leaving you with
.
is left by itself since it is already in a natural position.
Whenever you have a negative exponential term, you must it in the denominator.
This leaves the expression of .
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