Coordinate Geometry - SSAT Upper Level Quantitative

Card 0 of 1512

Question

What is the slope of the given linear equation?

2x + 4y = -7

Answer

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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Question

Find the slope of the line 6X – 2Y = 14

Answer

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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Question

What is the slope of the line:

$\frac{14}{3}x=\frac{1}{6}y-7$

Answer

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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Question

If 2x – 4y = 10, what is the slope of the line?

Answer

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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Question

What is the slope of the line with equation 4_x_ – 16_y_ = 24?

Answer

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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Question

What is the slope of a line that is parallel to the line 11x + 4y - 2 = 9 – 4x ?

Answer

We rearrange the line to express it in slope intercept form.

Any line parallel to this original line will have the same slope.

$ewline 11x+4y-2=9-4x ewline 4y=11-15x ewline y=\frac{11}{4}-\frac{15}{4}x ewline y=-\frac{15}{4}x+\frac{11}{4}$

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Question

Find the slope of the line that passes through the points $(2, -3)\text{ and }(-3, -3)$

Answer

Use the following formula to find the slope:

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{-3-(-3)}{-3-2}=\frac{0}{-5}=0$

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Question

Given the graph of the line below, find the equation of the line.

Answer

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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Question

Define an operation $\wedge$ as follows:

For all complex numbers ,

$a \wedge b = a \div (b+ 2i)$

Evaluate $8 \wedge 4$

Answer

$a \wedge b = a \div (b+ 2i) = \frac{a}{b+2i}$

$8 \wedge 4 = \frac{8}{4+2i}$

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

$8 \wedge 4 = \frac{8(4-2i)}{(4+2i)(4-2i)}$

$= \frac{8\cdot 4-8\cdot 2i}{4^{2}+2^{2}}$

$= \frac{32-16i}{16+4}$

$= \frac{32-16i}{20}$

$= \frac{32 }{20} - \frac{ 16}{20}i$

$= \frac{8}{5} -\frac{ 4}{5}i$

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Question

Raise to the power of 4.

Answer

$(4i)^{4} = 4^{4} \cdot i ^{4} = 256 \cdot 1 = 256$

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Question

Give the equation of the line through and .

Answer

First, find the slope:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} =\frac{9-1}{2-8}=\frac{8}{-6}=-\frac{4}{3}$

Apply the point-slope formula:

$y-y_{1}= m (x-x_{1})$

$y-1= -\frac{4}{3} (x-8)$

$y-1= -\frac{4}{3} x- \left (-\frac{4}{3} \right )8$

$y-1= -\frac{4}{3} x+ \frac{32}{3}$

$y= -\frac{4}{3} x+ \frac{35}{3}$

Rewriting in standard form:

$\frac{4}{3} x + y= \frac{35}{3}$

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Question

A line can be represented by . What is the slope of the line that is perpendicular to it?

Answer

You will first solve for Y, to get the equation in form.

represents the slope of the line, which would be $-\frac{2}{3}$ .

A perpendicular line's slope would be the negative reciprocal of that value, which is $\frac{3}{2}$ .

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Question

Examine the above diagram. What is ?

Answer

Use the properties of angle addition:

$\left ( x - 13\right )+ (y + 25) +43= 180$

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Question

Give the equation of a line that passes through the point and has an undefined slope.

Answer

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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Question

Give the equation of a line that passes through the point and has slope 1.

Answer

We can use the point slope form of a line, substituting $m = 1, x_{1} = 4, y _{1} = 7$ .

$y- y_{1} = m (x-x_{1})$

or

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Question

Find the equation the line goes through the points and .

Answer

First, find the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{0-3}=\frac{-2}{-3}=\frac{2}{3}$

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:

$y=\frac{2}{3}x+2$

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Question

A line passes through the points and . Find the equation of this line.

Answer

To find the equation of a line, we need to first find the slope.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-6}{-2-(-4)}=\frac{-8}{2}=-4$

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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Question

What is the equation of a line that passes through the points and ?

Answer

First, we need to find the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{-7-1}{-2-8}=\frac{-8}{-10}=\frac{4}{5}$

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.

$1=8(\frac{4}{5})+b$

Solve for .

$1=\frac{32}{5}+b$

$b=-\frac{27}{5}$

Now, put the slope and -intercept together to get $y=\frac{4}{5}x-\frac{27}{5}$

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Question

Are the following two equations parallel?

Answer

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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Question

Reduce the following expression:

$\frac{10a^{3}b^{5}c^{-2}}{2a^{2}}$

Answer

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 $\frac{5ab^{5}}{c^{2}}$ .

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