Other Lines - SSAT Upper Level Quantitative

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.

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 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.

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

Use the following formula to find the slope:

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.

First, find the slope:

Apply the point-slope formula:

Rewriting in standard form:

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 .

Use the properties of angle addition:

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

We can use the point slope form of a line, substituting .

or

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:

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.

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

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.

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 .

The equation of a line is

y=mx + b where m is the slope

Rearrange the equation to match this:

7x + 28y = 84

28y = -7x + 84

y = -(7/28)x + 84/28

y = -(1/4)x + 3

m = -1/4

To test the coordinates, plug the x-coordinate into the line equation and solve for y.

y = 1/3x -7

Test (3,-6)

y = 1/3(3) – 7 = 1 – 7 = -6 YES!

Test (3,7)

y = 1/3(3) – 7 = 1 – 7 = -6 NO

Test (6,-12)

y = 1/3(6) – 7 = 2 – 7 = -5 NO

Test (6,5)

y = 1/3(6) – 7 = 2 – 7 = -5 NO

Test (9,5)

y = 1/3(9) – 7 = 3 – 7 = -4 NO

Since the vertical coordinates of each point are given by y, solve each equation for y and plug in 6 for x, as follows:

Taking the difference of the resulting y -values give the vertical distance between the points (6,27) and (6,48), which is 21.