Equations of Lines - SSAT Upper Level Quantitative

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

First solve for the slope of the line, m using y=mx+b

m = (y2 – y1) / (x2 – x1)

= (15 – 14) / (_–_5 _–_3)

= (1 )/( _–_8)

=_–_1/8

y = –(1/8)x + b

Now, choose one of the coordinates and solve for b:

14 = –(1/8)3 + b

14 = _–_3/8 + b

b = 14 + (3/8)

b = 14.375

y = –(1/8)x + 14.375

P1 (0, 6) and P2 (4, 0)

First, calculate the slope: m = rise ÷ run = (y2 – y1)/(x2 – x1), so m = –3/2

Second, plug the slope and one point into the slope-intercept formula:

y = mx + b, so 0 = –3/2(4) + b and b = 6

Thus, y = –3/2x + 6

If P1(1, 3) and P2(3, 6), then calculate the slope by m = rise/run = (y2 – y1)/(x2 – x1) = 3/2

Use the slope and one point to calculate the intercept using y = mx + b

Then convert the slope-intercept form into standard form.

If we rearrange the second equation it is the same as the first equation. They are the same line.

Our first step will be to determing the slope of the line that connects the given points.

Our slope will be . Using slope-intercept form, our equation will be . Use one of the give points in this equation to solve for the y-intercept. We will use (2,6).

Now that we know the y-intercept, we can plug it back into the slope-intercept formula with the slope that we found earlier.

This is our final answer.

The slope intercept form states that y=mx+b. In order to convert the equation to the slope intercept form, isolate y on the left side:

8x-2y=12

-2y=-8x+12

y=4x-6

The answer is .

A line can only be represented in the form or , for appropriate constants , , and . A graph must have an equation that can be put into one of these forms to be a line.

represents a parabola, not a line. Lines will never contain an term.

In finding the equation of the line given its slope and a point through which it passes, we can use the slope-intercept form of the equation of a line:

, where is the slope of the line and is its -intercept.

Plug the given conditions into the equation to find the -intercept.

Multiply:

Subtract from each side of the equation:

Now that you have solved for , you can write out the full equation of the line: