Other Lines - SSAT Upper Level Quantitative

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.

Since we have –2x and +2x in the equations, it makes sense to add the equations together to give 8y = 16 yielding y = 2. Then we substitute y = 2 into one of the original equations to get x = –2. So the solution to the system of equations is (–2, 2)

when plugged in for and make the linear equation true, therefore those coordinates fall on that line.

Because this equation is true, the point must lie on the line. The other given answer choices do not result in true equalities.

We are looking for an ordered pair that makes the given equation true. To solve, plug in the various answer choices to find the true equality.

Because this equality is true, we can conclude that the point lies on this line. None of the other given answer options will result in a true equality.

In the multiple choice format, you can just plug in these points to see which satisfies both equations. and work for the first but not the second, and and work for the second but not the first. Only works for both.

Alternatively (or if you were in a non-multiple choice scenario), you could set the equations equal to each other and solve for one of the variables. So, for instance,

and

so

Now you can solve and get . Plug this back into either of the original equations and get .

Plug the x-coordinate of an answer choice into the equation to see if the y-coordinate matches with what comes out of the equation.

For ,

To find if a point is on the line, plug in the x-coordinate of the answer choice into the given equation. If the resulting value for the y-coordinate matches that of the answer choice, then that point is on the line.

For ,

To find which point lies on the line, plug in the x-coordinate value of an answer choice into the equation. If the y-coordinate value that comes out of the equation matches that of the answer choice, then the point is on the line.

For ,

So then, lies on the line.