GMAT Math

I understand that if the slopes are negative reciprocals then they will be perpendicular and intersect.

I also understand that if two lines have the same slope that the only way they can intersect is if they are on top of each. In this case, you would plug the coordinates into y = mx + b and if the equation checks out, then they intersect.

However, what if you have two line segments with different (not negative reciprocal) slopes?

Say for example line segment SR = S--> (5,2) and R --> (4,4) and line segment PQ = P--> (1,3) and Q--> (7,4)

My question boils down to what info do you need to know in order to determine whether two lines intersect?

I think this is very valuable info to know for the gmat.

Thanks!

I think I have the answer.

If two lines intersect, they will always be perpendicular. Therefore, if slopes are negative reciprocals, they will intersect. If they are not negative reciprocals, they will never intersect (except for the parallel line scenario)

Basically, you can determine whether lines intersect if you know the slopes of two different lines.

Please correct me if I'm wrong.

Thanks

If two lines intersect they will have a common point, the point of intersection. You have to simultaneously solve the eq of the two lines to get the common points. This is a necessary condition.
But in GMAT you need to know whether the data in the Q stem is sufficient.
For example the points you have taken (5,2) (4,4) and (1,3) and (7,4) if plotted on a graph will show the rough point of intersection is app ( x is slightly more than 4, y is slightly more than 3)
to be sure you need eq of the line SR and PQ.
coming to ur 2nd rhetorical Q
two line may be such that the product of their slopes are - 1 but be in diff quadrant so they will surely meet, but where it is difficult to guess.
Only is the lines are parallel where their slopes are = they will never meet.

Thanks, although I think I'm right about lines that intersect are always perpendicular so therefore all you need to find out are the slopes.

My question is how do you find the points of intersection if you have the two line segments?

If you are given the points (5,2) and (4,4) you have enough info to generate an equation of the line.

What i wrote is actually false. Just because two lines intersect, doesn't mean they're perpendicular. Not sure what I was thinking there.

Anyway, that still leaves my original point unanswered. How do you know if two lines intersect? What info do you need?

A detailed explanation would be great.

First of all any two lines that are not parallel (does not have the same slope) will intersect at some point.
Basically you need to take the two equations of the lines and equate them to see if you have a solution.
So if you have the four points you mentioned, you need to build the two equations of the lines (First find the slope m: Y1-Y2/(x1-x2) and then find the Y intersection by assign one point to the equation: Y=y=mx+b )
After you have two equations y= mx+b , y=kx+d
Solve for X : mx+b = kx+d
and then assign in one of the equations to get the Y

In the GMAT you might get a question asking to find if two curves intersect. for example y=x^2+2x+3 and y=3x^2+5x+5
The same principle holds here also.

Hope it helps

The only straight lines that will NEVER intersect are distinct parallel lines. If the slopes are identical and the y-intercepts different they will not intercept.

Quadratics are a different matter entirely. There are many quadratics that do not intersect, but quadratics have different slopes at multiple points along the curve so comparing slopes doesn't work.

pkw209 wrote:I understand that if the slopes are negative reciprocals then they will be perpendicular and intersect.

I also understand that if two lines have the same slope that the only way they can intersect is if they are on top of each. In this case, you would plug the coordinates into y = mx + b and if the equation checks out, then they intersect.

However, what if you have two line segments with different (not negative reciprocal) slopes?

Say for example line segment SR = S--> (5,2) and R --> (4,4) and line segment PQ = P--> (1,3) and Q--> (7,4)

My question boils down to what info do you need to know in order to determine whether two lines intersect?

I think this is very valuable info to know for the gmat.

Thanks!

First, let's clarify the difference between a line segment and a line. A line segment has 2 endpoints. A line has no endpoints; it goes on forever in each direction.

Two lines will intersect if they have different slopes. Two lines will not intersect (meaning they will be parallel) if they have the same slope but different y intercepts.

Here's a question about intersection:

If line M passes through (5,2) and (8,8), and line N line passes through (5,3) and (7,11), at what point do line M and line N intersect?

Line M: (5,2) and (8,8)
Slope = (8-2)/(8-5) = 2
Partial equation: y = 2x + b
To determine b, plug into the partial equation one of the given points (either is fine, but let's use x=5, y=2):
2 = 2*5 + b
b = -8
Thus, the full equation: y = 2x-8.

Line N: (5,3) and (7,11)
Slope = (11-3)/(7-5) = 4
Partial equation: y = 4x + b
To determine b, plug into the partial equation one of the given points (either is fine, but let's use x=5, y=3):
3 = 4*5 + b
b = -17
Thus, the full equation: y = 4x-17

Now that we know the equation of each line, we can determine where the lines intersect by setting the equations equal to each other.

2x-8 = 4x-17
9 = 2x
x = 9/2. This is the x coordinate of the point of intersection.

To determine the y coordinate, plug x=9/2 into one of the equations (let's use y=4x-17):

y = 4(9/2) - 17
y = 1

Thus, the point of intersection is (9/2, 1).

Notice that these coordinates also work in the other equation, y=2x-8:

1 = 2*(9/2) - 8
1 = 1.

Hope this helps!

Here's another question:

In the XY-coordinate plane, which of the following does not intersect y = 2x^2 + 3x + 1?
A. y = x^2 + 2x + 2
B. y = 2x^2 + 5x + 2
C. y = 2x^2 + 7x + 2
D. y = 4x^2 + 6x + 3
E. y = 4x^2 + 8x + 4

I would go with D. The x terms are direct multiples of the terms in the original equation and the y-intercept puts is above and inside the first equation.

Nice observation Tani,
But do you have a more deterministic approach. I mean, if the equation instead of D was:
D. y = 4x^2+5x+3

What would you do? I have my own way but I want to see if there is simple way (I think mine is a little out of the scope of GMAT)

eladoren wrote:Here's another question:

In the XY-coordinate plane, which of the following does not intersect y = 2x^2 + 3x + 1?
A. y = x^2 + 2x + 2
B. y = 2x^2 + 5x + 2
C. y = 2x^2 + 7x + 2
D. y = 4x^2 + 6x + 3
E. y = 4x^2 + 8x + 4

To find the point(s) of intersection, the process is similar to the one I described in my earlier post about intersecting lines.

Since at the point(s) of intersection the y value in each equation must be the same, we set the two equations equal to each other. Answer choice A, for example:

2x^2 +3x + 1 = x^2 + 2x + 2

To solve, we combine line terms:

x^2 + x -1 = 0.

A quadratic equation in the form ax^2 + bx + c will have no real solutions if its discriminant (b^2 - 4ac) is negative. If the equation above has no real solutions, there will no points of intersection between the 2 parabolas. Recognizing that in the equation above a=1, b=1, and c=-1, we get:

b^2- 4ac = 1^2 - (4*1*-1) = 1+4 = 5.

Since the discriminant is positive, the equation above will have 2 real solutions and the parabolas will have 2 points of intersection.

The process above involves 2 essential steps:

-- subtracting one quadratic expression from the other
-- determining whether the discriminant of the resulting expression is negative

So to solve the problem above, we need only subtract the given equation from each answer choice and determine which results in a negative discriminant.

Answer choice D:
4x^2 + 6x + 3 - (2x^2 + 3x + 1) = 2x^2 + 3x + 2
b^2 - 4ac = 3^2 - 4*2*2 = 9 - 16 = -7.
Since the discriminant is negative, there will be no points of intersection.

Hope this helps! Please note that I've never seen a question of this sort on the GMAT. Parabolas generally are beyond the scope of the test; the quadratic formula is not tested.

Thanks GMATGuruNY !
This is the same process I did (looking on the discriminante), I thought there is some other way because I think this is out of scope of GMAT.
BUT I have news for you this is actual (up-to-date) GMAT problem (not the exact numbers to avoid copyrights issues...)
Also the GMAT has started to ask questions that require you the knowledge of vieta's formulas. Of course, you can always solve questions of that sort without knowing these formulas but it might take you long instead of 10 seconds.

Thanks
Elad