One important thing to appreciate about GMAT math is that there is more to understand than simply the mathematics itself--there's also 
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Posts: 4,475
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
63% (02:44) correct
37%
(02:48)
wrong
based on 340
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Points J (3, 1) and K (– 1, – 3) are two vertices of an isosceles triangle. If L is the third vertex and has a y-coordinate of 6, what is the x-coordinate of L?
(A) –3
(B) –4
(C) –5
(D) –6
(E) –7
For a collection of coordinate geometry practice problems, as well as the OE of this problem, see:
https://magoosh.com/gmat/2014/gmat-prac ... -geometry/
Mike
_________________
(A) –3
(B) –4
(C) –5
(D) –6
(E) –7
For a collection of coordinate geometry practice problems, as well as the OE of this problem, see:
https://magoosh.com/gmat/2014/gmat-prac ... -geometry/
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Posts: 4,475
sudhirgupta93
Dear sudhirgupta93,
I'm happy to respond.
One important thing to appreciate about GMAT math is that there is more to understand than simply the mathematics itself--there's also the mind of the test maker. GMAT questions, especially harder ones, are designed specifically so that if one sees one particular shortcut, that unfolds very quickly to a solution. The insightful way takes less than 30 seconds. The brute force calculations, such as you are discussing, would take 10+ minutes. That's always the wrong choice on the GMAT. The insight to this problem is the fact that any two points of the form (a, b) and (-b, -a) are reflections of each other over the line y = - x. A mirror line is a perpendicular bisector of any segment connecting a point with its image. This very naturally makes this two points the endpoints of the base of an isosceles triangle, because the vertex could be anywhere on the mirror line. See:
GMAT Math: Special Properties of the Line y = x
Without that insight, we have no idea, and we would have to spend 10-15 minutes doing a ton of algebra. That approach is a strategic disaster and will not prepare you for doing math on the GMAT. Most harder GMAT Quant questions are about having the right insight, the insight that simplifies the problem.
See the blog article:
How to do GMAT Math Faster
Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
General Discussion
Use the distance formula.
The line between (-1,-3) and (3,1) forms the "base" of the isosceles triangle. We just need to find the distance to the apex of the triangle. The two legs, or distances will be equal, so set them equal to each other.
d=sqrt((3-x)^2+(1-6)^2)=sqrt((-1-x)^2+(-3-6)^2)
x=-6
Answer:
(D) –6
The line between (-1,-3) and (3,1) forms the "base" of the isosceles triangle. We just need to find the distance to the apex of the triangle. The two legs, or distances will be equal, so set them equal to each other.
d=sqrt((3-x)^2+(1-6)^2)=sqrt((-1-x)^2+(-3-6)^2)
x=-6
Answer:
(D) –6
mike34170
Joined: 10 Jun 2013
Last visit: 03 Feb 2016
Posts: 12
Concentration: General Management, Technology
Schools: HBS '18 Wharton '18 Stanford '18 CBS '18 LBS '18
GMAT Date: 06-26-2015
WE:Corporate Finance (Finance: Venture Capital)
Schools: HBS '18 Wharton '18 Stanford '18 CBS '18 LBS '18
Posts: 12
peachfuzz
I've handled it the same way and got the same result.
But how do we know that J and K form the base ? My guess is that if it doesn't, we could have other solutions
mike34170
Yeah, its because the y coord is fixed at 6. Any other value for y would not result in an isosceles triangle.
Can someone please explain why we consider JK as the base and not a side that could be equal to other side?
Also, do we have to plug in all answer choices to see which is the correct one by using distance formula or do we get some quadratic equation in terms of x if we equate the two distances KL and JL?
Also, do we have to plug in all answer choices to see which is the correct one by using distance formula or do we get some quadratic equation in terms of x if we equate the two distances KL and JL?
saikrishna1987
Joined: 02 Dec 2015
Last visit: 05 Nov 2018
Posts: 1
Location: India
Concentration: International Business, Human Resources
GMAT 1: 630 Q49 V27
GMAT 1: 630 Q49 V27
Posts: 1
Lets assume JK is not considered as base. Then, JL=JK
then, \sqrt{((3-x)^2)+25} = \sqrt{(16+16)}
this doesn't give us integral values for x which is not one of the options.
So, we can consider JL=KL.
The rest is known.
then, \sqrt{((3-x)^2)+25} = \sqrt{(16+16)}
this doesn't give us integral values for x which is not one of the options.
So, we can consider JL=KL.
The rest is known.
mikemcgarry
Thank you Mr Mike. That was a great insight. I wish we were taught coordinate geometry like that in our school..
mikemcgarry
mikemcgarry i just read through the blogpost and feel like someone just gave me a key to solve these isosceles questions in coordinate geometry. What a brilliant blogpost and kudos for the easy language.
Posts: 1,324
Try to explain this without a graph (but will draw one if anyone is interested)
Given Points:
J (3 , 1)
K (-1 , -3)
L (x , 6)
1st Observation) Side JK is the NON-Equal Side
For this kind of question, rather than just using the Distance Formula and rushing to plug all the answers in, I find it easier to Visualize the Horizontal and Vertical Distance from Point to Point
Vertical Distance from J to K = (1 - (-3)) = 4 Units
Horizontal Distance from J to K = (-1 - 3) = 4 Units
Length J to K = sqrt(16 + 16)
We know that the Y Coordinate of L = 6
Vertical Distance from J to L = (6 - 1) = 5 Units
Given that the Closest X Value from a Horizontal Distance is -3 ----- Horizontal Distance MIN = 6
sqrt(25 + 36) will NOT EQUAL Side Length of JK - sqrt(16 + 16) ---- and since the Horizontal Distance will only Increase, Side JL will NEVER Equal Side JK
Vertical Distance from K to L = (-3 - 6) = 9 Units
Again, looking at all the X Values that are given as Answer Choices, there is no way to Find a Horizontal Distance from Point K's X Coordinate of -1 to any of the Values that will lead us to saying ---- Side JK can ever equal Side KL
Most of this can be done through observation, so it's not as long as it seems. I'm writing it out in case anyone is following.
Thus, the Equal Side must be: Side JL = Side KL
2nd) Determining the Horizontal and Vertical Distance between the K and L --- and J and L
since we know that these 2 Sides must now be equal, the (Horizontal Distance + Vertical Distance) must be EQUAL in order for the Pythagoras Theorem to give us Equal Length Sides
Given J (3 , 1) and the Y - Coordinate of L is 6:
Vertical Distance = 5
Given K (-1 , -3) and the Y -Coordinate of L is 6:
Vertical Distance = 9
So the X-Coordinate that we pick must Satisfy the following:
5 + (Horizontal Distance from Point J to L measured by the X-Coordinates) = 9 + (Horizontal Distance from Point K to L measured by the X-Coordinates
Starting with B = -4
5 + (+1 to -4 = 5 Units) = 9 + (-1 to -4 = 3 Units)
5 + 5 = 9 + 3
Does not work.
At this Point, I would say we would need a further X Distance along the (-)Negative X Axis such that the Point is Further Away from Point J in order to satisfy the above Equation
Trying D = -6
5 + (+1 to -6 = 7 units) = 9 + (-3 to -6 = 3 Units)
5 + 7 = 9 + 3
12 = 12
WORKS!
the X - Coordinate of L must be = -6
Just to prove this worked (now that the problem is over) we can test the Distance Formula:
L(-6 , 6) ---- to ----- J (3 , 1) = sqrt( (9^2) + (5^2) ) = sqrt(81 + 25) = sqrt(106)
L(-6, 6) --- to ---- K (-1 , -3) = sqrt( (5^2) + (9^2) ) = sqrt (25 + 81) = sqrt(106
2 Sides of the Triangle are Equal, thus making it an Isosceles Triangle.
Takeaway: sometimes visualizing these harder problems is better than trying to plug-in to formulas. A lot of what I wrote could have been down with the Graphed Paper you get during the GMAT Exam and some Visualization.
-Answer -D-
Given Points:
J (3 , 1)
K (-1 , -3)
L (x , 6)
1st Observation) Side JK is the NON-Equal Side
For this kind of question, rather than just using the Distance Formula and rushing to plug all the answers in, I find it easier to Visualize the Horizontal and Vertical Distance from Point to Point
Vertical Distance from J to K = (1 - (-3)) = 4 Units
Horizontal Distance from J to K = (-1 - 3) = 4 Units
Length J to K = sqrt(16 + 16)
We know that the Y Coordinate of L = 6
Vertical Distance from J to L = (6 - 1) = 5 Units
Given that the Closest X Value from a Horizontal Distance is -3 ----- Horizontal Distance MIN = 6
sqrt(25 + 36) will NOT EQUAL Side Length of JK - sqrt(16 + 16) ---- and since the Horizontal Distance will only Increase, Side JL will NEVER Equal Side JK
Vertical Distance from K to L = (-3 - 6) = 9 Units
Again, looking at all the X Values that are given as Answer Choices, there is no way to Find a Horizontal Distance from Point K's X Coordinate of -1 to any of the Values that will lead us to saying ---- Side JK can ever equal Side KL
Most of this can be done through observation, so it's not as long as it seems. I'm writing it out in case anyone is following.
Thus, the Equal Side must be: Side JL = Side KL
2nd) Determining the Horizontal and Vertical Distance between the K and L --- and J and L
since we know that these 2 Sides must now be equal, the (Horizontal Distance + Vertical Distance) must be EQUAL in order for the Pythagoras Theorem to give us Equal Length Sides
Given J (3 , 1) and the Y - Coordinate of L is 6:
Vertical Distance = 5
Given K (-1 , -3) and the Y -Coordinate of L is 6:
Vertical Distance = 9
So the X-Coordinate that we pick must Satisfy the following:
5 + (Horizontal Distance from Point J to L measured by the X-Coordinates) = 9 + (Horizontal Distance from Point K to L measured by the X-Coordinates
Starting with B = -4
5 + (+1 to -4 = 5 Units) = 9 + (-1 to -4 = 3 Units)
5 + 5 = 9 + 3
Does not work.
At this Point, I would say we would need a further X Distance along the (-)Negative X Axis such that the Point is Further Away from Point J in order to satisfy the above Equation
Trying D = -6
5 + (+1 to -6 = 7 units) = 9 + (-3 to -6 = 3 Units)
5 + 7 = 9 + 3
12 = 12
WORKS!
the X - Coordinate of L must be = -6
Just to prove this worked (now that the problem is over) we can test the Distance Formula:
L(-6 , 6) ---- to ----- J (3 , 1) = sqrt( (9^2) + (5^2) ) = sqrt(81 + 25) = sqrt(106)
L(-6, 6) --- to ---- K (-1 , -3) = sqrt( (5^2) + (9^2) ) = sqrt (25 + 81) = sqrt(106
2 Sides of the Triangle are Equal, thus making it an Isosceles Triangle.
Takeaway: sometimes visualizing these harder problems is better than trying to plug-in to formulas. A lot of what I wrote could have been down with the Graphed Paper you get during the GMAT Exam and some Visualization.
-Answer -D-
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