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ABC is a right angle triangle, right angled at B Co ordinate

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ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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ABC is a right angle triangle, right angled at B Co ordinates of A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively. Length of line AC is 5 units, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the slope of line AC?

Pls note fig is drawn in 1st quadrant with Z > Y i.e point B is above A.

A. -4/3
B. 2/3
C. 1
D. 4/3
E. 2
[Reveal] Spoiler: OA

Last edited by Archit143 on 11 Nov 2012, 17:35, edited 1 time in total.
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 11 Nov 2012, 17:12
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Archit143 wrote:
ABC is a right angle triangle, right angled at B Co ordinates of A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively. Length of line AC is 5 units, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the slope of line AC?
A. -4/3
B. 2/3
C. 1
D. 4/3
E. 2

I'm happy to help with this. :-)

We know the hypotenuse AC = 5, and we know BC = 3, because B & C have the same y-coordinate (BC is horizontal) and they are separated in the horizontal direction by 3.

If a right triangle has a hypotenuse of 5 and one leg of 3, the other leg has to be 4. That is the inescapable conclusion of the Pythagorean Theorem. We must have a 3-4-5 triangle.

The trouble is ---- we don't know if Z > Y or Z < Y ---- the problem provides no information on that point. Thus, we have a horizontal segment BC of length 3, with a right angle at B, but we don't know whether the perpendicular segment AB goes up or down from B. We know AB must have a length of 4 and that it must make a right angle with B, but we don't know whether the direction from B to A is up or down. Therefore, the slope of AB could be either +4/3 or -4/3. Two answers are possible here.

I suspect either something was not copied correctly from the source, or that the source is faulty.

Let me know if you have any questions.

Mike :-)
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 11 Nov 2012, 17:28
mikemcgarry wrote:
Archit143 wrote:
ABC is a right angle triangle, right angled at B Co ordinates of A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively. Length of line AC is 5 units, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the slope of line AC?
A. -4/3
B. 2/3
C. 1
D. 4/3
E. 2

I'm happy to help with this. :-)

We know the hypotenuse AC = 5, and we know BC = 3, because B & C have the same y-coordinate (BC is horizontal) and they are separated in the horizontal direction by 3.

If a right triangle has a hypotenuse of 5 and one leg of 3, the other leg has to be 4. That is the inescapable conclusion of the Pythagorean Theorem. We must have a 3-4-5 triangle.

The trouble is ---- we don't know if Z > Y or Z < Y ---- the problem provides no information on that point. Thus, we have a horizontal segment BC of length 3, with a right angle at B, but we don't know whether the perpendicular segment AB goes up or down from B. We know AB must have a length of 4 and that it must make a right angle with B, but we don't know whether the direction from B to A is up or down. Therefore, the slope of AB could be either +4/3 or -4/3. Two answers are possible here.

I suspect either something was not copied correctly from the source, or that the source is faulty.

Let me know if you have any questions.

Mike :-)


Hi Mike

First of all thanks for replying to my post.... The actual question is to find the equation of AC which is not difficult after we get the slope.

Pls find below the actual question

In the figure to the left, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the equation of line AC?

The co ordinates remains the same as in the original question.

My doubt is
When we get the 3 - 4 -5 triangle with clear value as 3 , 4 and 5 not as ratio, also from the equation we know that change in y axis (delta Y ) is 4 and that of X axis is 3. why cant we conclude straight away that the slope is 4/3 .
Yes if we substitute the value of Z as y+ 4 than we get - 4/3

Why is it so..............
The solution assumes the value to be - 4/3 and calculates the equation of line.
Question; why is there two possible SLOPE of a single line.........
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 11 Nov 2012, 17:34
The line segment AB goes up in the.
The fig is drawn in 1st quadrant and Z > Y
I think that will help to arrive a conclusion

Sorry for the trouble i ll edit the original question.
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 28 Dec 2013, 11:05
Archit143 wrote:
ABC is a right angle triangle, right angled at B Co ordinates of A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively. Length of line AC is 5 units, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the slope of line AC?

Pls note fig is drawn in 1st quadrant with Z > Y i.e point B is above A.

A. -4/3
B. 2/3
C. 1
D. 4/3
E. 2


A is the answer to this one

Let me explain

We need to draw the triangle first. Now, x =3 so we are going to have a pythagorean triple 3-4-5.
Now are coordinate for A (3,z) C (6,y)

Slope will be (y-z )/ (6-3)

Do we know y-z? Yes we do from the height of the right triangle we get that difference of z-y = 4

Hence y-z will just be -4

So the slope is -4/3

Answer is A

Hope it helps
Cheers!

J :)
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 18 Jan 2014, 13:20
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From the option, it is easy to find out the answer.

Right angled triangle == Negative slope. Only one negative option.

So answer A

Method 2:

Since it is coordinate geometry , consider x = 0, y=0.

Then the coordinates A = (0,z) B = (0,0) C = (3,y)

Slope of the equation = (y2-y1)/(x2-x1) and the distance between AC = 5.

So (y-z)^2 = 16 or y-z = + or - 4.

Slope of the hypotenuse is -4/3 or 4/3 . But the hypotenuse will have negative slope, so the answer = -4/3

So the equation will be in the form of 4x+3y+K = 0.
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 18 Jan 2014, 21:41
mikemcgarry wrote:
Archit143 wrote:
ABC is a right angle triangle, right angled at B Co ordinates of A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively. Length of line AC is 5 units, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the slope of line AC?
A. -4/3
B. 2/3
C. 1
D. 4/3
E. 2

I'm happy to help with this. :-)

We know the hypotenuse AC = 5, and we know BC = 3, because B & C have the same y-coordinate (BC is horizontal) and they are separated in the horizontal direction by 3.

If a right triangle has a hypotenuse of 5 and one leg of 3, the other leg has to be 4. That is the inescapable conclusion of the Pythagorean Theorem. We must have a 3-4-5 triangle.

The trouble is ---- we don't know if Z > Y or Z < Y ---- the problem provides no information on that point. Thus, we have a horizontal segment BC of length 3, with a right angle at B, but we don't know whether the perpendicular segment AB goes up or down from B. We know AB must have a length of 4 and that it must make a right angle with B, but we don't know whether the direction from B to A is up or down. Therefore, the slope of AB could be either +4/3 or -4/3. Two answers are possible here.

I suspect either something was not copied correctly from the source, or that the source is faulty.

Let me know if you have any questions.

Mike :-)


It is 100% choice D because it is inferable that with the coordinate description (X,Z), (X,Y), (X+3,Y) that \(X=X\) and for Y to be parallel, the (X,Z) and (x+3,Y) relationship must be positive
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 19 Jan 2014, 15:18
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Archit143 wrote:
ABC is a right angle triangle, right angled at B Co ordinates of A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively. Length of line AC is 5 units, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the slope of line AC?

Pls note fig is drawn in 1st quadrant with Z > Y i.e point B is above A.

A. -4/3
B. 2/3
C. 1
D. 4/3
E. 2

Apparently Archit added this very important note after I wrote my initial response to the problem. As they say, a picture is worth a thousand words, and a GMAT math problem that includes a diagram often will make little or no sense without the diagram. The GMAT never gives you a diagram unless there's some crucial piece of information you need from the diagram: this is a very important point to appreciate.

With this added piece of information, it's perfectly clear that the answer is (A). If Z > Y, then the line must move DOWN from (X,Z) to (X+3, Y). That's the definition of a negative slope. With the analysis above, we saw that the slope would have to be either -4/3 or +4/3. With the information from the diagram it's clear that the slope is -4/3, answer = (A).

Please let me know if anyone has any further questions.
Mike :-)
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 17 Jan 2015, 08:40
Archit143 wrote:
ABC is a right angle triangle, right angled at B Co ordinates of A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively. Length of line AC is 5 units, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the slope of line AC?

Pls note fig is drawn in 1st quadrant with Z > Y i.e point B is above A.

A. -4/3
B. 2/3
C. 1
D. 4/3
E. 2


In triangle ABC line AB is parallel to Y-axis and line BC is parallel to X axis . Z>Y so line AC will be in 1 st quadrant
AC=5, BC= 3 and AB=\sqrt{(5^2-3^2)}=4
angle ACB, Tan(ACB)=4/3,NOTE : Slope of any line is always taken from positive x-axis in anticlockwise direction.
Slope=tan(180-ACB)=-tan (ACB)=-4/3

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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 20 Feb 2015, 07:36
mikemcgarry wrote:
Archit143 wrote:
ABC is a right angle triangle, right angled at B Co ordinates of A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively. Length of line AC is 5 units, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the slope of line AC?

Pls note fig is drawn in 1st quadrant with Z > Y i.e point B is above A.

A. -4/3
B. 2/3
C. 1
D. 4/3
E. 2

Apparently Archit added this very important note after I wrote my initial response to the problem. As they say, a picture is worth a thousand words, and a GMAT math problem that includes a diagram often will make little or no sense without the diagram. The GMAT never gives you a diagram unless there's some crucial piece of information you need from the diagram: this is a very important point to appreciate.

With this added piece of information, it's perfectly clear that the answer is (A). If Z > Y, then the line must move DOWN from (X,Z) to (X+3, Y). That's the definition of a negative slope. With the analysis above, we saw that the slope would have to be either -4/3 or +4/3. With the information from the diagram it's clear that the slope is -4/3, answer = (A).

Please let me know if anyone has any further questions.
Mike :-)


Sorry i did not understand quite well .
Pls note fig is drawn in 1st quadrant with Z > Y i.e point B is above A.
how can point B lie above A if Z>Y ? A, B are (X,Z), (X,Y)
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 29 Jun 2015, 07:59
if the same question were to be (X-3), then it would be positive slope 4/3.
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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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Re: ABC is a right angle triangle, right angled at B Co ordinate [#permalink]

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New post 20 Aug 2016, 10:56
mikemcgarry wrote:
Archit143 wrote:
ABC is a right angle triangle, right angled at B Co ordinates of A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively. Length of line AC is 5 units, x and y are greater than or equal to 0, the vertices of triangle ABC have coordinate values as shown, and the value of AC is as shown. If x = 3, then what could be the slope of line AC?

Pls note fig is drawn in 1st quadrant with Z > Y i.e point B is above A.

A. -4/3
B. 2/3
C. 1
D. 4/3
E. 2

Apparently Archit added this very important note after I wrote my initial response to the problem. As they say, a picture is worth a thousand words, and a GMAT math problem that includes a diagram often will make little or no sense without the diagram. The GMAT never gives you a diagram unless there's some crucial piece of information you need from the diagram: this is a very important point to appreciate.

With this added piece of information, it's perfectly clear that the answer is (A). If Z > Y, then the line must move DOWN from (X,Z) to (X+3, Y). That's the definition of a negative slope. With the analysis above, we saw that the slope would have to be either -4/3 or +4/3. With the information from the diagram it's clear that the slope is -4/3, answer = (A).

Please let me know if anyone has any further questions.
Mike :-)


The question is still wrong and confusing. The word RESPECTIVELY is not given any respect in this question.
A, B and C are (X,Z), (X,Y) and (X+3, Y) respectively
Pls note fig is drawn in 1st quadrant with Z > Y i.e point B is above A.

These statements do not hold true together.
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Re: ABC is a right angle triangle, right angled at B Co ordinate   [#permalink] 20 Aug 2016, 10:56
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