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Re: Co-ordinate Geometry- Area of a Triangle [#permalink]

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18 Feb 2011, 10:37

1

This post received KUDOS

Area of triangle= 1/2*base*height

Consider line segment joining (-2,1) and (3,1) as Base. Base=5. It is a line parallel to x-axis that intercepts at y=1.

Now; the y-coordinate of (x,y) can't be 1.

1) |y-1|=1 y-1=1 y=2 or y-1=-1 y=0

We know the y-coordinate of the third vertex. It's either 1 or 0.

Say y=1; Irrespective of what x is; the distance between base and the vertex will be 1.

Distance is always the perpendicular distance. from the base i.e. make a line of infinite length from vertex(x,y) parallel to the base. The perpendicular line from base to this line will be the distance.

So; height=1

Area = 1/2*base*height = 1/2*5*1=5/2

Likewise for y=0; base=5; height=1 Sufficient.

2) Knowing that it is a right triangle won't give us the area. We don't know the length of the sides. Neither do we know anything about the other two angles. Not sufficient.

(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

The vertices of triangle are (-2,1), (3,1) and (x,y) on a plane. What is the area of the triangle? a)|y-1|=1 b)The angle at vertex (x,y) is 90 degrees.

Can this be considered as a 700+ or 600-700 level question? I am new to GMAT and had no clue on this one.

Re: What is the area of the triangle? [#permalink]

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08 Mar 2012, 07:59

babusona wrote:

The vertices of triangle are (-2,1), (3,1) and (x,y) on a plane. What is the area of the triangle? a)|y-1|=1 b)The angle at vertex (x,y) is 90 degrees.

Can this be considered as a 700+ or 600-700 level question? I am new to GMAT and had no clue on this one.

Plot the given points on an xy graph. S1) You will see that y=0 or 2 and this means the height of the triangle is 1 for either case. Since we know the length of the base from points (-2,1) and (3,1), we can thus find the area of the triangle

S2) The value of angle at point (x,y) is not enough determine the height.

The vertices of triangle are (-2,1), (3,1) and (x,y) on a plane. What is the area of the triangle? a)|y-1|=1 b)The angle at vertex (x,y) is 90 degrees.

Can this be considered as a 700+ or 600-700 level question? I am new to GMAT and had no clue on this one.

Merging similar topics. Please refer to the solutions above and ask if anything remains unclear.

As for difficulty level: I'd say since the problem involves several concepts (geometry, coordinate geometry, absolute value) then it can be considered 650+ question.
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If vertices of a triangle have coordinates [#permalink]

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17 Sep 2012, 15:05

swatirpr wrote:

If vertices of a triangle have coordinates \((-2, 2), (3, 2), (x, y)\) , what is the area of the triangle?

1. \(|y - 2| = 1\) 2. angle at the vertex \((x, y)\) equals 90 degrees

Sorry to revive this old post.

So 1. \(|y - 2| = 1\) ===>\(y-2=1\)or\(y-2=-1\) ===> two solutions \(y=1\) or\(y=3\). Notice that that regardless of what value of x, the base will always be the distance between \((-2, 2) and (3, 2)\) and the height will always be 1. Therefore the area will always be the same.

2) We know the base and that the third vertex will form a right angle. Well if that is the case, the only right angles that can formed for a given line is one in which it is the diameter of a circle. The third point on this line will always form a right angled triangle. All right angled triangles have their hypotenuse as the diameter of a circle and the third vertex a point on the circle. As you can see, while we know that this third, unknown, vertex will be on this circle, we dont't know where. Therefore the right triangle will have different areas as the third point glides across the circle. The area will reach a maximum when the point forms a isoclese right triangle.

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Re: Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink]

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19 Jun 2014, 08:51

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Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink]

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03 Aug 2014, 06:34

Bunuel wrote:

sandhyash wrote:

Q - Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ?

1) |y-1| =1 2) Angle at Vertex (x,y) = 90 degrees

Source : MGMAT Test

Solution given : (A)

The explanation given in the solution is that :

From Statement (1) either y=0 or y=2, either way base is 5 and height =1. Hence area can be computed

Statement (2) is not enough

I did not understand this explanation. Please help !! Thanks in advance.

Below is almost identical question:

Vertices of a triangle have coordinates (-2, 2), (3, 2), (x, y). What is the area of the triangle?

(1) |y - 2| = 1 (2) angle at the vertex \((x, y)\) equals 90 degrees

Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).

(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.

Hey bunuel, brilliant explanation.

But, i have a silly doubt in this problem. arent we assuming that the base is fixed at (-2, 2), (3, 2). why cant the base be on the unknown point.

from 1st statement we get the value of y as 3 or 1. why cant the base be on (x,1) and (3,2) or (x,3) and (3,2)?

Q - Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ?

1) |y-1| =1 2) Angle at Vertex (x,y) = 90 degrees

Source : MGMAT Test

Solution given : (A)

The explanation given in the solution is that :

From Statement (1) either y=0 or y=2, either way base is 5 and height =1. Hence area can be computed

Statement (2) is not enough

I did not understand this explanation. Please help !! Thanks in advance.

Below is almost identical question:

Vertices of a triangle have coordinates (-2, 2), (3, 2), (x, y). What is the area of the triangle?

(1) |y - 2| = 1 (2) angle at the vertex \((x, y)\) equals 90 degrees

Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).

(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.

Hey bunuel, brilliant explanation.

But, i have a silly doubt in this problem. arent we assuming that the base is fixed at (-2, 2), (3, 2). why cant the base be on the unknown point.

from 1st statement we get the value of y as 3 or 1. why cant the base be on (x,1) and (3,2) or (x,3) and (3,2)?

We can consider any side of the triangle to be the base.
_________________

Re: Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink]

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08 Nov 2015, 07:55

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ?

If we modify the question, we only need to know the value of y, as the height is only affected by the value of y. From condition 1, |y-1|=1, y-1=-1,1, or y=0,2 Both makes the height 1, so this is sufficient and the answer is (A).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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