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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Joined: 09 Sep 2010 Posts: 12 Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink] ### Show Tags 18 Feb 2011, 09:13 1 11 00:00 Difficulty: 65% (hard) Question Stats: 57% (01:53) correct 43% (02:15) wrong based on 377 sessions ### HideShow timer Statistics 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. ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 50578 Re: Co-ordinate Geometry- Area of a Triangle [#permalink] ### Show Tags 18 Feb 2011, 09:55 11 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). Attachment: render.php (1) (1).png [ 17.42 KiB | Viewed 9513 times ] (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. _________________ ##### General Discussion Retired Moderator Joined: 20 Dec 2010 Posts: 1829 Re: Co-ordinate Geometry- Area of a Triangle [#permalink] ### Show Tags 18 Feb 2011, 09:37 2 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. Ans: "A" _________________ Intern Joined: 09 Sep 2010 Posts: 12 Re: Co-ordinate Geometry- Area of a Triangle [#permalink] ### Show Tags 18 Feb 2011, 18:51 I was stuck at understanding Statement (1) for a long time. That diagram really clears my doubt. Thanks guys Manager Joined: 16 May 2011 Posts: 189 Concentration: Finance, Real Estate GMAT Date: 12-27-2011 WE: Law (Law) Re: Co-ordinate Geometry- Area of a Triangle [#permalink] ### Show Tags 10 Jun 2011, 02:13 another simple and straightforward from Bunuel. must admit, i enjoy reading your posts. Manager Status: And the Prep starts again... Joined: 03 Aug 2010 Posts: 115 Re: Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink] ### Show Tags 08 Mar 2012, 07:25 Bunuel you genius... Best explanation as always. _________________ My First Blog on my GMAT Journey Arise, Awake and Stop not till the goal is reached Manager Joined: 12 Feb 2012 Posts: 125 If vertices of a triangle have coordinates [#permalink] ### Show Tags 17 Sep 2012, 14:05 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. Attachments gmat_delete2.PNG [ 10.16 KiB | Viewed 6688 times ] gmat_delete.PNG [ 8.9 KiB | Viewed 6685 times ] Intern Joined: 06 Apr 2011 Posts: 13 Re: Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink] ### Show Tags 18 Mar 2013, 13:04 Bunuel that was a simply marvelous No need for any calculations to solve the question Intern Joined: 15 Jul 2012 Posts: 34 Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink] ### Show Tags 03 Aug 2014, 05:34 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)? Math Expert Joined: 02 Sep 2009 Posts: 50578 Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink] ### Show Tags 12 Aug 2014, 01:44 saggii27 wrote: 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)? We can consider any side of a triangle to be the base. _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6500 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink] ### Show Tags 10 Nov 2015, 10:41 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 ? (1) |y-1| =1 (2) Angle at Vertex (x,y) = 90 degrees 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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