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If the vertices of a triangle have coordinates \((x, 1)\), \((5, 1)\), and \((5, y)\) where \(x \lt 5\) and \(y \gt 1\), what is the area of the triangle?

(1) \(x = y\)

(2) Angle at the vertex \((x, 1)\) is equal to angle at the vertex \((5, y)\)

Notice that vertex (x,1) will be somewhere on the green line segment and the vertex (5,y) will be somewhere on the blue line segment. So, in any case our triangle will be right angled, with a right angle at vertex (5, 1). Next, the length of the leg on the green line segment will be \(5-x\) and the length of the leg on the blue line segment will by \(y-1\). So, the area of the triangle will be: \(\text{area}=\frac{1}{2}*(5-x)*(y-1)\).

(1) \(x=y\). Since \(x \lt 5\) and \(y \gt 1\) then both \(x\) and \(y\) are in the range (1,5): \(1 \lt (x=y) \lt 5\). If we substitute \(y\) with \(x\) we'll get: \(\text{area}=\frac{1}{2}*(5-x)*(y-1)=\frac{1}{2}*(5-x)*(x-1)\), different values of \(x\) give different values for the area (even knowing that \(1 \lt x \lt 5\)). Not sufficient.

(2) Angle at the vertex \((x,1)\) is equal to angle at the vertex \((5,y)\). We have an isosceles right triangle: \(5-x=y-1\). Again if we substitute \(y-1\) with \(5-x\) we'll get: \(\text{area}=\frac{1}{2}*(5-x)*(y-1)=\frac{1}{2}*(5-x)*(5-x)\), different values of \(x\) give different values for the area. Not sufficient.

(1)+(2) \(x=y\) and \(5-x=y-1\). Solve for \(x\): \(x=y=3\), so \(\text{area}=\frac{1}{2}*(5-3)*(3-1)=2\). Sufficient.

I have a concern regarding this question. I think OA should be E not C.

How can x=y translates into 5−x=y−1 ??

Please see the attached pic, using which we can see that even when X = Y areas could be different.

Please assist.

Thanks

Hi WillGetIt,

With both conditions being true, you will get only a single isosceles triangle ((3,1), (5,1), (5,3)). Only these points will satisfy both requirements.

Already explained well. For the area, we need both height and base. So, here it is only possible, if we can know all vertices. Can you get the answer using (1) only, no, since many possibilities such as, (2,1),(5,2);(3,1),(5,3);(4,1),(5,4)..etc.Insufficient Forget about (1), think on (2), but still many possibilities when equal angles give us equal sides,such as (4,1),(5,2);(3,1),(5,3);(1,1),(5,1)..etc, Insufficient. So, A,B,D all cancelled since not at least one sufficiency. Now combine both, ...(2) requires that both distance from 5 will be equal. Make it narrower using (1)..so that finally x=y. Only one case is possible here which is (3,1),(5,3). Because we can calculate the area combinedly, C is the answer.

I think this is a high-quality question and I agree with explanation. Great question. Any similar co-ordinate geometry questions for practice outside GMATCLUB tests please suggest such questions are awesome man!

I think this is a high-quality question and I agree with explanation. Great question. Any similar co-ordinate geometry questions for practice outside GMATCLUB tests please suggest such questions are awesome man!

If the vertices of a triangle have coordinates \((x, 1)\), \((5, 1)\), and \((5, y)\) where \(x \lt 5\) and \(y \gt 1\), what is the area of the triangle?

(1) \(x = y\)

(2) Angle at the vertex \((x, 1)\) is equal to angle at the vertex \((5, y)\)

Responding to a pm:

Angle at (5, 1) will be 90 degrees.

(1) \(x = y\) The co-ordinates are (x, 1), (5, 1) and (5, x). Not sufficient alone since we don't know x.

(2) Angle at the vertex \((x, 1)\) is equal to angle at the vertex \((5, y)\)[/quote] This means the two sides are equal. So 5 - x = y - 1 Not sufficient alone.

Using both statements,

5 - x = x - 1 x = 3 = y

We get all co-ordinates so we will be able to find the area. Answer (C)
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I am able to understand line distance as y-1 is because of sqrt((5-5)^2+(y-1)^2) = y-1 but, why the other distance is 5-x and not x-5 as the calculation would be sqrt((x-5)^2/+(1-1)^2) = x-5

In this case, from the problem statement itself we can deduce that x & y are nothing but the two sides of the right angled triangle. So, doesn't x = y itself imply that the triangle is iso. rt angled triangle?In that case we already get to know that the angles are equal. Please let me know if I am missing something/understanding is incorrect.

In this case, from the problem statement itself we can deduce that x & y are nothing but the two sides of the right angled triangle. So, doesn't x = y itself imply that the triangle is iso. rt angled triangle?In that case we already get to know that the angles are equal. Please let me know if I am missing something/understanding is incorrect.

TIA.

The length of the leg on the green line segment will be 5−x and the length of the leg on the blue line segment will by y−1, NOT x and y. So, x = y does NOT mean that we have a right isosceles triangle.
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