In the rectangular coordinate system shown above, points O, P, and Q r

All points equidistant from O and Q lie on the line x = 2 so the fire station should lie on this line.
All points equidistant from O and P lie on the line y = 3 so the fire station should lie on this line too.
These two intersect at (2, 3) and that will be the point equidistant from all 3 points.

Answer (E)

Or

You can think of the question in terms of the perpendicular bisectors of triangle OPQ. Their point of intersection will be equidistant from all three vertices.
Again the perpendicular bisector of OQ will be x = 2 and of OP will be y = 3. They will intersect at (2, 3).
Answer (E)

Answer is E : (2,3)

the point we're asked is the Centroid of the triangle (scalene). Let it be (x,y). Then it should have same distance from all three points.So using the Pythagoras theorem of distance between 2 points:
distance from origin = distance from (4,0)
\(\sqrt{(x^2 + y^2)} = \sqrt{((4-x)^2 + y^2)}\)

distance from origin = distance from (0,6)
\(\sqrt{(x^2 + y^2)} = \sqrt{(x^2 + (6-y)^2)}\)

Solve these 2 for x and y and it will give (2,3)



(Share a kudo's , if you like the explanation)

The point should be the circumcentre of the triangle and in right angled triangle circumcentre lies on hypotenuse and is mid-point of it . Therefore it can be directly seen that mid point of the hypotenuse will be E.(2,3).

Is it a general fact, that bisectors of the two legs of a right triangle will cut the hypotenuse exactly in half, in other words, intersect the hypotenuse at its midpoint?

Yes, that is correct. The circumcenter of a right triangle (the point of intersection of perpendicular bisectors) will bisect the hypotenuse.

Another way to phrase this property is: in a right triangle, the median drawn to the hypotenuse has the measure half the hypotenuse.
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I followed the midpoint property of co-ordinate geometry and got the same result. Is there any problem ? Bunuel, VeritasPrepKarishma, chetan2u, JeffTargetTestPrep, GMATPrepNow.

All three points would be on the periphery of a circle while the fire station would be at the center.
So, (X-a)^2+(Y-b)^2=C^2, where (a,b) are the three points.
Solving three equations we get (X,Y) as (2,3).

Use Centroid of Triangle formula.

Centroid of Triangle is (x1+x2+x3)/3, (y1+y2+y3)/3

here, X and Y are points of triangle.

Well sunngupt11,

If you use centroid formula you get\(\frac {4}{3}\), \(\frac {6}{3}\)

Also centroid of triangle lies inside the triangle. If you look at the responses above many have solved this using circumradius or perpendicular bisectors of triangle.

Now the three perpendicular bisectors of triangle can meet inside the triangle or outside the triangle .

I guess what must have confused you is that centroid divides the media in a ratio of 2:1. but that dosen't mean all the medians of a triangle will be of equal length.
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knowing that it is a right triangle, is using the midpoint formula suitable?

(X1 + x2)/ 2, (y1+y2)/2

(0 + 4) / 2 , (0 + 6) / 2

2,3

Responding to a pm:


Not sure how you get (y-6)^2= (x-4)^2

On solving, I get y^2 = (y - 6)^2 which gives y = 3
and x^2 = (x - 4)^2 which gives x = 2

Point will be equidistant from all three developments if it lies on the perpendicular bisector of line joining PQ i.e, MID POINT OF PQ(2,3) as per the given options.

Can we find mid point?

(0+4)/2 and (6+0)/2

Can we approach the problem like below?

Since it is asked us to find a point in the coordinate system, which must be equal distances to the other points, can't we use the distance between two points or difference between 2 points? In differences, the only right answer will be E.

Solution:

By looking at the graph, the easiest way is to take the midpoint of PQ (since it is already equidistant to P and Q) as the locations of the fire station. Let it be F. So F = ((0 + 4)/2, (6 + 0)/2) = (2, 3). We see that this point is also equidistant from O = (0, 0). Therefore, (2, 3) must be the location where the fire station should be built.

Alternate Solution:

Recall that the set of points equidistant from two given points is a line that passes through the midpoint of the two given points and that is perpendicular to the line passing through the two points.

Using this fact, we see that every point on the line y = 3 is equidistant to P and O (since the midpoint of P and O is (0, 3)), and every point on the line x = 2 is equidistant to O and Q (since the midpoint of O and Q is (2, 0)). The intersection of the lines y = 3 and x = 2 is the point (2, 3), and this point is equidistant to P, O and Q.

Answer: E
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We have two coordinates located at (0,6) and (4,0). What point is equidistant?

What's the midpoint of 0 and 4? 2. The x-coordinate will be 2.

What's the midpoint of 6 and 0? 3. The y-coordinate will be 3.

Thus, we have (2,3). Answer is E.
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OFFICIAL GMAT EXPLANATION

Geometry Coordinate geometry

Any point equidistant from the points (0,0) and (4,0) must lie on the perpendicular bisector of the segment with endpoints (0,0) and (4,0), which is the line with equation x = 2. Any point equidistant from the points (0,0) and (0,6) must lie on the perpendicular bisector of the segment with endpoints (0,0) and (0,6), which is the line with equation y = 3. Therefore, the point that is equidistant from (0,0), (4,0), and (0,6) must lie on both of the lines x = 2 and y = 3, which is the point (2,3).

Alternatively, let (x, y) be the point equidistant from (0,0), (4,0), and (0,6). Since the distance between (x, y) and (0,0) is equal to the distance between (x, y) and (4,0), it follows from the distance formula that √x^2 + y^2 = √(x−4)^2 + y^2. Squaring both sides gives x^2 + y^2 = (x − 4)^2 + y^2. Subtracting y^2 from both sides of the last equation and then expanding the right side gives x^2 = x^2 − 8x + 16, or 0 = −8x + 16, or x = 2. Also, since the distance between (x, y) and (0,0) is equal to the distance between (x, y) and (0,6), it follows from the distance formula that √x^2 + y^2 = √x^2 + (y−6)^2. Squaring both sides of the last equation gives x^2 + y^2 = x^2 + (y − 6)^2. Subtracting x^2 from both sides and then expanding the right side gives y^2 = y^2 − 12y + 36, or 0 = −12y + 36, or y = 3.

it can be solved by using midpoint formula: (x1+x2)/2; (y1+y2)/2
(4+0)/2 ;(6+0)/2
Ans: 2,3