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02 Mar 2011, 06:10
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Difficulty:
55%
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Question Stats:
56% (01:42) correct
44%
(01:50)
wrong
based on 364
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Is point A closer to point (1,2) than to point (2,1) ?
(1) Point A lies on the line y = x
(2) Point A lies on the line y = -x
(1) Point A lies on the line y = x
(2) Point A lies on the line y = -x
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02 Mar 2011, 06:23
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Is point \(A\) closer to point \((1, 2)\) than to point \((2, 1)\)?
Refer to the diagram below.
(1) Point \(A\) lies on the line \(y = x\).
You can see that no matter where on blue line point A is, it will always be equidistant from the given points. Sufficient.
(2) Point \(A\) lies on the line \(y = -x\).
If A is on the red line we cannot say whether it's closer to point (1, 2) than to point (2, 1). Not sufficient.
Answer: A
graph.php.png [ 17.76 KiB | Viewed 10408 times ]
Refer to the diagram below.
(1) Point \(A\) lies on the line \(y = x\).
You can see that no matter where on blue line point A is, it will always be equidistant from the given points. Sufficient.
(2) Point \(A\) lies on the line \(y = -x\).
If A is on the red line we cannot say whether it's closer to point (1, 2) than to point (2, 1). Not sufficient.
Answer: A
Attachment:
graph.php.png [ 17.76 KiB | Viewed 10408 times ]
02 Mar 2011, 06:30
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Is point A closer to point (1,2) than to point (2,1) ?
1.Point A lies on the line y=x
2.Point A lies on the line y= -x
Equation of a line passing through two points (1,2) and (2,1)
\(\frac{1-2}{2-1}=\frac{y-2}{x-1}\)
\(y = -x + 3\)
Slope =-1
1. y=x;
Slope = 1
Thus this line is perpendicular to the line (y = -x+3)
Not only this is perpendicular, but also it bisects the line in two equal parts.
Thus, all points on this line will be equidistant from both points (1,2) and (2,1).
Sufficient.
How to be sure that it is perpendicular bisector;
Let us find the midpoint of (1,2) and (2,1)
(2+1)/2, (1+2)/2
3/2, 3/2
Does this point satisfy the equation y=x; indeed it does.
2. y = -x
This line passes through the origin and goes at 45 degrees in 2nd and 4th quadrant.
So; if we visualize, when x=2, y=-2;
Point 2,1 will be closer
if x=-2, y=2
Point 1,2 will be closer.
Not sufficient.
Ans: "A"
1.Point A lies on the line y=x
2.Point A lies on the line y= -x
Equation of a line passing through two points (1,2) and (2,1)
\(\frac{1-2}{2-1}=\frac{y-2}{x-1}\)
\(y = -x + 3\)
Slope =-1
1. y=x;
Slope = 1
Thus this line is perpendicular to the line (y = -x+3)
Not only this is perpendicular, but also it bisects the line in two equal parts.
Thus, all points on this line will be equidistant from both points (1,2) and (2,1).
Sufficient.
How to be sure that it is perpendicular bisector;
Let us find the midpoint of (1,2) and (2,1)
(2+1)/2, (1+2)/2
3/2, 3/2
Does this point satisfy the equation y=x; indeed it does.
2. y = -x
This line passes through the origin and goes at 45 degrees in 2nd and 4th quadrant.
So; if we visualize, when x=2, y=-2;
Point 2,1 will be closer
if x=-2, y=2
Point 1,2 will be closer.
Not sufficient.
Ans: "A"
