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In the rectangular coordinate system above, the line y = x i

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by BTGmoderatorDC » Wed Jul 31, 2019 5:35 pm

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Answers

A

B

C

D

E

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Difficulty—

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In the rectangular coordinate system above, the line y = x is the perpendicular bisector of segment AB (not shown), and the x-axis is the perpendicular bisector of segment BC (not shown). If the coordinates of point A are (2,3), what are the coordinates of point C ?

(A) (-3,-2)
(B) (-3,2)
(C) (2,-3)
(D) (3,-2)
(E) (2,3)

OA D

Source: Official Guide
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by Jay@ManhattanReview » Wed Jul 31, 2019 11:21 pm
BTGmoderatorDC wrote:Image

In the rectangular coordinate system above, the line y = x is the perpendicular bisector of segment AB (not shown), and the x-axis is the perpendicular bisector of segment BC (not shown). If the coordinates of point A are (2,3), what are the coordinates of point C ?

(A) (-3,-2)
(B) (-3,2)
(C) (2,-3)
(D) (3,-2)
(E) (2,3)

OA D

Source: Official Guide
We know that y = x bisects the line segment AB and the coordinates of A are (2, 3); thus, coordinates of B would be (3, 2) (x and y coordinates of A are swapped). Again, Since X-axis is the bisector of BC, and the coordinates of B are (3, 2), the coordinates of C would be (3, -2). Point C would be refection of pont B on X plane. Or, C would line on the IV quadrant.

The correct answer: D

Hope this helps!

-Jay
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by swerve » Thu Aug 01, 2019 4:14 pm
Since the line \(y=x\) is the perpendicular bisector of segment \(AB\), then point \(B\) is the mirror reflection of point \(A\) around the line \(y=x\), so its coordinates are \((3, 2)\). In any mirror reflection around the line \(y = x\), the \(x-\)coordinate and the \(y-\)coordinate of a point become interchanged.

The same way, since the \(x-\)axis is the perpendicular bisector of segment \(BC\) then the point \(C\) is the mirror reflection of point \(B\) around the \(y-\)axis, so its coordinates are \((3, -2)\). In any mirror reflection around the \(x-\)axis, the \(x-\)coordinate remains the same, and the sign of the \(y-\)coordinate changes.

Therefore, the correct answer is __D__
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