SOLUTION: Please help me work this problem. I have the answer but don't know how to work it. Please explain in very plain English terms! Point A (-4,1) is in the standard (x,y) coordi

Question 30467: Please help me work this problem. I have the answer but don't know how to work it. Please explain in very plain English terms!
Point A (-4,1) is in the standard (x,y) coordinate plane. What must be the coordinates of Point B so that the line x = 2 is the perpendicular bisector of AB ? (note: there is a line over AB) Thank you for your help...Karen
Found 3 solutions by mbarugel, longjonsilver, venugopalramana:
Answer by mbarugel(146)   (Show Source): You can put this solution on YOUR website!
First of all, let's see what should the form of the segment AB be. We know that the line x = 2 is perpendicular to AB. Since x = 2 is a vertical line, then the segment AB must be horizontal (ie, parallel to the x-axis)
Given that the segment AB is horizontal, and that the point A is at (-4, 1), then the y-coordinate of B must be 1, just as the y-coordinate of A. If this weren't so, then the segment would not be horizontal. So the coordinates of point B must be (X, 1) (we still need to find X).
Now, we know that x=2 is not only perpendicular to AB, but it's also its bisector, meaning that it 'splits' the segment AB into two equally sized segments. We know that one of the extremes of the segment (the point A) is at X=-4, and that its middle point is at X=2. Therefore, it's clear that the other extreme of the segment must be at X=8 (there are 6 units from -4 to 2, and 6 units from 2 to 8; so 2 is the middle point of -4 and 8).
So we conclude that point B is located at (8, 1).
I hope this helps!
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Answer by longjonsilver(2297)   (Show Source): You can put this solution on YOUR website!
this is a very simple question really, just hidden in a lot of jargon.

Imagine the vertical line at x=2.
Imagine the point (-4,1)

Sketch a little hand-drawn graph of these.

Now, for the line x=2 to be perpendicular (ie at right angles) to the line AB, AB must be horizontal...since x=2 is vertical.

As this horizontal line passes through (-4,1), the equation of the line AB is y=1. However, the question doesn't ask for that. It asks for the point on the line AB (equation y=1) so that x=2 is a bisector.

Bisector means it cuts the line AB into 2 equal parts.

So, point (-4,1) is how far from x=2?... x=-4 and x=2? the seperation is 6.

So, the point B, on the other side must also be a distance of 6 away from the x=2 line. This means it mus be ay x=8. So the point is (8,1)

jon.

jon.

Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!
Point A (-4,1) is in the standard (x,y) coordinate plane. What must be the coordinates of Point B so that the line x = 2 is the perpendicular bisector of AB ? (note: there is a line over AB) Thank you for your help...Karen
LINE X=2 IS PARALLEL TO Y AXIS SO ITS PERPENDICULAR WOULD BE PARALLEL TO X AXIS
HENCE ITS EQUATION SHALL BE Y=K......THAT IS EQN.OF AB SHALL BE Y=CONSTANT
FURTHER X=2 BISECTS AB.HENCE MID POINT OF AB SHALL BE ON X=2
THAT IS X COORDINATE OF MIDPOINT SHALL BE EQUAL TO 2.IF WE TAKE POINT B AS (P,Q)
THEN X COORDINATE OF MID POINT OF AB SHALL BE = (-4+P)/2=2
-4+P=4
P=8
WE SPROVED EARLIER THAT EQN OF AB SHALL BE IN THE FORM Y=CONSTANT..OR ITS SLOPE IS ZERO SINCE Y=CONSTANT MEANS ...Y=0*X+CONSTANT
SLOPE OF AB =(Q-1)/(P-(-4))=0...SO Q-1=0...SO Q=1...
HENCE POINT B IS (8,1)
YOU WILL SEE ITS EQN. IS Y=1..TO UNDERSTAND HOW THEY LOOK LIKE SEE THE GRAPH BELOW

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