Question 30467
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

{{{ graph( 300, 300, -10, 10, -10, 10, ((x-2)/0.00001),1) }}}