If x + 3y = 16 is the perpendicular bisector of AB and A(5, 7), then B =
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it is given that x + 3y = 16 is the perpendicular bisector of AB where A (5, 7)
we have to find the co-ordinates of point B.
as x + 3y = 16 is perpendicular on AB, slope of lines must be -1.
i.e., slope of {x + 3y = 16} × slope of AB = -1
⇒-1/3 × slope of AB = -1
⇒slope of line AB = 3
let point B = (x, y)
then, slope of AB = (y - 7)/(x - 5)
(y - 7)/(x - 5) = 3
⇒y - 7 = 3x - 15
⇒y - 3x + 8 = 0...........(1)
now middle point of AB =[ (x + 5)/2 , (y + 7)/2] this point lies at one end of the line x + 3y = 16
so, putting it on x + 3y = 16
⇒(x + 5) + 3(y + 7) = 32
⇒x + 3y = 6 .........(2)
from equations (1) and (2) we get,
y = 1, x = 3
therefore co-ordinate of point B = (3,1)
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