Question

if the line x-4y-6=0 is the perpendicular bisector of the line segment PQ and the coordinate of P are (1,3),find the co-ordinates of Q.

Answer 1

x-4y-6 = 0

Express in the form of y=mx+c, where m is the slope

x-4y-6=0
-4y = -x+6
y=1/4 x - 3/2

When two lines are perpendicular, the product of their slope =  -1
Slope of the given line = 1/4
Let slope of PQ =  m2
1/4 x m2 = -1
m2= -4

Let (x, y) be a general point on PQ
Using this point to get the equation of PQ, since we know the slope of PQ

(y-3)/(x-1) = -4
y-3 = -4(x-1)
y-3 = -4x - 4
y = -4x -1

The mid-point is the point of intersection of the two lines:

So, at that point
1/4 x - 3/2 = -4x -1
1/4x + 4x = -1 + 3/2
4  1/4 x = 1/2
x = 2/17

y= -4x -1 = -4(2/17) - 1
y = -1  8/17

The mid-point is (2/17,  -1 8/17)

Let point Q be (a , b)

Getting the mid-point using point P

mid-point = {(a+1)/2  , (b+3)/2} = (2/17,  -1 8/17)

⇒(a+1)/2 = 2/17
a+1 = 2 x 2/17 = 4/17
a = -13/17

⇒ (b+3)/2 = -1  8/17
b+3 = -2  16/17
b= -5  16/17

So Q is (-13/17,  -5 16/17)