Question

the perpendicular bisector of the line segment joining the points A(1;5) and B(4 ;6) cuts y- axis at .

Answer 1

The equation of line joining the points A(x1, y1) and B(x2, y2) is given by,

(y-y1)(x2-x1) =(x-x1)(y2-y1)

When we substitute the coordinates of A and B in this equation, we get,

(y-5)(4-1)=(x-1)(6-5)
3(y-5)=x-1
3y-15=x-1

=> x-3y+14=0
y=(x+14)/3 [ y=mx+c ]
Slope of this line = 1/3

Slope of the perpendicular bisector of this line is given by -1/(1/3) = - 3

Hence the equation of perpendicular bisector is
y = (-3)x+c (1)

Now this perpendicular bisector passes through the mid point of A(1,5) and B(4,6)

Mid point of A, B is given by O[(x1+x2) /2,(y1+y2)/2] => O( 5/2,11/2)

Substitute O(x, y) in equation (1)
11/2 = - 15/2 + c
=> c = 8

Thus the equation of perpendicular bisector is

y = - 3x + 8 (2)

The point of intersection of this line with y axis is given by putting x=0 in equation (2),

We get,
y = 8

Hence the perpendicular bisector of this line intersects y axis at the point (0,8)