Question

Where does the perpendicular bisector of the line segment joining the points $A(1,5)$ and $B(4,6)$ cut the y-axis?

Answer 1

slope of AB =$\frac{G-5}{4-1}=\frac{1}{3}$

slope of line perpendicular to AB = - ${\left(\frac{1}{3}\right)}^{-1}$

= - 3

let c is mid point of AB then

c=$=(\frac{4+1}{2},\frac{6+5}{2})$

$=(\frac{5}{2},\frac{11}{2})$

thus equation of line passes through c with slope -3 is

$(y-\frac{11}{2})=-3\cdot (x-5/2)$

at x=0 y=13.

thus it cuts y axis at y=13.