The problem lies in showing k=3
y=2x-12 or y-2x+12=0
point of intersection with the x axis is (6,0)
now distance between point (0,k) and line is given by
[k-2(0)+12]/(5)^(1/2).................1
but using distance formula we get RO=[36+k^2]^(1/2)................2
squaring and equating equations 1 and 2 we get
k^2+144+24k=180+5k^2 or
4k^2-24k+36=0 or
k^2-6k+9=0 or
(k-3)^2=0
k=3
second method
since RQ and QP are perpendicular therefor product of their slopes will be -1
point Q=(6,0)
slope of RQ=-k/6
slope of QP=2
therfor
(-k/6)x2=-1 or
k=3
FOR PERPENDICULAR BISECTOR slope=-(1/2) and one point is the mid point of QP
now use y=mx+c (slope(m) already given, just put the coordinates of midpoint to get the value of c)