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)