Line #PQ# passes through the points, #P(0,k), and, Q(h,2).#
#:." the slope, say "m_1," of "PQ" is "(k-2)/(0-h)=(2-k)/h.......(1).#
The eqn. of the #bot"-bisector"# of #PQ# is #4y-2x=17....(2).#
We see that, this has the slope #m_2=2/4=1/2.......................(3).#
Clearly, #m_1*m_2=-1.#
#:. (2-k)/h*1/2=-1 rArr 2-k+2h=0...................(4).#
Also, the mid-point #M# of #PQ# lies on the p.b. of #PQ.#
But, #M=M((0+h)/2,(k+2)/2),# and, this lies on #(2).#
#:. 4*(k+2)/2-2*h/2=17, i.e., 2k+4-h=17, or, #
# 2k-h-13=0..........................................................(5).#
Solving #(4) and (5)" for "h and k,# we get,
# h=3, k=8.#
Enjoy Maths.!