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Answer 1 · 2017-08-23T16:51:19

See below.

The place for the perpendicular bisector to #bar(PQ)# can be stated as

#norm(u-P)=norm(u-Q)# where #u = (x,y)# Now

#norm(u)^2-2 << u,P>> +norm(P)^2=norm(u)^2-2 << u, Q >> + norm(Q)^2#

or

#2 << u, Q-P >> +norm(P)^2-norm(Q)^2 = 0# or

#2hx+(4-2k)y-4 - h^2 + k^2=0# but this equation must be proportional to

#2x-4y+17=0#

so we have

#lambda(2hx+(4-2k)y-4 - h^2 + k^2)=2x-4y+17#

Comparing coefficients and solving we have

#{( k^2 lambda- 4 lambda - h^2 lambda =17),(2 k lambda-4lambda=4),(2hlambda=2):}#

so we have #h =3# and #k = 8# and #lambda = 1/3#

Answer 2 · 2017-08-23T18:01:20

# h=3, k=8.#

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.!