We have #2# Methods to solve this Problem.
#1^(st)# Method :-
Let #Q(4,9) and R(8,5).#
#A=QR=sqrt{(4-8)^2+(9-5)^2}=4sqrt2.#
#"Area of "Delta=1/2"(Base)(Altitude)"#
#:. 34=1/2(A)"(Altitude)"rArr"Altitude="{(34)(2)}/(4sqrt2)=17/sqrt2.#
This means the the #3^(rd) "Vertex"#, call it #P(x,y),# is situated at a
#bot-"distance"# of #17/sqrt2# from the line-sgmt. #QR=A.#
At the same, #Delta PQR" is isisceles with, "PQ [=B=C] = PR.#
Therefore, #P# lies on the #bot-#bisector of #QR#, at a distance of
#17/sqrt2.#
Let #M# is the mid-point of #QR.#
#:. M=M(6,7), PM=17/sqrt2, and, PM bot QR.#
Slope of #QR=-1 rArr "Slope of "PM=1=tan(pi/4).#
Now, to find #P,#we use the following very useful Result:-
Result :- On a line #l,# passing through #(x_0,y_0)# & making an
angle #theta# with the #+ve# X-axis, the point on #l# at a dist.
#r# from #(x_0,y_0)# is given by #(x_0+-rcostheta,y_0+-rsintheta).#
Here, we have #l=PM, (x_0,y_0)=M(6,7), theta=pi/4, and, r=PM=17/sqrt2.#
#:.," for "P(x,y), x=6+-(17/sqrt2)cos(pi/4)#
#=6+-(17/sqrt2)(1/sqrt2)#
#:. x=29/2, or, -5/2.#
Similarly, #y=31/2, or, -3/2.#
Hence, the possible co-ords. of the #3^(rd)# vertex are #(29/2,31/2)#
or, #(-5/2,-3/2).#
Enjoy Maths.!