This one looks fun; wonder why no one did it. It's probably easier with trigonometry but I'll avoid that.

The vague wording is annoying. We're told isosceles but not told definitively it's #AB=AC.# We have to consider both cases.

This would be easier to visualize (and hopefully to solve) if one of the lines is parallel to an axis. Let's try it in a transformed space that's a 45 degree rotation:

#(( x'),( y')) = ( (cos 45^circ, -sin 45^circ),(sin 45^circ, cos 45^circ) ) ((x),(y)) = 1/sqrt{2} ( (1, -1),(1, 1) ) ((x),(y))#

Let's drop the square root of two and remember to scale the area appropriately.

#x'=x-y#

#y'=x+y#

#x=(x'+y')/2 quad quad y=(y'-x')/2#

#x+y=5# becomes #y'=5#

#7x-y=3# becomes #7 ((x'+y')/2) - (y'-x')/2 = 3 #

# 7x'+7y'-y'+x'=6#

#8x' + 6y' = 6#

Substituting, they meet at #A(-3,5) quad# [this is in the transformed space; they meet at #(1,4)# in the original space.]

Let's translate the vertex we know to the center:

#x''=x'+3#

#y''=y'-5#

Let's drop the primes till the end for sanity. Our transformed sides are:

#y=0 quad # aka the x axis

#4x+3y=0 quad # a line through the origin#

That's easier to think about. We have one vertex at the origin O. We have one on the x axis, let's call it #P(p,0).# We have one on the line #4x+3y=0# we can call #Q(3q,-4q).#

The area is #|1/2(p(-4q)) |=|2pq|.# We set it up so #p# and #q# have the same sign. We scaled by #sqrt{2}# so area scaled by #2#, so #10# in the transformed space,

#10 = 2 pq#

#pq = 5#

We have three possibilities for isosceles triangles, #OP=OQ#, #OP=PQ#, #OQ=PQ#. Each of these has two solutions, second and fourth quadrant for #Q.# We'll work each out squared.

#|OP|^2=|OQ|^2 #

#p^2 = (3q)^2 + (4q)^2 = (5q)^2#

We knew that Pythagorean Triple was coming.

#p = pm 5q #

We already determined #p# and #q# have the same sign.

#p=5q#

#5=pq=5q^2#

#q=pm 1 quad p=5q=pm 5#

That's about as a simple as we could hope for. This probably corresponds to the intended solution.

#|OP|^2=|PQ|^2#

#p^2 = (p-3q)^2 + (4q)^2#

#(5/q)^2 = ((5/q)-3q)^2 + (4q)^2#

#25 = (5-3q^2)^2 + 16q^4#

#0 = -30q^2 + 25q^4#

We can rule out #q=0#

# q = pm sqrt(6/5), p = pm 5 \sqrt{5/6}#

# |OQ|^2=|PQ|^2#

#(5q)^2 = (p-3q)^2 + (4q)^2#

# (5q)^2 = (5/q-3q)^2 + (4q)^2#

#q = pm \sqrt{5/6}, p=pm 5 sqrt{6/5} #

We can summarize our solutions as

# q=pm 1, pm \sqrt{5/6}, pm \sqrt{6/5}#

Let's figure out how to map each #q# back to an equation in the original space.

#x''=x'+3 = x-y+3#

#y''=y'-5=x+y-5#

#x=(x'+y')/2 = (x''-3 + y''+5)/2 = (x''+y''+2)/2#

#y=(y'-x')/2 = (y''+5 - (x'' - 3))/2 = (y'' - x'' + 8)/2#

We see #A''(0,0) to A(1,4)# so that worked

#P''(p,0)=P''(5/q,0) to P( 5/{2q}+1, -5/{2q}+4 ) #

#Q(3q,-4q) to Q(-q/2+1, -7/2 q + 4) #

The line PQ is

# (y -(-7/2 q + 4))(5/{2q} - (-q/2)) = (x - (-q/2+1))(-5/{2q} -(-7/2q)) #

# (q^2 + 5) (2 y + 7 q - 8) = (7 q^2 - 5) (2 x + q - 2) #

# 2 (7 q^2 - 5) x - 2 (q^2 + 5) y = (q^2 + 5)( 7 q - 8) -(7 q^2 - 5) (q-2) #

Now we write our six equations for our six #q#s: # pm 1, pm \sqrt{5/6}, pm \sqrt{6/5}#

# x-3y=-1#

#x-3y = -21#

# x - 7y = 4 sqrt(30) - 27 #

# x -7y = - 4 sqrt(30) - 27 #

# 17 x -31 y = - 20 sqrt(30) - 107 #

#17 x - 31y = 20 sqrt(30) - 107 #

Check:

graph{0=(x+y -5)(7x-y -3)(x-3y - -1)(x-3y - -21)(x - 7y -( 4 sqrt(30) - 27))( x -7y -( - 4 sqrt(30) - 27))( 17 x -31 y - ( - 20 sqrt(30) - 107))( 17 x - 31y - ( 20 sqrt(30) - 107 )) [-8.045, 11.955, -1.12, 8.88]}

x+y=5, 7x-y=3, x-3y=-1, x-3y = -21, x - 7y = 4 sqrt(30) - 27, x -7y = - 4 sqrt(30) - 27, 17 x -31 y = - 20 sqrt(30) - 107, 17 x - 31y = 20 sqrt(30) - 107

Interesting intersections. Looks right.