Straight Lines

The points A(-2, 3) and C(9, 4) are two vertices of an isosceles triangle ABC in which AC = BC. The side AB lies on x+y=1. Find coordinates of B.

I’ve tried so much - I’ve worked out the length of AC and got root122, worked out the equation of AC as 11y=x+35. I’ve also tried letting B be (a,b) and working out an equation for the line BC and finding the equation where x+y=1 intersects this.

Am I missing something?

Because B lies on x + y = 1, you can write its coordinates as (x, 1 - x).

You know the distance AC, which you have worked out correctly.

You know the distance BC is the same as this. Write an equation for the distance BC using the above version of the coordinates of B, and see there that gets you.

So
CB^2 = ((4-(1-x))^2 + (9-x)^2
CB^2 = (x+3)^2 + (9-x)^2
CB^2= 2x^2-12x+90

Then there would I go from here? Would I make the quadratic equal to 122 and solve?

Thanks

Yes!