The Base of an Isosceles triangle is the line from (4,-3) to (-4,5) Find the locus of the third vertex?

The equation of the perpendicular bisector of (x₁,y₁) and (x₂,y₂) is (x₂-x₁)x + (y₂-y₁)y = (x₂²-x₁²+y₂²-y₁²)/2
Here this is 8x - 8y = -16/2, which simplifies to x-y=-1,
but not at the midpoint as three collinear points do not form a triangle, so:
Locus of third point: x-y=-1 {x≠0}

Graph: https://www.desmos.com/calculator/njpztuilmk

Let (x, y) be the coordinates of the third vertex.

As it is an isosceles triangle, (x, y) is equidistance to (4, -3) and (-4, 5).
√[(x - 4)² + (y + 3)²] = √[(x + 4)² + (y - 5)²]
(x - 4)² + (y + 3)² = (x + 4)² + (y - 5)²
x² - 8x + 16 + y² + 6y + 9 = x² + 8x + 16 + y² - 10y + 25
-8x + 16 + 6y + 9 = 8x + 16 - 10y + 25
(-8x - 8x) + (6y + 10y) + (16 + 9 - 16 - 25) = 0
-16x + 16y - 16 = 0
x - y + 1 = 0

The locus of the third vertex :
x - y + 1 = 0

The third vertex must lie on the perpendicular bisector of the base.

(x - 4)² + (y + 3)² - (x + 4)² - (y - 5)² = 0
-8x - 8x + 6y + 10y + 16 + 9 - 16 - 25 = 0
-16x + 16y - 16 = 0
x - y + 1 = 0

The locus lies on that line, but it is not the entire line. The point (0, 1) is the midpoint of the side joining the given vertices. Collinear points cannot be vertices of a triangle, so this point must be excluded.

The locus of the third point:

(x, y) satisfying x - y + 1 = 0, excluding point (0, 1)

let vertex is A ( h,k) & point B( 4 .- 3 ) point C( -4, 5)
so AB= sqrt[(h-4)^2+ (k+3)^2]
AC = sqrt [ (h+4)^2 + ( k-5)^2]
so AB = AC
so (h-4)^2 + ( k+3)^2 = (h+4)^2 + (k- 5)^2
or ( h-4)^2 - ( h+4)^2 = ( k-5)^2 - (k+3)^3
or [ h-4+ h+4] [ h-4-h-4] = [k-5+k+3] [k-5-k-3]
or2h*-8 = (2k-2)**-8
or 2h = 2k- 2
or 2h- 2k+ 2 =0
so locus of point A( h, k ) is
2x - 2y + 2 = 0
or x - y+1 =0 answer

Here's what you do:

- Find the slope of the line thru (4, -3) and (-4, 5)

- Find the negative reciprocal of that number.

- Find the midpoint of the line segment from (4, -3) to (-4, 5)

Then the locus is the line thru the midpoint whose slope is

the negative reciprocal.