Pls help Analytic Geometry Problem?

Let (x₁, y₁) be the coordinates of the third vertex.

Length of AB
= √[(2 + 2)² + (4 - 3)²
= √17

Equation of the line AB
(y - 4) / (x - 2) = (3 - 4) / (-2 - 2)
(y - 4) / (x - 2) = 1 / 4
x - 2 = 4(y - 4)
x - 2 = 4y - 16
x - 4y + 14 = 0

Distance between (x₁, y) and the line AB
= |x₁ - 4y₁ + 14| / √[1² + (-4)²]
= |x₁ - 4y₁ + 14| / √17

Area of the triangle :
(1/2) × (√17) × (|x₁ - 4y₁ + 14| / √17) = 2
|x₁ - 4y₁ + 14| = 4
|x₁ - 4y₁ + 14|² = 4²
x₁² + 16y₁² + 196 - 8x₁y₁ + 28x₁ - 112y₁ = 16
x₁² + 16y₁² - 8x₁y₁ + 28x₁ - 112y₁ + 180 = 0

The locus of the third vertex :
x² + 16y² - 8xy + 28x - 112y + 180 = 0

We know the base -- as line and length. We are missing the altitude.
The length of the base is
√( (2 - (-2) )^2 + (4 - 3)^2)
= √17
So, if the altitude is h, then
h * √ 17 / 2 = 2;
so h = 4 / √17
The gradient of AB is (2 - (-2) ) / (4 - 3) = 4

OOPS! I HAVE THE x- AND y-VALUES REVERSED in the gradient!
Other than that (!!), the principles of procedure remain correct.

So the gradient of a perpendicular line is -1/4 .
If BC is perpendicular to AB, then the equation of BC is
y - 3 = (-1/4) (x + 2)
Now if x changes by 1 along this line, then y changes by -1/4;
so the distance between the two points is √17 / 4,
which is (√17 / 4) / (4 / √17) times the distance that we want;
i.e., a factor 17/16 times that needed.
So we change x by 16/17, and y by -4/17,
e.g., from (2, 4) to (2 16/17, 3 13/17) , = (50/17, 64/17) -- call it D, on one side;
and to (18/17, 72/17) -- call it E, on the other.
The equation of the line through D parallel to AB is
y - 64/17 = 4 (x - 50/17) ,
or y = 4x - (200 - 64) / 17
i.e., y = 4x - 8 .
The corresponding line through E is
y - 72/17 = 4 (x - 18/17)
or y = 4x - (4 * 18 - 72) / 17
i.e., y = 4x .
Thus the required locus is
y = 4x or y = 4x - 8 ;
or y = 4x - 4 ± 4