$AB$ is a chord of the circle $x^ {2}+y^ {2}-7x-4=0$. If $(1,-1)$ is the mid point of the chord $AB$ then the area of the triangle formed by $AB$ and the coordinate aces is:

Using the midpoint chord formula where T = S1, we substitute the midpoint (1, -1) into the circle's equation x^2 + y^2 - 7x - 4 = 0 to find the line containing the chord. This gives the equation x(1) + y(-1) - 7((x + 1)/2) - 4 = 1 - 1 - 7(1) - 4. Simplifying this results in the line 5x + 2y - 3 = 0. The intercepts are 3/5 on the x-axis and 3/2 on the y-axis, so the area of the right triangle formed with the axes is 1/2 * (3/5) * (3/2) = 9/20.

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