\( 1 \leftarrow \begin{array}{l}\text { Determine the value of } A \text { so that the line whose equation is } A x+y-7=0 \text { is perpendicular to the line containing the points } \\ (2,-5) \text { and }(-2,6) \text {. } \\ A=\square \text { (Type an integer or a simplified fraction.) }\end{array} \)

To find the value of \( A \) such that the line \( A x + y - 7 = 0 \) is perpendicular to the line through the points \( (2, -5) \) and \( (-2, 6) \), we first need to determine the slope of the line connecting these two points. The slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is calculated using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-5)}{-2 - 2} = \frac{11}{-4} = -\frac{11}{4}. \] Now, the slope of the line \( A x + y - 7 = 0 \) can be rewritten in slope-intercept form \( y = -A x + 7 \). The slope of this line is \( -A \). For two lines to be perpendicular, the product of their slopes must equal \(-1\). Therefore, we have: \[ (-A) \cdot \left(-\frac{11}{4}\right) = -1. \] Simplifying this equation: \[ \frac{11A}{4} = -1. \] Multiplying both sides by 4 gives: \[ 11A = -4. \] Now, divide both sides by 11 to solve for \( A \): \[ A = -\frac{4}{11}. \] Thus, the value of \( A \) is: \[ \boxed{-\frac{4}{11}}. \]