To find the slope of the diagonal GE in square EFGH, we start with the given information about diagonal FH. The equation of the line along which diagonal FH lies is given by:
[tex]\[ y - 3 = -\frac{1}{3}(x + 9) \][/tex]
This equation is in point-slope form, [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] represents the slope of the line. From the equation [tex]\( y - 3 = -\frac{1}{3}(x + 9) \)[/tex], we can extract the slope [tex]\( m \)[/tex]:
[tex]\[
m = -\frac{1}{3}
\][/tex]
In a square, the diagonals intersect each other at right angles (90 degrees). When two lines are perpendicular, the product of their slopes is [tex]\(-1\)[/tex]. Let [tex]\( m_{GE} \)[/tex] be the slope of diagonal GE. Therefore, the slope of FH and the slope of GE must satisfy the following relationship:
[tex]\[
m_{FH} \cdot m_{GE} = -1
\][/tex]
We already know the slope of FH, [tex]\( m_{FH} = -\frac{1}{3} \)[/tex]. Substituting the values we have:
[tex]\[
\left(-\frac{1}{3}\right) \cdot m_{GE} = -1
\][/tex]
Solving for [tex]\( m_{GE} \)[/tex]:
[tex]\[
m_{GE} = \frac{-1}{-\frac{1}{3}} = 3
\][/tex]
Thus, the slope of diagonal GE is:
[tex]\[
\boxed{3}
\][/tex]
