Write the equation of the line through the given point. Use slope-intercept form. $(-9,6) ;$ perpendicular to $y=-\frac{6}{-x} x+1$

Step 1 ：First, we need to find the slope of the given line. The equation of the line is in the form \(y = mx + b\), where \(m\) is the slope. In this case, the equation is \(y = -6/x + 1\). This is not in the standard form, so we need to rewrite it.

Step 2 ：Multiplying both sides by \(-x\), we get \(-xy = 6 + x\). Rearranging, we get \(xy = -6 - x\). Dividing both sides by \(x\), we get \(y = -6/x - 1\).

Step 3 ：So, the slope of the given line is \(-6/x\).

Step 4 ：The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. So, the slope of the line we are trying to find is \(-1/(-6/x) = x/6\).

Step 5 ：Now, we can use the point-slope form of the equation of a line to find the equation of the line we are looking for. The point-slope form is \(y - y1 = m(x - x1)\), where \((x1, y1)\) is a point on the line and \(m\) is the slope.

Step 6 ：Substituting the given point \((-9, 6)\) and the slope we found, we get \(y - 6 = (x/6)(x - (-9))\).

Step 7 ：Simplifying, we get \(y - 6 = x^2/6 + 3x/2\).

Step 8 ：Adding 6 to both sides, we get \(y = x^2/6 + 3x/2 + 6\).

Step 9 ：So, the equation of the line through the given point and perpendicular to the given line is \(\boxed{y = x^2/6 + 3x/2 + 6}\).