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Find the equation of a line perpendicular to $2 x-2 y=-6$ that passes through the point $(9,6)$.

Step 1 ：Understand the problem: We are asked to find the equation of a line that is perpendicular to the given line and passes through the given point.

Step 2 ：Find the slope of the given line: The equation of the given line is \(2x - 2y = -6\). We can rewrite this in slope-intercept form (y = mx + b) to find the slope. \(2x - 2y = -6\) becomes \(-2y = -2x - 6\), which simplifies to \(y = x + 3\). So, the slope of the given line is 1.

Step 3 ：Find the slope of the line we are trying to find: Since the line we are trying to find is perpendicular to the given line, its slope is the negative reciprocal of the slope of the given line. The negative reciprocal of 1 is -1.

Step 4 ：Use the point-slope form of a line to find the equation of the line: The point-slope form of a line is \(y - y_1 = m(x - x_1)\), where m is the slope and \((x_1, y_1)\) is a point on the line. We know that the slope of the line we are trying to find is -1, and it passes through the point (9,6). Plugging these values into the point-slope form gives us: \(y - 6 = -1(x - 9)\), which simplifies to \(y - 6 = -x + 9\), and further simplifies to \(y = -x + 15\).

Step 5 ：Check the solution: The line \(y = -x + 15\) is perpendicular to the given line and passes through the point (9,6), so it is the correct solution.

Step 6 ：\(\boxed{y = -x + 15}\) is the equation of the line.