What is the equation of a line perpendicular to y = -x + 7 through (3, 5)?

The equation of a line perpendicular to \( y = -x + 7 \) through \( (3, 5) \) is \( y = x + 2 \).

To find the equation of a line perpendicular to \( y = -x + 7 \), we first need to determine the gradient (slope) of the given line. The gradient of \( y = -x + 7 \) is -1. For two lines to be perpendicular, the product of their gradients must be -1. Therefore, the gradient of the perpendicular line will be the negative reciprocal of -1, which is 1.

Next, we use the point-slope form of the equation of a line, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the gradient and \( (x_1, y_1) \) is a point on the line. Here, \( m = 1 \) and the point given is \( (3, 5) \).

Substituting these values into the point-slope form, we get:
\[ y - 5 = 1(x - 3) \]

Simplifying this equation:
\[ y - 5 = x - 3 \]
\[ y = x + 2 \]

Thus, the equation of the line perpendicular to \( y = -x + 7 \) and passing through the point \( (3, 5) \) is \( y = x + 2 \). This method ensures that the new line has the correct gradient and passes through the specified point.