QUESTION 1 1.1 The line PQ has equation \( y=3 x-8 \) and point \( P \) has coordination \( (6,10) \). find the equation of the line that passes through \( P \) and is perpendicular to \( P Q \).

To find the equation of the line that passes through point \( P(6,10) \) and is perpendicular to the line \( PQ \), we first need to determine the slope of line \( PQ \). The slope of line \( PQ \) is 3 (from the equation \( y = 3x - 8 \)). The slope of a line that is perpendicular to another line is the negative reciprocal. Hence, the slope of the perpendicular line is \( -\frac{1}{3} \). Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m = -\frac{1}{3} \) and \( (x_1, y_1) = (6, 10) \), we have: \[ y - 10 = -\frac{1}{3}(x - 6). \] Now, rearranging this equation, we get: \[ y = -\frac{1}{3}x + 12. \] Thus, the equation of the line that is perpendicular to \( PQ \) and passes through point \( P \) is \( y = -\frac{1}{3}x + 12 \).