Use the given conditions to write an equation for the line in point-slope form and in slope-intercept form. Passing through \( (6,-3) \) and perpendicular to the line whose equation is \( y=\frac{1}{3} x+2 \) Write an equation for the line in point-slope form. \( \square \) (Simplify your answer. Use integers or fractions for any numbers in the equation.)

To write the equation of the line in point-slope form, you first need to find the slope of the line that is given. The equation \( y=\frac{1}{3} x+2 \) has a slope of \( \frac{1}{3} \). Since the new line is perpendicular to this line, its slope will be the negative reciprocal, which is \( -3 \). Using the point-slope form \( y - y_1 = m(x - x_1) \) with the point \( (6, -3) \) and slope \( -3 \), we plug in the values: \[ y - (-3) = -3(x - 6) \] This simplifies to: \[ y + 3 = -3(x - 6) \] Now, let's rewrite it to put a box around our point-slope form: Point-slope form: \( y + 3 = -3(x - 6) \) For the slope-intercept form, we can further simplify: Expand and simplify: \[ y + 3 = -3x + 18 \] \[ y = -3x + 15 \] So the slope-intercept form is: Slope-intercept form: \( y = -3x + 15 \)