\( 1 \leftarrow \) Use the given conditions to write an equation for the line in point-slope form and general form. Passing through \( (2,-7) \) and perpendicular to the line whose equation is \( x-2 y-5=0 \) The equation of the line in point-slope form is \( \square \). (Type an equation. Use integers or fractions for any numbers in the equation.) The equation of the line in general form is \( \square=0 \). (Type an expression using \( x \) and \( y \) as the variables. Simplify your answer. Use integers or fractions for any numbers in the expression.)

To find the equation of the line in point-slope form, we first need the slope of the given line, which is \(x - 2y - 5 = 0\). Rearranging it into slope-intercept form \(y = mx + b\) gives us: \[ 2y = x - 5 \implies y = \frac{1}{2}x - \frac{5}{2} \] The slope \(m\) of this line is \(\frac{1}{2}\). Since the desired line is perpendicular, its slope will be the negative reciprocal, which is \(-2\). Now we can use the point-slope form \(y - y_1 = m(x - x_1)\): \[ y - (-7) = -2(x - 2) \implies y + 7 = -2(x - 2) \] Thus, the equation of the line in point-slope form is: \[ y + 7 = -2(x - 2) \] For the general form, we need to simplify and rearrange this equation. Distributing: \[ y + 7 = -2x + 4 \] Now, bringing all terms to one side gives us: \[ 2x + y + 3 = 0 \] So, the equation of the line in general form is: \[ 2x + y + 3 = 0 \] Finally, the answers are: The equation of the line in point-slope form is \(y + 7 = -2(x - 2)\). The equation of the line in general form is \(2x + y + 3 = 0\).