What is the distance, in units, from the point \( (-3,8) \) to the line \( -4 y=8-3 x \) ?

To find the distance from the point \((-3,8)\) to the line given by the equation \(-4y = 8 - 3x\), we first rewrite the equation in slope-intercept form \(y = mx + b\). Starting with \(-4y = 8 - 3x\), we divide everything by \(-4\): \[ y = \frac{3}{4}x - 2 \] This indicates that the line has a slope of \(\frac{3}{4}\) and a y-intercept of \(-2\). Next, we can use the formula for the distance \(d\) from a point \((x_0, y_0)\) to a line in the form \(Ax + By + C = 0\), which is given by: \[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \] We first rewrite the line equation \(y = \frac{3}{4}x - 2\) into standard form. This gives us: \[ -\frac{3}{4}x + y + 2 = 0 \quad \text{or equivalently} \quad 3x - 4y + 8 = 0 \] In this equation, \(A = 3\), \(B = -4\), and \(C = 8\). Now, substituting the point \((-3, 8)\) into the distance formula: \[ d = \frac{|3(-3) - 4(8) + 8|}{\sqrt{3^2 + (-4)^2}} = \frac{|-9 - 32 + 8|}{\sqrt{9 + 16}} = \frac{|-33|}{\sqrt{25}} = \frac{33}{5} \] Thus, the distance from the point \((-3,8)\) to the line \(-4y = 8 - 3x\) is \[ \frac{33}{5} \text{ units}. \]