\( \ell \) contains the points \( \left(-\frac{9}{2},-1\right) \) and \( \left(3, \frac{11}{4}\right) \). Find the distance between \( \ell \) and the point \( R\left(\frac{3}{2},-\frac{11}{2}\right) \) und your answer to the nearest tenth.

To find the distance between the line \( \ell \) containing the points \( \left(-\frac{9}{2},-1\right) \) and \( \left(3, \frac{11}{4}\right) \) and the point \( R\left(\frac{3}{2},-\frac{11}{2}\right) \), we first need the equation of the line. 1. Calculate the slope \( m \) of the line \( \ell \): \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{11}{4} - (-1)}{3 - \left(-\frac{9}{2}\right)} = \frac{\frac{11}{4} + \frac{4}{4}}{3 + \frac{9}{2}} = \frac{\frac{15}{4}}{\frac{15}{2}} = \frac{1}{2} \] 2. Use point-slope form of a line \( y - y_1 = m(x - x_1) \) with point \( \left(-\frac{9}{2}, -1\right) \): \[ y - (-1) = \frac{1}{2} \left(x + \frac{9}{2}\right) \] Simplifying this, we have: \[ y + 1 = \frac{1}{2}x + \frac{9}{4} \implies y = \frac{1}{2}x + \frac{5}{4} \] 3. Rewrite this in standard form \( Ax + By + C = 0 \): \[ \frac{1}{2}x - y + \frac{5}{4} = 0 \implies x - 2y + \frac{5}{2} = 0 \implies x - 2y + 2.5 = 0 \] This means \( A = 1, B = -2, C = 2.5 \). 4. Now use the point-to-line distance formula: \[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \] with \( (x_0, y_0) = \left(\frac{3}{2}, -\frac{11}{2}\right) \): \[ d = \frac{|1 \cdot \frac{3}{2} - 2 \cdot \left(-\frac{11}{2}\right) + 2.5|}{\sqrt{1^2 + (-2)^2}} \] 5. Calculate: \[ = \frac{\left|\frac{3}{2} + 11 + 2.5\right|}{\sqrt{1 + 4}} = \frac{\left|\frac{3}{2} + 13.5\right|}{\sqrt{5}} = \frac{\left|\frac{3 + 27}{2}\right|}{\sqrt{5}} = \frac{\left|\frac{30}{2}\right|}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5} \] 6. Approximate \( 3\sqrt{5} \) to the nearest tenth: \[ 3 \sqrt{5} \approx 3 \cdot 2.236 = 6.708 \approx 6.7 \] Thus, the distance between the line \( \ell \) and the point \( R\left(\frac{3}{2},-\frac{11}{2}\right) \) is approximately \( \boxed{6.7} \).