[1 mak) A line segrnent joins the points \( C(-5.6) \) and \( D(7,2) \). (a) Colculate the midpoint of the Ine segment CD \[ \begin{array}{l} 17\left(\frac{-1+1+1}{2}+\frac{1+12}{2}\right) \\ m_{2}=-5+7=\frac{2}{2}= \\ N_{y}=\frac{\operatorname{Len} 2}{2}=-7= \\ m=(-1,4) \end{array} \] \( \qquad \) (2 man (b) Find the gradient of the line segment \( C D \). \( \qquad \) (2 ms (c) Another line, \( A B \), is parallel to \( C D \) and passes through the point \( (0,1) \). Write down the equation of the line \( A B \).

To calculate the midpoint of the line segment CD, use the formula: \[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] Plugging in the coordinates of points C and D: \[ \text{Midpoint} = \left( \frac{-5 + 7}{2}, \frac{6 + 2}{2} \right) = \left( \frac{2}{2}, \frac{8}{2} \right) = (1, 4) \] Now, for the gradient of the line segment CD, use: \[ \text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} \] Substituting the coordinates: \[ \text{Gradient} = \frac{2 - 6}{7 - (-5)} = \frac{-4}{12} = -\frac{1}{3} \] For the line AB that is parallel to CD, since parallel lines have the same gradient, the gradient of AB will also be \(-\frac{1}{3}\). Using the point-slope form of the equation of a line, you can write: \[ y - y_1 = m(x - x_1) \] Given that \( A(0, 1) \): \[ y - 1 = -\frac{1}{3}(x - 0) \Rightarrow y = -\frac{1}{3}x + 1 \] And there you have it; the equation for line AB is \( y = -\frac{1}{3}x + 1 \).