11 has coordinates \( (4,1) \) and \( N \) has coordinates \( (-2,-7) \). (a) Find the length of \( M N \). (b) Find the gradient of \( M N \). (c) Find the equation of the perpendicular bisector of \( M N \). (c) (b) (b)

To find the length of \( M N \), we can use the distance formula given by \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Plugging in the coordinates \( M(4, 1) \) and \( N(-2, -7) \), we get: \[ d = \sqrt{(-2 - 4)^2 + (-7 - 1)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10. \] The length of \( M N \) is \( 10 \) units. Now, to find the gradient of the line \( M N \), we use the formula for the gradient (slope) defined as \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Using the points \( M(4, 1) \) and \( N(-2, -7) \): \[ m = \frac{-7 - 1}{-2 - 4} = \frac{-8}{-6} = \frac{4}{3}. \] The gradient of \( M N \) is \( \frac{4}{3} \). Next, we find the equation of the perpendicular bisector of \( M N \). First, the midpoint \( P \) of \( M N \) can be calculated as: \[ P = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{4 + (-2)}{2}, \frac{1 + (-7)}{2} \right) = \left( 1, -3 \right). \] Since the gradient of \( M N \) is \( \frac{4}{3} \), the gradient of the perpendicular bisector is the negative reciprocal: \[ m_{\text{perpendicular}} = -\frac{3}{4}. \] Using point-slope form, the equation for the perpendicular bisector through point \( P(1, -3) \) becomes: \[ y - (-3) = -\frac{3}{4}(x - 1). \] Simplifying this gives: \[ y + 3 = -\frac{3}{4}x + \frac{3}{4} \quad \Rightarrow \quad y = -\frac{3}{4}x - \frac{9}{4}. \] Thus, the equation of the perpendicular bisector of \( M N \) is \( y = -\frac{3}{4}x - \frac{9}{4} \).