1. **Problem statement:** Given points P(-3,4), M(a,2) which is the midpoint of PQ, and Q lies on the x-axis (so Q's y-coordinate is 0). We need to find: 2.1 The gradient of PQ. 2.2 The equation of the perpendicular bisector of PQ. 2.3 The coordinates of Q using the midpoint formula. 2.4 The length of PM to two decimal places. 2.5 The equation of the line parallel to the y-axis passing through P. --- 2.1 **Gradient of PQ:** The gradient formula is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ where $P(x_1,y_1)$ and $Q(x_2,y_2)$. Since Q lies on the x-axis, $Q = (x_Q, 0)$. So, $$m_{PQ} = \frac{0 - 4}{x_Q - (-3)} = \frac{-4}{x_Q + 3}$$. --- 2.2 **Equation of the perpendicular bisector of PQ:** The perpendicular bisector passes through midpoint M and has slope the negative reciprocal of $m_{PQ}$. Slope of perpendicular bisector: $$m_{perp} = -\frac{1}{m_{PQ}} = -\frac{1}{\frac{-4}{x_Q + 3}} = \frac{x_Q + 3}{4}$$. Equation using point-slope form: $$y - 2 = m_{perp}(x - a)$$. --- 2.3 **Coordinates of Q using midpoint formula:** Midpoint formula: $$M = \left(\frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}\right)$$. Given $M = (a, 2)$, $P = (-3,4)$, and $Q = (x_Q, 0)$: $$a = \frac{-3 + x_Q}{2}$$ $$2 = \frac{4 + 0}{2} = 2$$ (confirms midpoint y-coordinate) From the first equation: $$2a = -3 + x_Q \Rightarrow x_Q = 2a + 3$$. So, $Q = (2a + 3, 0)$. --- 2.4 **Length of PM:** Distance formula: $$PM = \sqrt{(x_M - x_P)^2 + (y_M - y_P)^2}$$ Substitute $P(-3,4)$ and $M(a,2)$: $$PM = \sqrt{(a + 3)^2 + (2 - 4)^2} = \sqrt{(a + 3)^2 + (-2)^2} = \sqrt{(a + 3)^2 + 4}$$. --- 2.5 **Equation of line parallel to y-axis through P:** Lines parallel to the y-axis have equations of the form: $$x = k$$ Since it passes through $P(-3,4)$, the equation is: $$x = -3$$. --- **Summary:** - Gradient of PQ: $$m_{PQ} = \frac{-4}{x_Q + 3}$$ with $x_Q = 2a + 3$. - Perpendicular bisector: $$y - 2 = \frac{x_Q + 3}{4}(x - a)$$. - Coordinates of Q: $$Q = (2a + 3, 0)$$. - Length of PM: $$PM = \sqrt{(a + 3)^2 + 4}$$. - Equation of line parallel to y-axis through P: $$x = -3$$.