1. **Problem statement:** (a) Find the coordinates where the graph of $y = 5x - 3$ crosses the y-axis. (b) Given points $A(1,7)$ and $B(5,15)$, find the equation of the perpendicular bisector of line $AB$ in the form $y = mx + c$. 2. **Part (a) - y-intercept:** The y-axis is where $x=0$. Substitute $x=0$ into $y=5x-3$: $$y = 5(0) - 3 = -3$$ So, the point is $(0, -3)$. 3. **Part (b) - Perpendicular bisector of $AB$:** - Find midpoint $M$ of $AB$: $$M = \left(\frac{1+5}{2}, \frac{7+15}{2}\right) = (3, 11)$$ - Find slope of $AB$: $$m_{AB} = \frac{15 - 7}{5 - 1} = \frac{8}{4} = 2$$ - Slope of perpendicular bisector $m_{\perp}$ is negative reciprocal: $$m_{\perp} = -\frac{1}{2}$$ - Use point-slope form with midpoint $M$: $$y - 11 = -\frac{1}{2}(x - 3)$$ - Simplify: $$y - 11 = -\frac{1}{2}x + \frac{3}{2}$$ $$y = -\frac{1}{2}x + \frac{3}{2} + 11 = -\frac{1}{2}x + \frac{25}{2}$$ 4. **Final answers:** (a) The graph crosses the y-axis at $(0, -3)$. (b) The equation of the perpendicular bisector is: $$y = -\frac{1}{2}x + \frac{25}{2}$$