Question:

Step 1: Determine the equation of the altitude from vertex A to side BC. The coordinates of the vertices are $A(9, 3)$, $B(7, –1)$, and $C(1, –1)$. The slope of side BC is $m_{BC} = \frac{-1 - (-1)}{1 - 7} = \frac{0}{-6} = 0$. Since BC is a horizontal line, the altitude from A to BC is a vertical line passing through A. The equation of this altitude is $x = 9$. Step 2: Determine the equation of the altitude from vertex C to side AB. The slope of side AB is $m_{AB} = \frac{-1 - 3}{7 - 9} = \frac{-4}{-2} = 2$. The altitude from C to AB is perpendicular to AB, so its slope is $m_{\perp AB} = -\frac{1}{2}$. Using the point-slope form with $C(1, -1)$: $y - (-1) = -\frac{1}{2}(x - 1)$ $y + 1 = -\frac{1}{2}x + \frac{1}{2}$ $y = -\frac{1}{2}x + \frac{1}{2} - 1$ $y = -\frac{1}{2}x - \frac{1}{2}$ Step 3: Find the orthocentre by intersecting the altitudes. The orthocentre is the intersection of the altitudes. Substitute $x = 9$ (from the first altitude) into the equation of the second altitude: $y = -\frac{1}{2}(9) - \frac{1}{2}$ $y = -\frac{9}{2} - \frac{1}{2}$ $y = -\frac{10}{2}$ $y = -5$ The coordinates of the orthocentre are $(9, -5)$.