<p>Consider \(\triangle ABC\), \(A(5,-1)\), \(B(\alpha,-7)\), \(C(-2,\beta)\). Let \((-6,-4)\) is image of orthocentre of \(\triangle ABC\) in the point mirror \(M\) which is mid-point of the side \(BC\). Also \((p,q)\) is circumcentre of triangle \(ABC\), then:</p>
<p>the value of \(\beta^2 - \alpha^2 + 5\beta - \alpha\) is 12.</p>
<p>the value of \(2p + 1\) is 0.</p>
<p>the value of \(2q + 5\) is \(-6\).</p>
<p>the value of \(q^2 - \dfrac{p}{2}\) is \(\dfrac{13}{2}\).</p>
Step-by-Step Solution
<div class="solution">
<p><strong>Step 1:</strong> To find the orthocentre of \(\triangle ABC\), we first need to find the equations of the altitudes of \(\triangle ABC\). The altitude from vertex \(A\) is perpendicular to \(BC\), the altitude from vertex \(B\) is perpendicular to \(AC\), and the altitude from vertex \(C\) is perpendicular to \(AB\).</p>
<p><strong>Step 2:</strong> The slope of line \(BC\) is given by \(\frac{\beta - (-7)}{-2 - \alpha} = \frac{\beta + 7}{-2 - \alpha}\). The slope of the altitude from \(A\) is the negative reciprocal of this slope, which is \(\frac{2 + \alpha}{\beta + 7}\). The equation of the altitude from \(A\) can be written as \(y - (-1) = \frac{2 + \alpha}{\beta + 7}(x - 5)\). Similarly, we can find the equations of the altitudes from \(B\) and \(C\).</p>
<p><strong>Step 3:</strong> The orthocentre \(H\) is the point of intersection of the three altitudes. However, since \((-6,-4)\) is the image of the orthocentre in the point mirror \(M\), which is the midpoint of \(BC\), we can use this information to find the coordinates of \(H\). The midpoint \(M\) of \(BC\) is given by \(\left(\frac{\alpha - 2}{2}, \frac{\beta - 7}{2}\right)\). Since \(M\) is the midpoint of \(BC\), the image of \(H\) in \(M\) is \((-6,-4)\), so \(H\) is at \(\left(2 \cdot \frac{\alpha - 2}{2} - (-6), 2 \cdot \frac{\beta - 7}{2} - (-4)\right) = (\alpha - 2 + 6, \beta - 7 + 4) = (\alpha + 4, \beta - 3)\).</p>
<p><strong>Step 4:</strong> The circumcentre \((p,q)\) of \(\triangle ABC\) is the point of intersection of the perpendicular bisectors of the sides of \(\triangle ABC\). The midpoint of \(BC\) is \(\left(\frac{\alpha - 2}{2}, \frac{\beta - 7}{2}\right)\), and the slope of \(BC\) is \(\frac{\beta + 7}{-2 - \alpha}\). The slope of the perpendicular bisector of \(BC\) is the negative reciprocal of this slope, which is \(\frac{2 + \alpha}{\beta + 7}\). The equation of the perpendicular bisector of \(BC\) can be written as \(y - \frac{\beta - 7}{2} = \frac{2 + \alpha}{\beta + 7}\left(x - \frac{\alpha - 2}{2}\right)\). We can find the equations of the perpendicular bisectors of \(AB\) and \(AC\) similarly.</p>
<p><strong>Step 5:</strong> Since the circumcentre \((p,q)\) is the point of intersection of the perpendicular bisectors, we can solve the system of equations formed by the perpendicular bisectors to find \((p,q)\). However, given the complexity of the problem and the information provided, let's analyze the options and use the properties of the orthocentre and circumcentre to find a relationship between \(\alpha\), \(\beta\), \(p\), and \(q\).</p>
<p><strong>Step 6:</strong> From the given options, let's examine the relationship between \(\alpha\), \(\beta\), \(p\), and \(q\). We are given that \((-6,-4)\) is the image of the orthocentre in the midpoint \(M\) of \(BC\). Using the midpoint formula, we can relate \(\alpha\) and \(\beta\) to the coordinates of \(M\). Since \(M\) is the midpoint of \(BC\), its coordinates are \(\left(\frac{\alpha - 2}{2}, \frac{\beta - 7}{2}\right)\). The image of the orthocentre \(H(\alpha + 4, \beta - 3)\) in \(M\) is \((-6,-4)\), so we can set up the equation \((\alpha + 4) - 2\left(\frac{\alpha - 2}{2}\right) = -6\) and \((\beta - 3) - 2
Correct Answer: A,B,C,D