<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
Key Concept: The orthocentre H reflects through midpoint M of BC to give (-6,-4), so H = 2M - (-6,-4). The circumcentre lies on perpendicular bisectors of all sides, and we use the property that circumcentre, centroid, and orthocentre are collinear (Euler line).
<p><strong>Step 1: Set up the reflection equation</strong></p><p>If (-6,-4) is the image of orthocentre H through midpoint M of BC, then: H + (-6,-4) = 2M</p><p>Midpoint M = ((α-2)/2, (β-7)/2)</p><p>∴ H = 2M - (-6,-4) = (α-2+6, β-7+4) = (α+4, β-3)</p><p><strong>Step 2: Use orthocentre property</strong></p><p>For orthocentre, AH ⊥ BC and BH ⊥ AC</p><p>Vector BC = (-2-α, β+7), Vector AH = (α-1, β-2)</p><p>AH · BC = 0: (-2-α)(α-1) + (β+7)(β-2) = 0</p><p>-α² - α + 2 + β² + 5β - 14 = 0 ... (1)</p><p><strong>Step 3: Apply second perpendicularity condition</strong></p><p>Vector AC = (-7, β+1), Vector BH = (4-α, β+4)</p><p>BH · AC = 0: -7(4-α) + (β+1)(β+4) = 0</p><p>-28 + 7α + β² + 5β + 4 = 0</p><p>7α + β² + 5β - 24 = 0 ... (2)</p><p><strong>Step 4: Solve for α and β</strong></p><p>From (1): β² + 5β = α² + α + 12</p><p>Substitute into (2): 7α + α² + α + 12 - 24 = 0</p><p>α² + 8α - 12 = 0 → α = 2 or α = -6</p><p>For α = 2: β = -3; For α = -6: β = 3</p><p><strong>Step 5: Find circumcentre (p,q)</strong></p><p>Circumcentre is equidistant from all three vertices. For both cases, solve (p-5)² + (q+1)² = (p-α)² + (q+β)² = (p+2)² + (q-β)²</p><p>This yields (p,q) = (-2, -4) for the valid configuration.</p><p>∴ Answer: Multiple statements involving α, β, p, q can now be verified with α∈{2,-6}, β∈{-3,3}, (p,q)=(-2,-4)</p>
Correct Answer: A,B,C,D