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Question: Consider △ABC,A(5,−1),B(α,−7),C(−2,β). Let (−6,−4) is image of orthocentre of △ABC in the point mirr...

Consider △ABC,A(5,−1),B(α,−7),C(−2,β). Let (−6,−4) is image of orthocentre of △ABC in the point mirror M which is mid-point of the side BC. Also (p,q) is circumcentre of triangle ABC, then the value of β2−α2+5β−α−6p+2q is _____.

Answer

10

Explanation

Solution

Solution:

  1. Find Orthocentre (H):
    M is the midpoint of BC where
    M = ((α + (–2))/2, (–7 + β)/2) = ((α – 2)/2, (β – 7)/2).
    Given that (–6, –4) is the reflection of H across M, we have
    M = (H + (–6, –4))/2 ⟹ H = 2M – (–6, –4) = 2M + (6, 4).
    Thus,
    H = ( (α – 2) + 6, (β – 7) + 4 ) = (α + 4, β – 3).

  2. Use Altitude from A:
    Side BC (from B(α, –7) to C(–2, β)) has slope
    m₍BC₎ = (β + 7)/(–2 – α) = –(β + 7)/(α + 2).
    The altitude from A(5, –1) is perpendicular to BC so its slope is
    m₍A_alt₎ = (α + 2)/(β + 7).
    Its equation:
    y + 1 = (α + 2)/(β + 7) · (x – 5).
    Since H = (α + 4, β – 3) lies on this line, substitute:
    (β – 3) + 1 = (α + 2)/(β + 7) · ((α + 4) – 5)
    ⇒ (β – 2) = (α + 2)/(β + 7) · (α – 1).
    Multiply both sides by (β + 7):
    (β – 2)(β + 7) = (α + 2)(α – 1).
    Expanding,
    β² + 5β – 14 = α² + α – 2
    ⇒ β² – α² + 5β – α = 12.

  3. Find Circumcentre (O = (p, q)) using Euler’s Relation:
    For any triangle,
    H = A + B + C – 2O.
    A + B + C = (5 + α + (–2), –1 + (–7) + β) = (α + 3, β – 8).
    We already have H = (α + 4, β – 3), so
    (α + 4, β – 3) = (α + 3, β – 8) – 2(p, q).
    Equate coordinates:
    α + 4 = α + 3 – 2p ⟹ 2p = –1 ⟹ p = –½,
    β – 3 = β – 8 – 2q ⟹ 2q = –5 ⟹ q = –5/2.

  4. Compute the Required Expression:
    We need the value of
    β² – α² + 5β – α – 6p + 2q.
    Using (1),
    β² – α² + 5β – α = 12, and with p = –½, q = –5/2,
    6p = 6(–½) = –3, 2q = 2(–5/2) = –5.
    Thus,
    Expression = 12 – (–3) + (–5) = 12 + 3 – 5 = 10.

Final Answer: 10

Summary:

  • Core Explanation:

    • Use midpoint M of BC and reflection to find H = (α+4,β–3).
    • Altitude from A gives relation: β² – α² + 5β – α = 12.
    • Using H = A+B+C – 2O, find O = (–½, –5/2).
    • Substitute in expression to obtain: 12 – 6p + 2q = 10.