Question

Let A(0,0), B(15,0) and C(6,18) are vertices of ∆ABC, where H is point of intersection of altitudes of ∆ABC. If orthocentre of ∆BHC is (p, q) then p² + q² + 2p + 2q + 1 is equal to

• A. 1
• B. 3
• C. 4

Answer 1

Answer: (A)

1

Answer 2

Let H be the orthocenter of $\triangle ABC$. A key property in triangle geometry states that the orthocenter of the triangle formed by two vertices of $\triangle ABC$ and its orthocenter (e.g., $\triangle BHC$) is the third vertex of $\triangle ABC$ (i.e., A). Given that H is the orthocenter of $\triangle ABC$, the orthocenter of $\triangle BHC$ is vertex A. The problem states that the orthocenter of $\triangle BHC$ is (p, q). Therefore, (p, q) = A. The coordinates of vertex A are given as (0, 0). So, p = 0 and q = 0. We need to evaluate the expression $p^2 + q^2 + 2p + 2q + 1$. Substituting p = 0 and q = 0 into the expression: $0^2 + 0^2 + 2(0) + 2(0) + 1 = 0 + 0 + 0 + 0 + 1 = 1$. The expression can also be rewritten as $(p+1)^2 + q^2$. Substituting p=0 and q=0 gives $(0+1)^2 + 0^2 = 1^2 + 0 = 1$.