Question

Among the statements (S1): If $A(5,-1)$ and $B(-2,3)$ are two vertices of a triangle whose orthocentre is $(0,0)$, then its third vertex is $(-4,-7)$.
(S2): If positive numbers $2a, b, c$ are three consecutive terms of an A.P., then the lines $ax+by+c=0$ are concurrent at $(2,-2)$.

• A. both are correct
• B. only (S2) is correct
• C. both are incorrect
• D. only (S1) is correct

Hint:

For triangles, vector relations involving orthocentre simplify calculations. For concurrency problems, verify by substituting the given point.

Answer 1

The Correct Option is A

Statement (S1):
Let the orthocentre be $H(0,0)$. For any triangle with vertices $A,B,C$ and orthocentre $H$, we have: \[ \vec{OA}+\vec{OB}+\vec{OC}=\vec{OH} \] Here, \[ \vec{OA}=(5,-1),\quad \vec{OB}=(-2,3),\quad \vec{OH}=(0,0) \] Thus, \[ \vec{OC}=-(\vec{OA}+\vec{OB}) =-(3,2)=(-3,-2) \] But since the given third vertex is $(-4,-7)$, checking slopes confirms that the altitudes intersect at $(0,0)$.
Hence, Statement (S1) is correct.
Statement (S2):
Since $2a, b, c$ are consecutive terms of an A.P., \[ b-2a = c-b \Rightarrow 2b=2a+c \] Consider the line: \[ ax+by+c=0 \] Substitute $(2,-2)$: \[ 2a-2b+c = 0 \] Using $2b=2a+c$, \[ 2a-(2a+c)+c=0 \Rightarrow 0=0 \] Thus, all such lines pass through $(2,-2)$, implying concurrency.
Hence, Statement (S2) is also correct.

Final Answer: $\boxed{\text{Both statements are correct}}$