Among the statements: (S1): If A(5,-1) and B(-2,3) are two vertices of a triangle, whose o…

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 $(-4,-7)$

and

(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 $(2,-2)$,

Accepted Answer

Statement 1: This is the same configuration as Question 173. Solving with $mAH	imes mBC=-1$ and $mBH	imes mAC=-1$ gives the third vertex $(-4,-7)$. Statement 1 is correct.

Statement 2: Since $2a,b,c$ are in A.P.: $2b=2a+c \implies 2a-2b+c=0$.

Comparing with $ax+by+c=0$: this matches the form $ax+by+c=0$ when $x=2,\ y=-2$ (i.e. substituting $x=2,y=-2$ into $ax+by+c$ gives $2a-2b+c$, which is exactly zero by the A.P. condition).

So every such line $ax+by+c=0$ passes through the fixed point $(2,-2)$ --- the lines are concurrent there. Statement 2 is correct.

Both statements are correct.

Solutions · 1