Question

In $\triangle PQR$, vertex P is (1, 1) and orthocentre O is (2, 4). Also given that the side PQ and QR are represented by ax + by + c = 0 where $3a^2 + 4b^2 + 5c^2 = 12ab + 8c^2 - 8b^2$ and $b \neq \frac{a-c}{2}$.

1. The coordinates of the vertex Q is

• A. (2, 1)
• B. (1,-2)
• C. (-1, 2)
• D. (-1,-2)

Answer 1

Answer: (B)

(1, -2)

Answer 2

The given condition $3a^2 + 4b^2 + 5c^2 = 12ab + 8c^2 - 8b^2$ simplifies to $(a - 2b)^2 - c^2 = 0$, which yields $(a - 2b - c)(a - 2b + c) = 0$. This means either $a = 2b + c$ or $a = 2b - c$.

For the line $ax + by + c = 0$:

If $a = 2b + c$, the line passes through $(-1, 2)$.

If $a = 2b - c$, the line passes through $(1, -2)$.

The given constraint $b \neq \frac{a-c}{2}$ eliminates the case $a = 2b + c$. Therefore, both lines PQ and QR must pass through $(1, -2)$, which means Q = $(1, -2)$.