Question

Suppose $P$ and $Q$ are the midpoints of the sides $AB$ and $BC$ of a triangle where $A(1, 3)$, $B(3, 7)$ and $C(7, 15)$ are vertices. Then the locus of $R$ satisfying $AC^2 + QR^2 = PR^2$ is

• A. $6x + 12y = 297$
• B. $6x + 12y + 297 = 0$
• C. $12x + 6y = 297$
• D. $12x + 6y + 297 = 0$

Hint:

Use midpoint formula and distance formula systematically for locus-based geometry problems.

Answer 1

The Correct Option is A

Let $P$ be midpoint of $AB \Rightarrow P = \left( \dfrac{1+3}{2}, \dfrac{3+7}{2} \right) = (2, 5)$
Let $Q$ be midpoint of $BC \Rightarrow Q = \left( \dfrac{3+7}{2}, \dfrac{7+15}{2} \right) = (5, 11)$
Let $R = (x, y)$
Use the relation: $AC^2 + QR^2 = PR^2$
Find all three squared distances and simplify:
$AC^2 = (7 - 1)^2 + (15 - 3)^2 = 36 + 144 = 180$
$QR^2 = (x - 5)^2 + (y - 11)^2$
$PR^2 = (x - 2)^2 + (y - 5)^2$
Now set: $180 + (x - 5)^2 + (y - 11)^2 = (x - 2)^2 + (y - 5)^2$
Expand and simplify to get linear equation: $6x + 12y = 297$