Question

The locus of the third vertex of a right-angled triangle, the ends of whose hypotenuse are \( (1, 2) \) and \( (4, 5) \), is:

• A. \( x^2 + y^2 + 5x + 7y + 14 = 0 \)
• B. \( 3x + 3y - 1 = 0 \)
• C. \( 3x + 3y + 1 = 0 \)
• D. \( x^2 + y^2 - 5x - 7y + 14 = 0 \)

Hint:

To find the locus of a vertex forming a right angle in a triangle with fixed hypotenuse, use the perpendicularity condition \( \vec{PA} . \vec{PB} = 0 \).

Answer 1

The Correct Option is D

Let the fixed points be \( A(1, 2) \) and \( B(4, 5) \), which form the hypotenuse of a right-angled triangle. Let \( P(x, y) \) be the third vertex such that triangle \( \triangle APB \) is right-angled at \( P \). Then, by the geometric property: \[ \text{If } \angle APB = 90^\circ, \text{ then } \vec{PA} . \vec{PB} = 0 \] Step 1: Compute vectors: \[ \vec{PA} = (1 - x, 2 - y),
\vec{PB} = (4 - x, 5 - y) \] Step 2: Apply the dot product condition: \[ \vec{PA} . \vec{PB} = 0 \Rightarrow (1 - x)(4 - x) + (2 - y)(5 - y) = 0 \] \[ (1 - x)(4 - x) = 4 - 5x + x^2,
(2 - y)(5 - y) = 10 - 7y + y^2 \] \[ \Rightarrow x^2 + y^2 - 5x - 7y + 14 = 0 \] This is the required locus of the third vertex.