Q161 / 270 · NDA 2013 (Phase 2) · Mathematics
A point $P$ moves such that its distances from $(1,2)$ and $(-2,3)$ are equal. Then the locus of P is
- Correct answerAstraight line
- Bparabola
- Cellipse
- Dhyperbola
Solution
Let point $P$ is $(\alpha, \beta)$
According to the question,
$\sqrt{(\alpha-1)^2+(\beta-2)^2}=\sqrt{(\alpha-(-2))^2+(\beta-3)^2}$
Squaring both sides,
$(\alpha-1)^2+(\beta-2)^2 =(\alpha+2)^2+(\beta-3)^2$
$\alpha^2+\beta^2-2 \alpha-4 \beta+5=\alpha^2+\beta^2+4 \alpha-6 \beta+13$
or, $6 \alpha-2 \beta+8=0$
$\Rightarrow 3 \alpha-\beta+4=0$
Locus of point P is a straight line whose equation is: $3 x-y+4=0$
Hence, the correct answer is option: 'A'.