Question

The straight line through A(a, b) intersects the line through B(c,d) at 'P' at right angles. T
locus of P is
A) (x - a) (x - c) + (y - b)(y - d) = 0 B) (x – a)(x – c) - (y - b)(y - d) = 0
c) (x - b)(x - d) + (y - a)(y - d) = 0 D) (x - b) (x – d) + (y - a)(y- c) = 0​

Answer 1

$\textbf{Given:}$

$\text{The straight line through A(a, b) intersects}$

$\text{the line through B(c,d) at 'P' at right angles}$

$\textbf{To find:}$

$\text{Locus of P}$

$\textbf{Solution:}$

$\text{Let the coordinates of P be (h,k) }$

$\text{As per given data,}$

$AP\,{\perp}\,BP$

$\implies\textbf{Slope of AP}{\times}\textbf{Slope of BP}=-1$

$\implies\dfrac{k-b}{h-a}{\times}\dfrac{k-d}{h-c}=-1$

$\implies\dfrac{(k-b)(k-d)}{(h-a)(h-c)}=-1$

$\implies\,(k-b)(k-d)=-(h-a)(h-c)$

$\implies\,(h-a)(h-c)+(k-b)(k-d)=0$

$\therefore\textbf{The locus of P is}\bf\;\;(x-a)(x-c)+(y-b)(y-d)=0$

$\textbf{Answer:}$

$\textbf{Option (A) is correct}$

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