Question

The equation of the locus of the points equidistant from the points A(-a,0)and B(a,0)​

Answer 1

Step-by-step explanation:

you have to find the distance x,y a point

Answer 2

$Let \: P(x,y) \: be \: equidistant \:from \:the$

$points \: A(-a,0) \: and \: B(a,0)$

$\therefore AP = BP .$

$So, AP^{2} = BP^{2}$

/* By using Distance Formula : */

$i.e., [x - (-a)]^{2} + ( y - 0)^{2} = (x-a)^{2}+ (y-0)^{2}$

$\implies (x+a)^{2} +\cancel { y^{2} }= (x-a)^{2}+ \cancel {y^{2}}$

$\implies x^{2} + 2ax + a^{2} = x^{2} -2ax+a^{2}$

$\implies 4ax = 0$

$\implies x = 0$

Therefore.,

$\red{ Required \: equation : }$

$\green { \:x = 0 }$

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