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Find the equation to the locus of a poin...

Find the equation to the locus of a point equidistant from the points `A(1,3)a n dB(-2,1)dot`

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To find the equation of the locus of a point that is equidistant from the points A(1, 3) and B(-2, 1), we can follow these steps: ### Step 1: Define the Point Let the point P be represented by the coordinates (h, k). ### Step 2: Write the Distance Formulas The distance from point A to point P (AP) and from point B to point P (BP) can be expressed using the distance formula: - Distance AP = √[(h - 1)² + (k - 3)²] ...
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Knowledge Check

  • If the equation of the locus of a point equidistant from the points (a_1,b_1) and (a_2,b_2) is (a_1-a_2)x+(b_2+b_2)y+c=0 , then the value of C is

    A
    `a_1^2-a_2^2+b_1^2-b_2^2`
    B
    `sqrt(a_1^2+b_1^2-a_2^2-b_2^2)`
    C
    `(1)/(2)(a_1^2+a_2^2+b_1^2+b_2^2)`
    D
    `(1)/(2)(a_1^2+b_2^2+a_1^2+b_2^2)`

  • If the equation of the locus of a point equidistant from the points (a_(1),b_(1) , and (a_(2),b_(2)) is (a_(1)-a_(2))x+(b_(1)-b_(2))y+c=0 , then the value of c is

    A
    `1/2(a_(2)^(2)+b_(2)^2)-(a_(1)^(2)-b_(1)^(2))`
    B
    `(a_(1)^(2)-a_(2)^(2)+b_(1)^(2)-b_(2)^(2))`
    C
    `1/2(a_(1)^(2)+a_(2)^(2)+b_(1)^(2)+b_(2)^(2))`
    D
    `sqrt((a_(1)^(2)+b_(1)^(2)-a_(2)^(2)-b_(2)^(2)))`

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