1. The distance between P(2,-3) and Q(10,y) is 10 units. ⇒ √((10-2)²+[y-(-3)]²) = 10 ⇒ √(64+y²+9+6y) = 10 Squaring both sides 64+y²+9+6y = 100 ⇒ y²+6y-27 = 0 ⇒ y²+9y-3y-27 = 0 ⇒ y(y+9)-3(y+9) = 0 ⇒ (y+9)(y-3) = 0 ⇒ (y+9) = 0 or (y-3) = 0 ⇒ y = -9 or y = 3

    The distance between P(2,-3) and Q(10,y) is 10 units.
    ⇒ √((10-2)²+[y-(-3)]²) = 10
    ⇒ √(64+y²+9+6y) = 10
    Squaring both sides
    64+y²+9+6y = 100
    ⇒ y²+6y-27 = 0
    ⇒ y²+9y-3y-27 = 0
    ⇒ y(y+9)-3(y+9) = 0
    ⇒ (y+9)(y-3) = 0
    ⇒ (y+9) = 0 or (y-3) = 0
    ⇒ y = -9 or y = 3

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  2. Point P(x,y) is equidistant from A(3,6) and B(-3,4). Therefore, PA = PB ⇒ √((3-x)²+(6-y)²) = √((-3-x)²)+(4-y)²) ⇒ √(9+x²-6x+36+y²-12y) = √(9+x²+6x+16+y²-8y) Squaring both sides 9+x²-6x+36+y²-12y = 9+x²+6x+16+y²-8y ⇒ -12x-4y = -20 ⇒ 3x+y=5 Here is video explanation (~ ̄▽ ̄)~

    Point P(x,y) is equidistant from A(3,6) and B(-3,4). Therefore, PA = PB
    ⇒ √((3-x)²+(6-y)²) = √((-3-x)²)+(4-y)²)
    ⇒ √(9+x²-6x+36+y²-12y) = √(9+x²+6x+16+y²-8y)
    Squaring both sides
    9+x²-6x+36+y²-12y = 9+x²+6x+16+y²-8y
    ⇒ -12x-4y = -20
    ⇒ 3x+y=5

    Here is video explanation (~ ̄▽ ̄)~

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