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
Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
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.
See less⇒ √((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
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).
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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See this for video explanation 😀😎
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Here is the Video Explanation for you ✌️
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