Point P(x,y) is equidistant from A(3,6) and B(-3,4). Therefore, PA = PB β β((3-x)Β²+(6-y)Β²) = β((-3-x)Β²)+(4-y)Β²) β β(9+xΒ²-6x+36+yΒ²-12y) = β(9+xΒ²+6x+16+yΒ²-8y) Squaring both sides 9+xΒ²-6x+36+yΒ²-12y = 9+xΒ²+6x+16+yΒ²-8y β -12x-4y = -20 β 3x+y=5 Here is video explanation (ο½οΏ£β½οΏ£)ο½
Point P(x,y) is equidistant from A(3,6) and B(-3,4). Therefore, PA = PB
β β((3-x)Β²+(6-y)Β²) = β((-3-x)Β²)+(4-y)Β²)
β β(9+xΒ²-6x+36+yΒ²-12y) = β(9+xΒ²+6x+16+yΒ²-8y)
Squaring both sides
9+xΒ²-6x+36+yΒ²-12y = 9+xΒ²+6x+16+yΒ²-8y
β -12x-4y = -20
β 3x+y=5
Find the values of y for which the distance between the points P(2, β 3) and Q(10, y) is 10 units.
The distance between P(2,-3) and Q(10,y) is 10 units. β β((10-2)Β²+[y-(-3)]Β²) = 10 β β(64+yΒ²+9+6y) = 10 Squaring both sides 64+yΒ²+9+6y = 100 β yΒ²+6y-27 = 0 β yΒ²+9y-3y-27 = 0 β y(y+9)-3(y+9) = 0 β (y+9)(y-3) = 0 β (y+9) = 0 or (y-3) = 0 β y = -9 or y = 3
The distance between P(2,-3) and Q(10,y) is 10 units.
See lessβ β((10-2)Β²+[y-(-3)]Β²) = 10
β β(64+yΒ²+9+6y) = 10
Squaring both sides
64+yΒ²+9+6y = 100
β yΒ²+6y-27 = 0
β yΒ²+9y-3y-27 = 0
β y(y+9)-3(y+9) = 0
β (y+9)(y-3) = 0
β (y+9) = 0 or (y-3) = 0
β 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 (β 3, 4).
Point P(x,y) is equidistant from A(3,6) and B(-3,4). Therefore, PA = PB β β((3-x)Β²+(6-y)Β²) = β((-3-x)Β²)+(4-y)Β²) β β(9+xΒ²-6x+36+yΒ²-12y) = β(9+xΒ²+6x+16+yΒ²-8y) Squaring both sides 9+xΒ²-6x+36+yΒ²-12y = 9+xΒ²+6x+16+yΒ²-8y β -12x-4y = -20 β 3x+y=5 Here is video explanation (ο½οΏ£β½οΏ£)ο½
Point P(x,y) is equidistant from A(3,6) and B(-3,4). Therefore, PA = PB
β β((3-x)Β²+(6-y)Β²) = β((-3-x)Β²)+(4-y)Β²)
β β(9+xΒ²-6x+36+yΒ²-12y) = β(9+xΒ²+6x+16+yΒ²-8y)
Squaring both sides
9+xΒ²-6x+36+yΒ²-12y = 9+xΒ²+6x+16+yΒ²-8y
β -12x-4y = -20
β 3x+y=5
Here is video explanation (ο½οΏ£β½οΏ£)ο½
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