Example 1 :
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Solution :
Distance between P and Q = 10 units.
Formula to find distance between two points :
√[(x2 - x1)2 + (y2 - y1)2]
Here x1 = 2, y1 = -3, x2 = 10 and y2 = y.
Then,
√[(10-2)2 + (y-(-3))2] = 10
√[82 + (y + 3)2] = 10
64 + y2 + 6y + 9 = 100
y2 + 6y + 9 + 64 - 100 = 0
y2 + 6 y - 27 = 0
(y + 9)(y - 3) = 0
y + 9 = 0 y - 3 = 0
y = -9 y = 3
Example 2 :
If Q(x, 0) is equidistant from P(2, 5) and R(-2, 9), find the values of x.
Solution :
From the given information,
PQ = QR -----(1)
Formula to find the distance between two points is
√(x2 - x1)2 + (y2 - y1)2
To find PQ, substitute x1 = 5, y1 = -3, x2 = x, y2 = 0.
PQ = √[(2-x)2 + (5-0)2]
PQ = √[(2-x)2 + 52]
PQ = √[(2-x)2 + 25]
To find QR, substitute x1 = x, y1 = 0, x2 = -2, y2 = 9.
RQ = √[(-2-x)2 + (9-0)2]
RQ = √[(2 + x)2 + (9)2]
RQ = √[(2 + x)2 + 81]
(1)-----> PQ = RQ
√[(2-x)2 + 25] = √[(2 + x)2 + 81]
Square both sides.
(2-x)2 + 25 = (2+x)2 + 81
4 - 4x + x2 + 25 = 4 + 4x + x2 + 81
- 4x + 29 = 4x + 85
-56 = 8x
-7 = x
Example 3 :
Find a relation between x and y such that the points (x, y) is equidistant from the points (3, 6) and (-3, 4).
Solution :
Point (x,y) is equidistant from the points (3, 6) and (-3, 4)
Then,
√[(x-3)2 + (y-6)2] = √[(x+3)2 + (y-4)2]
Square both sides.
(x-3)2 + (y-6)2 = (x+3)2 + (y-4)2
x2 - 6x + 9 + y2 - 12y + 36 = x2 + 6x + 9 + y2 - 8y + 16
Simplify.
- 6x + 9 - 12y + 36 = 6x + 9 - 8y + 16
- 6x - 12y + 45 = 6x - 8y + 25
-12x - 4y + 20 = 0
Divide both sides by -4.
3x + y - 5 = 0
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