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We will discuss here how to solve the problems on distance formula.
The distance between two points A (x1, y1) and B (x2, y2) is given by the formula
AB = β(x1βx2)2+(y1βy2)2
1. If the distance between the points (5, - 2) and (1, a) is 5, find the values of a.
Solution:
We know, the distance between (x1, y1) and (x2, y2)
is β(x1βx2)2+(y1βy2)2
Here, the distance = 5, x1 = 5, x2 = 1, y1 = -2 and y2 = a
Therefore, 5 = β(5β1)2+(β2βa)2
βΉ 25 = 16 + (2 + a)2
βΉ (2 + a)2 = 25 - 16
βΉ (2 + a)2 = 9
Taking square root, 2 + a = Β±3
βΉ a = -2 Β± 3
βΉ a = 1, -5
2. The co-ordinates of points on the x-axis which are at a distance of 5 units from the point (6, -3).
Solution:
Let the co-ordinates of the point on the x-axis be (x, 0)
Since, distance = β(x2βx1)2+(y2βy1)2
Now taking (6, -3) = (x1, y1) and (x, 0) = (x2, y2), we get
5 = β(xβ6)2+(0+3)2
Squaring both sides we get
βΉ 25 = (x β 6)2 + 32
βΉ 25 = x2 β 12x + 36 + 9
βΉ 25 = x2 β 12x + 45
βΉ x2 β 12x + 45 β 25 = 0
βΉ x2 β 12x + 20 = 0
βΉ (x β 2)(x β 10) = 0
βΉ x = 2 or x = 10
Therefore, the required points on the x-axis are (2, 0) and (10, 0).
3. Which point on the y-axis is equidistance from the points (12, 3) and (-5, 10)?
Solution:
Let the required point on the y-axis (0, y).
Given (0, y) is equidistance from (12, 3) and (-5, 10)
i.e., distance between (0, y) and (12, 3) = distance between (0, y) and (-5, 10)
βΉ β(12β0)2+(3βy)2 = β(β5β0)2+(10βy)2
βΉ 144 + 9 + y2 β 6y = 25 + 100 + y2 β 20y
βΉ 14y = -28
βΉ y = -2
Therefore, the required point on the y-axis = (0, -2)
4. Find the values of a such that PQ = QR, where P, Q and R are the points whose coordinates are (6, - 1), (1, 3) and (a, 8) respectively.
Solution:
PQ = β(6β1)2+(β1β3)2
= β52+(β4)2
= β25+16
= β41
QR = β(1βa)2+(3β8)2
= β(1βa)2+(β5)2
= β(1βa)2+25
Therefore, PQ = QR
βΉ β41 = β(1βa)2+25
βΉ 41 = (1 - a)2 + 25
βΉ (1 - a)2 = 41 - 25
βΉ (1 - a)2 = 16
βΉ 1 - a = Β±4
βΉ a = 1 Β±4
βΉ a = -3, 5
5. Find the points on the y-axis, each of which is at a distance of 13 units from the point (-5, 7).
Solution:
Let A (-5, 7) be the given point and let P (0, y) be the required point on the y-axis. Then,
PA = 13 units
βΉ PA2 = 169
βΉ (0 + 5)2 + (y - 7)2 = 169
βΉ 25 + y2 - 14y + 49 = 169
βΉ y2 β 14y + 74 = 169
βΉ y2 β 14y β 95 = 0
βΉ (y - 19)(y + 5) = 0
βΉ y β 19 = 0 or, y + 5 = 0
βΉ y = 19 or, y = -5
Hence, the required points are (0, 19) and (0, -5)
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