We will discuss here how to solve the problems on distance formula.
The distance between two points A (x\(_{1}\), y\(_{1}\)) and B (x\(_{2}\), y\(_{2}\)) is given by the formula
AB = \(\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\)
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 (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\))
is \(\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\)
Here, the distance = 5, x\(_{1}\) = 5, x\(_{2}\) = 1, y\(_{1}\) = -2 and y\(_{2}\) = a
Therefore, 5 = \(\sqrt{(5 - 1)^{2} + (-2 - a)^{2}}\)
⟹ 25 = 16 + (2 + a)\(^{2}\)
⟹ (2 + a)\(^{2}\) = 25 - 16
⟹ (2 + a)\(^{2}\) = 9
Taking square root, 2 + a = ±3
⟹ a = -2 ± 3
⟹ 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 = \(\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\)
Now taking (6, -3) = (x\(_{1}\), y\(_{1}\)) and (x, 0) = (x\(_{2}\), y\(_{2}\)), we get
5 = \(\sqrt{(x - 6)^{2} + (0 + 3)^{2}}\)
Squaring both sides we get
⟹ 25 = (x – 6)\(^{2}\) + 3\(^{2}\)
⟹ 25 = x\(^{2}\) – 12x + 36 + 9
⟹ 25 = x\(^{2}\) – 12x + 45
⟹ x\(^{2}\) – 12x + 45 – 25 = 0
⟹ x\(^{2}\) – 12x + 20 = 0
⟹ (x – 2)(x – 10) = 0
⟹ 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)
⟹ \(\sqrt{(12 - 0)^{2} + (3 - y)^{2}}\) = \(\sqrt{(-5 - 0)^{2} + (10 - y)^{2}}\)
⟹ 144 + 9 + y\(^{2}\) – 6y = 25 + 100 + y\(^{2}\) – 20y
⟹ 14y = -28
⟹ 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 = \(\sqrt{(6 - 1)^{2} + (-1 - 3)^{2}}\)
= \(\sqrt{5^{2} + (-4)^{2}}\)
= \(\sqrt{25 + 16}\)
= \(\sqrt{41}\)
QR = \(\sqrt{(1 - a)^{2} + (3 - 8)^{2}}\)
= \(\sqrt{(1 - a)^{2} + (-5)^{2}}\)
= \(\sqrt{(1 - a)^{2} + 25}\)
Therefore, PQ = QR
⟹ \(\sqrt{41}\) = \(\sqrt{(1 - a)^{2} + 25}\)
⟹ 41 = (1 - a)\(^{2}\) + 25
⟹ (1 - a)\(^{2}\) = 41 - 25
⟹ (1 - a)\(^{2}\) = 16
⟹ 1 - a = ±4
⟹ a = 1 ±4
⟹ 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
⟹ PA\(^{2}\) = 169
⟹ (0 + 5)\(^{2}\) + (y - 7)\(^{2}\) = 169
⟹ 25 + y\(^{2}\) - 14y + 49 = 169
⟹ y\(^{2}\) – 14y + 74 = 169
⟹ y\(^{2}\) – 14y – 95 = 0
⟹ (y - 19)(y + 5) = 0
⟹ y – 19 = 0 or, y + 5 = 0
⟹ y = 19 or, y = -5
Hence, the required points are (0, 19) and (0, -5)
● Distance and Section Formulae
10th Grade Math
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