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 coordinates of points on the xaxis which are at a distance of 5 units from the point (6, 3).
Solution:
Let the coordinates of the point on the xaxis 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 xaxis are (2, 0) and (10, 0).
3. Which point on the yaxis is equidistance from the points (12, 3) and (5, 10)?
Solution:
Let the required point on the yaxis (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 yaxis = (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 yaxis, 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 yaxis. 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
From Problems on Distance Formula to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.