Problems on Distance Formula

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)





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.