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)

● Distance and Section Formulae

• Distance Formula
• Distance Properties in some Geometrical Figures
• Conditions of Collinearity of Three Points
• Problems on Distance Formula
• Distance of a Point from the Origin
• Distance Formula in Geometry
• Section Formula
• Midpoint Formula
• Centroid of a Triangle
• Worksheet on Distance Formula
• Worksheet on Collinearity of Three Points
• Worksheet on Finding the Centroid of a Triangle
• Worksheet on Section Formula

10th Grade Math

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