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Problems on Distance Formula

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 = √(x1βˆ’x2)2+(y1βˆ’y2)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 √(x1βˆ’x2)2+(y1βˆ’y2)2

Here, the distance = 5, x1 = 5, x2 = 1, y1 = -2 and y2 = a

Therefore, 5 = √(5βˆ’1)2+(βˆ’2βˆ’a)2

⟹ 25 = 16 + (2 + a)2

⟹ (2 + a)2 = 25 - 16

⟹ (2 + a)2 = 9

Taking square root, 2 + a = Β±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 = √(x2βˆ’x1)2+(y2βˆ’y1)2

Now taking (6, -3) = (x1, y1) and (x, 0) = (x2, y2), we get

5 = √(xβˆ’6)2+(0+3)2

Squaring both sides we get

⟹ 25 = (x – 6)2 + 32

⟹ 25 = x2 – 12x + 36 + 9

⟹ 25 = x2 – 12x + 45

⟹ x2 – 12x + 45 – 25 = 0

⟹ x2 – 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)

⟹ √(12βˆ’0)2+(3βˆ’y)2 = √(βˆ’5βˆ’0)2+(10βˆ’y)2

⟹ 144 + 9 + y2 – 6y = 25 + 100 + y2 – 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 = √(6βˆ’1)2+(βˆ’1βˆ’3)2

     = √52+(βˆ’4)2

     = √25+16

     = √41

QR = √(1βˆ’a)2+(3βˆ’8)2

     = √(1βˆ’a)2+(βˆ’5)2

     = √(1βˆ’a)2+25

Therefore, PQ = QR

⟹ √41 = √(1βˆ’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

⟹ PA2 = 169

⟹ (0 + 5)2 + (y - 7)2 = 169

⟹ 25 + y2 - 14y + 49 = 169

⟹ y2 – 14y + 74 = 169

⟹ y2 – 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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