# FIND MISSING COORDINATE USING DISTANCE FORMULA

Let A(x1, y1) and B(x2, y2) be two endpoints of the line segment. To find the length of the line segment or distance between two points, we use the formula given below.

d = √(x2 – x1)2 + (y2 – y1)2

Find a given that :

Problem 1 :

P(2, 3) and Q(a, -1) are 4 units apart

Solution :

d = √(x2 – x1)2 + (y2 – y1)2

Let x1 = 2, y1 = 3, x2 = a, y2 = -1

Here d = 4

4 = √(a – 2)2 + (-1 – 3)2

4 = √(a – 2)2 + (-4)2

4 = √(a – 2)2 + 16

Taking square on both sides.

42 = (a – 2)2 + 16

(a – 2)2 = 0

a - 2 = 0

a = 2

So, the missing coordinates are a = 2.

Problem 2 :

P(-1, 1) and Q(a, -2) are 5 units apart

Solution :

d = √((x2 – x1)2 + (y2 – y1)2)

Let x1 = -1, y1 = 1, x2 = a, y2 = -2

Here d = 5

5 = √(a + 1)2 + (-2 – 1)2

5 = √(a + 1)2 + (-3)2

5 = √(a + 1)2 + 9

Taking squares on both sides.

25 = (a + 1)2 + 9

Subtracting 9 on both sides,

16 = (a + 1)2

Again take squares on both sides.

(a + 1) = √16

a + 1 = ± 4

 a + 1 = 4a = 4 - 1a = 3 a + 1 = -4a = -4 - 1a = -5

So, the missing coordinates are a = 3 or a = -5.

Problem 3 :

X (a, a) is √8 units from the origin

Solution :

X(a, a) and O(0, 0)

d = √((x2 – x1)2 + (y2 – y1)2)

Let x1 = a, y1 = a, x2 = 0, y2 = 0

Here d = √8

√8 = √(0 - a)2 + (0 – a)2

√8 = √a2 + a2

√8 = √2a2

Taking squares on both sides.

8 = 2a2

8/2 = a2

4 = a2

a = ±2

Problem 4 :

A (0, a) is equidistant from P(3, -3) and  Q(-2, 2)

Solution :

AP = AQ

d = √((x2 – x1)2 + (y2 – y1)2)

√((3 – 0)2 + (-3 – a)2)= √((-2 – 0)2 + (2 – a)2)

√(9 + (-3)2 + a2 – 2(-3)(a))) = √(4 + 22 + a2 – 2(2)(a))

√(9 + 9 + a2 + 6a) = √(4 + 4 + a2 – 4a)

√18 + a2 + 6a = √(8 + a2 – 4a)

Taking square on both sides.

18 + a2 + 6a = 8 + a2 – 4a

18 + a2 + 6a - 8 - a2 + 4a = 0

10 + 10a = 0

10a = -10

a = -1

Problem 5 :

Find b given that A(3, -2) and B(b, 1) are √13 units apart.

Solution :

d = √((x2 – x1)2 + (y2 – y1)2)

Let x1 = 3, y1 = -2, x2 = b, y2 = 1

Here d = √13

√13 = √((b – 3)2 + (1 + 2)2)

√13  = √(b – 3)2 + (3)2

√13  = √(b – 3)2 + 9

Taking square on both sides.

13  = (b – 3)2 + 9

Subtracting 9 on both sides.

13 - 9 = (b – 3)2

4 = (b – 3)2

b - 3 = √4

b - 3 = √4

b - 3 = ±2

 b - 3 = 2b = 2 + 3b = 5 b - 3 = -2b = -2 + 3b = 1

So, the missing coordinates are b = 1 or b = 5.

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