FIND THE MISSING VALUE  WHEN DISTANCE IS GIVEN

Use the given distance d between the two points to find the value of x or y.

Problem 1 :

(0, 3), (x, 5); d = 2√10

Solution :

d = √(x₂ - x₁)² + (y₂ - y₁)²

Let x₁ = 0, x₂ = x, y₁ = 3, y₂ = 5

Here d = 2√10

2√10 = √(x - 0)² + (5 - 3)²

2√10 = √x² + 2²

2√10 = √x² + 4

Taking squares on both sides,

(2√10)² = (√x² + 4)²

4 × 10 = x² + 4

40 = x² + 4

x² = 40 - 4

x² = 36

x = 6

So, the missing value of x is 6.

Problem 2 :

(-3, -1), (2, y); d = √41

Solution :

d = √(x₂ - x₁)² + (y₂ - y₁)²

Let x₁ = -3, x₂ = 2, y₁ = -1, y₂ = y

Here d = √41

√41 = √(2 + 3)² + (y + 1)²

√41 = √5² + (y + 1)²

√41 = √(25 + (y + 1)²)

Taking squares on both sides,

(√41)² = (√25 + (y + 1)²)²

41 = 25 + (y + 1)²

41 = 25 + y² + 1 + 2y

y² + 2y + 26 - 41 = 0

y² + 2y - 15 = 0

(y + 5) (y - 3) = 0

y = -5 (or) y = 3

So, the missing value of y is -5 or 3.

Problem 3 :

(x, 7), (-4, 1); d = 6√2

Solution :

d = √(x₂ - x₁)² + (y₂ - y₁)²

Let x₁ = x, x₂ = -4, y₁ = 7, y₂ = 1

Here d = 6√2

6√2 = √(-4 - x)² + (1 - 7)²

6√2 = √(-4 - x)² + (-6)²

6√2 = √(-4 - x)² + 36

Taking squares on both sides,

(6√2)² = (√(-4 - x)² + 36)²

36 × 2 = (-4 - x)² + 36

72 = x² + 8x + 16 + 36

x² + 8x + 52 - 72 = 0

x² + 8x - 20 = 0

(x - 2) (x + 10) = 0

x = 2 or x = -10

So, the missing value of x is 2 or -10.

Problem 4 :

(1, y), (8, 13); d = √74

Solution :

d = √(x₂ - x₁)² + (y₂ - y₁)²

Let x₁ = 1, x₂ = 8, y₁ = y, y₂ = 13

Here d = √74

√74 = √(8 - 1)² + (13 - y)²

√74 = √7² + (13 - y)²

√74 = √(49 + (13 - y)²)

Taking squares on both sides,

(√74)² = (√49 + (13 - y)²)²

74 = 49 + (13 - y)²

74 = 49 + 169 + y² - 26y

y² - 26y + 218 - 74 = 0

y² - 26y + 144 = 0

(y - 8) (y - 18) = 0

y = 8 or y = 18

So, the missing value of y is 8 or 18.

Problem 5 :

Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units.

Solution :

d = √(x₂ - x₁)² + (y₂ - y₁)²

Let x₁ = -3, x₂ = a, y₁ = -14, y₂ = -5

9 = √(a + 3)² + (-5 + 14)²

9² = (a + 3)² + 9²

(a + 3)² = 0

a + 3 = 0

a = -3

So, the value of a is -3.

Problem 6 :

If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find distance PQ.

Solution :

Given that distance between PA and QA are equal.

PA = QA

√(3 - 2)² + (8 + 4)² = √(-10 - 2)² + (-4 - y)²

√1² + 12² = √(-12)² + (-4 - y)²

√(1 + 144) = √144 + (-4 - y)²

Removing square roots on both sides, 

145 = 144 + (-4 - y)²

145 - 144 = (-4 - y)²

1 = (-4 - y)²

√1 = (-4 - y)

So, the possible values of y are -3 and -5.

Problem 7 :

The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, –9) and has diameter 10√2 units.

Solution :

Center of the circle is (2a, a – 7).

One of the points lie on the circle is (11, -9)

Diameter of the circle = 10√2 units.

radius = 5√2 units.

Distance between center and one of the points lie on the circle = radius

√(2a - 11)² + (a - 7 + 9)² =  5√2

(2a - 11)² + (a + 2)² =  (5√2)²

(2a)² - 2(2a)(11) + 11² + a²  + 2(a) (2) + 2² = 25(2)

4a² - 44a + 121 + a² + 4a + 4 = 50

5a² - 40a + 125 - 50 = 0

5a² - 40a + 75 = 0

a² - 8a + 15 = 0

a² - 3a - 5a + 15 = 0

a(a - 3) - 5(a - 3) = 0

(a - 3)(a - 5) = 0

a = 3 and a = 5