Question

find the value of a if the distanc between the point A(-3,-14) and B(a,-5)is. 9 units​

Answer 1

Given points:-

• A(-3, -14) and B(a, -5)
• Distance between these two points is 9 units

To Find:-

• The value of a

Solution:-

We know, The formula used to find distance between two points Is known as distance formula.

The distance formula is as follows:-

• √(x₂ - x₁)² + (y₂ - y₁)²

The given points are:-

• A(-3, -14) and B(a, -5)

Here,

x₁ = -3 and x₂ = a

y₁ = -14 and y₂ = -5

Putting values in the distance formula:-

AB = √[a - (-3)]² + [(-5) - (-14)]²

= AB = √(a + 3)² + [14 - 5]²

⇒ AB = √a² + 6a + 9 + (9)²

⇒ AB = √a² + 6a + 9 + 81

⇒ AB = √a² + 6a + 90

Now,

As, it is given that distance between these two points is 9 units it means AB = 9

Hence,

9 = √a² + 6a + 90

Squaring both sides,

= (9)² = (√a² + 6a + 90)²

⇒ 81 = a² + 6a + 90

⇒ a² + 6a + 90 - 81 = 0

⇒ a² + 6a + 9 = 0

⇒ a² + 3a + 3a + 9 = 0

⇒ a(a + 3) + 3(a + 3) = 0

⇒ (a + 3)(a + 3) = 0

Either,

a + 3 = 0

⇒ a = -3

Or

a + 3 = 0

⇒ a = -3

∴ The value of a is -3

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Answer 2

$\sf{Answer}$

Step-by-step-explanation:-

Topic :-

Cordinate - Geometry

Given :-

• The distance between A & B is 9 units
• Its cordinates are
• A = (-3 , -14)
• B = ( a , -5)

To find :-

Value of a

Formula implemented:-

AB = $\sf\sqrt{(x1 - x2)² + (y1-y2)²}$

(x1, y1) are cordinates of 1st point

(x2,y2) are cordinates of 2nd point

Solution:-

Plugging valuea in formula

• x1 = -3
• y1 = -14
• x2 = a
• y2 = -5

Distance between A& B is

AB = $\sf\sqrt{(-3-a)² + ( -14 +5)²}$

9 = $\sf\sqrt{(-(3+a)²+(-9)²}$

9 = $\sf\sqrt{(3+a)²+81}$

Squaring on both sides

9² = $\sf{(3+a)² + 81}$

81 = (3+a)² + 81

0 = (3+a)²

Expand the following

9 + a² + 6a = 0

a² + 6a + 9 = 0

Splitting the middle term

a² + 3a + 3a + 9 = 0

a(a +3)+3(a + 3) = 0

(a+3)(a+3) = 0

a = -3

So, the required value of a is -3