Question

If the distance between two points (5,2); (3,a) is √8 units then a =​

Answer 1

Given :-

• If the distance between two points (5,2)(3,a) is √8 units

To find :-

• Value of a

Solution :-

According to distance formula

→ AB = √(x2 - x1)² + (y2 - y1)²

Let,

• A = (5, 2)
• B = (3, a)

According to the given condition

• AB = √8

→ AB = √(x2 - x1)² + (y2 - y1)²

→ √8 = √(3 - 5)² + (a - 2)²

→ √8 = √(-2)² + a² + 4 - 4a

→ √8 = √4 + a² + 4 - 4a

→ √8 = √8 + a² - 4a

Squaring both side

→ 8 = 8 + a² - 4a

→ a² - 4a + 8 - 8 = 0

→ a² - 4a = 0

→ a(a - 4) = 0

Either

→ a = 0

Or

→ a - 4 = 0

→ a = 4

Hence,

• Value of a is 4

Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• The distance between two points

(5,2)(3,a) is √8 units.

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Value of "a "

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

Let,

P = (5, 2)

Q = (3, a)

$\underline{\:\textsf{As we know that:}}$

$\star \boxed{ \green{\bold{ d = \sqrt{ ({x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} } }}}$

Where,

• d = PQ

• x₁ = 5

• x₂ = 5

• y₁ = a

• y₂ = 2

$\underline{\:\textsf{ Now, put the given values in the formula :}}$

→$\sf{\sqrt{(x_2-x_1)^2+(y_2-y_2)^2}}$

→$\sf \sqrt{(3-5)^2+(a-2)^2}$

→$\sf \sqrt{(-2)^2+(a-2)^2}$

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$\bf\underline{\green{\:\:\:\:\:\:\:\:A.T.Q:-\:\:\:\:\:\:\:}}$

→√8 = PQ

→ √8 = √(-2)² + a² + 4 - 4a

→ √8 = √4 + a² + 4 - 4a

→ √8 = √8 + a² - 4a

→ a² - 4a + 8 - 8 = 0

→ a² - 4a = 0

→ a(a - 4) = 0

Hence,

$\bf{ \longrightarrow a = 0}$

OR,

$\bf{ \longrightarrow a = 4}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Value of a is \textbf{ 4 or, 0 }}}.$

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