Question

Find the value of a if the distance between the points ( a,2) ( 3,4) is 2 root 2

Answer 1

$\huge\underline\mathbb{\red Q\pink{U}\purple{ES} \blue{T} \orange{IO}\green{N :}}$

Find the value of a if the distance between the points ( a,2) ( 3,4) is 2√2.

$\huge\underline\mathbb{\red S\pink{O}\purple{LU} \blue{T} \orange{IO}\green{N :}}$

★ Given points,

• A(a, 2)
• B(3, 4)

★ To find,

• Value of a.

★ Distance between A & B :

• AB = 2√2

★ Let,

• x1 = a, y1 = 2
• x2 = 3, y2 = 4

By using Distance formula...

$\tt\purple{ Distance \: formula : \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2 }- y_{1})}^{2} } }$

• Substitute the values.

$\sf\:⟹ 2\sqrt{2} = \sqrt{ {(3 - a)}^{2} + {(4 - 2)}^{2} } </p><p>$

$\sf\:⟹ 2\sqrt{2} = \sqrt{ {(3 - a)}^{2} + {(2)}^{2} }$

• Squaring on both sides.

$\sf\:⟹ (2\sqrt{2})^{2} = ( \sqrt{ {(3 - a)}^{2} + {(2)}^{2} })^{2}</p><p>$

$\sf\:⟹ 8 = (3 - a)^{2} + (2)^{2}$

• (a - b)² = a² + b² - 2ab

$\sf\:⟹ 8 = (3)^{2} + (a)^{2} - 2(3)(a) + 4$

$\sf\:⟹ 8 = 9 + a^{2} - 6a + 4$

$\sf\:⟹ 8 = a^{2} - 6a + 13$

$\sf\:⟹ a^{2} - 6a + 13 - 8 = 0$

$\sf\:⟹ a^{2} - 6a + 5 = 0$

$\sf\:⟹ a^{2} - a - 5a + 5 = 0$

$\sf\:⟹ a(a - 1) - 5(a - 1) = 0$

$\sf\:⟹ (a - 1)(a - 5) = 0$

$\sf\:⟹ a - 1 = 0 ; a - 5 = 0$

$\sf\:⟹ a = 0 + 1 ; a = 0 + 5$

$\sf\:⟹ a = 1 ; a = 5$

$\underline{\boxed{\bf{\purple{∴ Hence, \: the \: value \: of \: a = 1 \: (or) \: 5}}}}$