Question

find the value of a when the distance between the points (3, a) and (4,1) is V10.
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Answer 1

Solution :-

Distance between the points (3, a) and (4, 1) = √10

Using Distance formula

$\tt d = \sqrt{ (x_2 - x_1)^{2} + (y_2 - y_1) ^{2} }$

Here,

• x1 = 3
• y1 = a
• x2 = 4
• y2 = 1

Substituting the values

$\tt \implies \sqrt{ (4 - 3)^{2} + (1 - a) ^{2} } = \sqrt{10}$

$\tt \implies \sqrt{ 1^{2} + {1}^{2} + {a}^{2} - 2(1)(a) } = \sqrt{10}$

[ Because, (a - b)² = a² - 2ab + b² ]

$\tt \implies \sqrt{ 1 + 1 + {a}^{2} - 2a } = \sqrt{10}$

$\tt \implies \sqrt{ 2 + {a}^{2} - 2a } = \sqrt{10}$

Squaring on both sides

$\tt \implies (\sqrt{2 + {a}^{2} - 2a }) ^{2} = ( \sqrt{10} )^{2}$

$\tt \implies 2 + {a}^{2} - 2a = 10$

$\tt \implies {a}^{2} - 2a + 2 - 10 = 0$

$\tt \implies {a}^{2} - 2a - 8 = 0$

Splitting the middle term

$\tt \implies {a}^{2} - 4a + 2a - 8 = 0$

$\tt \implies a(a - 4) + 2(a - 4) = 0$

$\tt \implies (a + 2)(a - 4)= 0$

$\tt \implies a + 2 = 0 \ \ \ or \ \ \ a - 4= 0$

$\tt \implies a= - 2 \ \ \ or \ \ \ a = 4$

Therefore the value of a is - 2 or 4.

Answer 2

$\huge{\boxed{\boxed{\tt{Answer:}}}}$

${\boxed{\boxed{\tt{Given\:points\:are:}}}}$

➹(3,a) and (4,1)

${\boxed{\boxed{\tt{Find:}}}}$

➹Find the value of a when the distance between the points (3, a) and (4,1) is V10.

${\boxed{\boxed{\tt{Using\:distance\:formula:}}}}$

➹${\boxed{\boxed{\tt{ \sqrt{(4 - 3) {}^{2} \: + \: (1 - a) {}^{2} \: = \sqrt{10} } }}}}</p><p>$

Squaring on both sides, using the formula:

➹${\boxed{\boxed{\tt{(4 - 3) {}^{2} \: + \: (1 - a) {}^{2} = 10}}}}$

Solving the question, we get:

➹${\boxed{\boxed{\tt{1 + 1 + a {}^{2} - 2a = 10 }}}}$

➹${\boxed{\boxed{\tt{2 + a {}^{2} - 2a = 10 }}}}$

➹${\boxed{\boxed{\tt{a {}^{2} - 2a - 8 = 0}}}}$

➹${\boxed{\boxed{\tt{a {}^{2} + 2a - 4a - 8 = 0 }}}}$

➹${\boxed{\boxed{\tt{a(a + 2) - 4(a + 2 )= 0}}}}$

➹${\boxed{\boxed{\tt{(a - 4)(a + 2) = 0}}}}$

Therefore, values of a are -2 and 4.