Question

Find the value of’ k ‘ for which the distance between the two points A (3 , -1 )and
B (11 , k) is 10 units
plz solve it

Answer 1

Given :

The distance between the two points A (3 , -1 )and B (11 , k) is 10 units

To find :

• Value of " k "

Solution :

★ Coordinate of two points is given ★

• A = (3, - 1)
• B = (11, k)
• AB = 10 units

Applying distance formula

→ AB = √(x₂ - x₁)² + (y₂ - y₁)²

• x₁ = 3
• x₂ = 11
• y₁ = - 1
• y₂ = k

Substitute all the values

→ 10 = √(11 - 3)² + (k - (-1))²

→ 10 = √(8)² + (k + 1)²

• Apply identity
• (a + b)² = a² + b² + 2ab

→ 10 = √64 + k² + 1 + 2k

• Squaring both the sides

→ (10)² = (√65 + k² + 2k)²

→ 100 = 65 + k² + 2k

→ 100 - 65 = k² + 2k

→ 35 = k² + 2k

→ k² + 2k - 35 = 0

• Split middle term

→ k² + 7k - 5k - 35 = 0

→ k(k + 7) - 5(k + 7) = 0

→ (k + 7)(k - 5) = 0

Either

→ k + 7 = 0

→ k = - 7

Or

→ k - 5 = 0

→ k = 5

•°• Value of " k " = - 7 or 5

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Answer 2

${\large{\bold{\rm{\underline{Understanding \; the \; question}}}}}$

★ This question says that we have to find the value of k for which the distance between the two points A(3,-1)and B(11,k) is 10 units.

${\large{\bold{\rm{\underline{Given \; that}}}}}$

★ The distance between the two points A(3,-1)and B(11,k) is 10 units.

${\large{\bold{\rm{\underline{To \; find}}}}}$

★ Value of k

${\large{\bold{\rm{\underline{Solution}}}}}$

★ Value of k = -7 or 5

${\large{\bold{\rm{\underline{Using \; concept}}}}}$

★ Algebraic identity.

★ Middle term splitting method.

★ Distance formula.

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

~ According to the question, the cordinate of the two points are given,

●↝ A(3,-1)

●↝ B(11,k)

●↝ AB = 10 units

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~ Now let's use distance formula..!

${\boxed{\boxed{\green{\sf{Distance \: = \: \sqrt{(x_{2}-x_{1})}^{2} + (y_{2} - y_{1})^{2}}}}}}$

${\bf{Here,}}$

${\sf{\bull \leadsto x_{2} \: is \: 11}}$

${\sf{\bull \leadsto x_{1} \: is \: 3}}$

${\sf{\bull \leadsto y_{2} \: is \: k}}$

${\sf{\bull \leadsto y_{1} \: is \: -1}}$

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~ Now let's put the values..!

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

${\tt{:\implies AB \: = \: \sqrt{(x_{2}-x_{1})}^{2} + (y_{2} - y_{1})^{2}}}$

${\tt{:\implies AB \: = \: \sqrt{(11-3)}^{2} + (k-(-1))^{2}}}$

${\tt{:\implies AB \: = \: \sqrt{(8)}^{2} + (k+)^{2}}}$

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~ Now we have to use an algebraic identity here. Let's use the given identity..!

${\boxed{\boxed{\green{\sf{(a+b)^{2} = a^{2} + b^{2} + 2ab}}}}}$

~ Now let's put the values..!

${\tt{:\implies 10 \: \sqrt{64} + k^{2} + 1 + 2k}}$

${\tt{:\implies 10^{2} = (\sqrt{65} + k^{2} + 2k)^{2}}}$

${\tt{:\implies 100 = 65 + k^{2} + 2k}}$

${\tt{:\implies 100 - 65 = k^{2} + 2k}}$

${\tt{:\implies 35 = k^{2} + 2k}}$

${\tt{:\implies k^{2} + 2k - 35 = 0}}$

_________________

~ Henceforth, we get a quadratic equation as k² + 2k - 35 = 0. To solve this let us use middle term splitting method..!

${\tt{:\implies k^{2} + 7k - 5k - 35 = 0}}$

${\tt{:\implies k(k+7) - 5(k+7) = 0}}$

${\tt{:\implies (k+7)(k-5) = 0}}$

_________________

~ Now let's solve it..!

${\bf{Either,}}$

${\tt{:\implies k + 7 = 0}}$

${\tt{:\implies k = 0 - 7}}$

${\tt{:\implies k = -7}}$

${\bf{Or}}$

${\tt{:\implies k - 5 = 0}}$

${\tt{:\implies k = 0 + 5}}$

${\tt{:\implies k = 5}}$

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${\frak{Henceforth, \: -7 \: or \: 5 \: is \: the \: value \: of \: k}}$

${\large{\bold{\rm{\underline{Information \; about \; topic}}}}}$

Knowledge about Quadratic equations -

★ Sum of zeros of any quadratic equation is given by ➝ α+β = -b/a

★ Product of zeros of any quadratic equation is given by ➝ αβ = c/a

★ A quadratic equation have 2 roots

★ ax² + bx + c = 0 is the general form of quadratic equation

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✴ Algebraic identities –

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A+B)^{2} \: = \: = A^{2} \: + \: 2AB \: + B^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A-B)^{2} \: = \: = A^{2} \: - \: 2AB \: + B^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto A^{2} \: - B^{2} \: = \: (A+B) \: (A-B)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A+B)^{2} \: = (A-B)^{2} \: +4AB}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A-B)^{2} \: = (A+B)^{2} \: -4AB}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A+B)^{3} \: = A^{3} + \: 3AB \: (A+B) \:+ B^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A-B)^{3} \: = A^{3} - \: 3AB \: (A-B) \: + B^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto A^{3} \: + B^{3} = \: (A+B) (A^{2} - AB + B^{2}}}}$

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✴ Factorised identities -

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\; =\;a^{2}\:+\:b^{2}\:+\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)^{2} \: = \: a^{2} + 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a-b)^{2} \: = a^{2} - 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)(a-b) \: = \: a^{2} - b^{2}}\end{gathered}$

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