Question

If the point P (k–1,2) is equidistant from the points A(3,k) and B(k,5), find the values of k.

Answer 1

$\sf\red{\underline{\underline{Solution.}}}$

$\sf{We \: have,}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{ P \implies{(k - 1, 2)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{ A \implies{(3, \: k)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{ B \implies(k, \: 5)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \red{ PA = PB}$

$\: \: \: \: \: \: \: \: \: \: \rm ∵ P \: is \: equidistant \: from \: \sf\red {a}\: and \: \sf \red{ b}$

$\implies \: \: \: \: \: \: \: \: \: \: \: \sf \red{PA {}^{2} = PB {}^{2} }$

$\sf ∴ (3 - (k - 1 ) ) {}^{2} + (k - 2) {}^{2}$

$\: \: \: \: \: \ \: \: \sf = (k - (k - 1)) {}^{2} + (5 - 2) {}^{2}$

$\implies \: \: \: \: \: \: \sf (3 - k + 1) {}^{2} + (k - 2) {}^{2}$

$\implies \: \: \bf = (k - k + 1) {}^{2} + 3 {}^{2}$

$\implies \: \: \sf (4 - k) {}^{2} + (k - 2) {}^{2} = 1 + 9$

$\implies\sf16 + k {}^{2} - 8 k + k {}^{2} + 4 - 4k =10$

$\implies \ \: \: \: \sf \:2k {}^{2} - 12k + 10 = 0$

$\bf \red{Dividing \: by \: 2,}$

$\rm k {}^{2} - 6k + 5 = 0$

$\implies \: \: \: \sf k {}^{2} - 5k - k + 5 = 0$

$\implies \: \: \sf k( k - 5) - 1(k - 5) = 0$

$\implies \: \: \sf (k - 5)(k - 1) = 0$

$\rm ∴ Either \: k - 5 = 0 \: or \: k - 1 = 0$

$\sf i.e., \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: k = 5 \: or \: k = 1.$

Answer 2

$\Large\underline\textsf{\red{AnsWer:}}$

$\star$ $\normalsize\sf \blue{The \: value \: of \: k \: is \: 5 }$

$\star$ $\normalsize\sf \blue {The \: value \: of \: k \: can \: also \: be \: 1}$

$\Large\underline\textsf{\red{Step \: by \: step \: explanation:}}$

$\normalsize\textbf{\underline{Points \: given :-}}$

$\star$$\normalsize\sf \purple{ \: \: P \: = \: (k -1,2)}$

$\star$$\normalsize\sf \purple{ \: \: A \: = \: (3, k)}$

$\star$$\normalsize\sf \purple{ \: \: B \: = \: (k,5)}$

$\normalsize\textbf{\underline{Also, \: it \: is \: given \: that :-}}$

$\normalsize\sf{ P \: is \: equidistant \: from \: points \: A \: and \: B, \: So;}$

$\star$ ${\boxed{\sf \orange{ PA \: = \: PB}}}$

$\normalsize\textbf{\underline{Now, \: Two \: concepts \: we \: use :-}}$

$\normalsize\sf{ 1) \: We \: use \: Distance \: formula \: to \: determine \: values.}$

$\normalsize\sf{2) We \: square \: both \: sides \: to \: cancel \: roots }$

$\normalsize\sf{ i.e. \: (PA)^2 \: = (PB)^2}$

$\normalsize\textbf{\underline{Let's \: solve:-}}$

$\star$$\normalsize\sf{Using \: \pink{Distance \: formula;} }$

${\boxed{\sf{Distance \: formula \: = \: \sqrt{(x_2 \: - \: x_1 )^2 \: + \: (y_2 \: - \: y_1)^2} }}}$

$\star$ $\normalsize\sf \pink{ Distance \: of \: PA: \: }$

➠ $\normalsize\sf{PA \: = \: \sqrt{ (3 -(k -1))^2 + ( k-2)^2} }$

$\star$ $\normalsize\sf \pink{Distance \: of \: PB:}$

➠ $\normalsize\sf{PB\: = \: \sqrt{ (k-(k-1))^2 +(5-2)^2} }$

$\star$ $\normalsize\sf \pink{Squaring \: both \: sides:}$

➠ $\normalsize\sf{ (PA)^2 \: = \: (PB)^2 }$

➠$\normalsize\sf{\sqrt{ [ (3 -(k-1))^2 +(k-2)^2]^2} }$

$\normalsize\sf{ \: \: \: \: \: \: \: \: \: \: \: = \: \sqrt{ [k-(k-1)^2 +(5-2)^2]^2} }$

$\footnotesize\sf{ \: \: \: \: \: \: \: \: \: (Cancel \: roots \: both \: sides)}$

➠$\normalsize\sf{(3-(k-1))^2 +(k-2)^2}$

$\normalsize\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: (k-(k-1)^2 +(5-2)^2}$

➠$\normalsize\sf{(3-k+1)^2 + (k-2)^2}$

$\normalsize\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: (k-k+1)^2 - (3)^2}$

➠$\normalsize\sf{(4-k)^2 + (k-2)^2 \: = \: 10 }$

$\footnotesize\sf{\: \: \: \: \:( using \: identity : (a-b)^2 \: = \: a^2 + b^2 - 2ab)}$

➠$\normalsize\sf{16 + k^2 - 8k + k^2 + 4 - 4k \: = \: 10}$

$\footnotesize\sf{ \: \: \: \: \: \: \: \: \: \: (Take \: 2 \: common) }$

➠$\normalsize\sf{ 2(k^2 - 6k + 5) \: = \: 0 }$

➠$\normalsize\sf{k^2 - 6k + 5 = 0 }$

$\footnotesize\sf{\: \: \: \: \: \: (Using \: middle \: term \: factorization)}$

➠$\normalsize\sf{k^2 - 5k -k + 5 = 0}$

➠$\normalsize\sf{k(k-5) - 1(k-5) = 0}$

➠$\normalsize\sf{(k-5)(k-1) = 0}$

✰$\normalsize\textsf{\underline{case \: 1}}$

➠$\normalsize\sf{ k - 5 = 0}$

➠$\normalsize\sf{ k = 0 + 5 = 5}$

✰$\normalsize\textsf{\underline{case \: 2}}$

➠$\normalsize\sf{k - 1 = 0}$

➠$\normalsize\sf{ k = 0 + 1 = 1}$

➠${\boxed{\sf \orange{k = 1,5}}}$