Question

If the points p(2,2) is equidistant from the points A(-2,k) and B(-2k,-4), find k. Also, find the length of AP​

Answer 1

Given that , The points P(2,2) is equidistant from the points A(-2,k) and B(-2k,-4) .

Exigency To Find : The value of k & the length of AP ?

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⠀⠀▪︎⠀We know that to calculate the Distance between two points if we have given with co – ordinates of the two points ( x₁ , y₁ ) and ( x₂ , y₂ ) we use Distance Formula and that's given by :

$\qquad \star \:\:\underline {\boxed {\pmb{\sf { \:Distance \: =\: \: \sqrt{ \bigg( x_2 - x_1 \bigg)^2 + \bigg( y_2 - y_1 \bigg)^2 \:}}}}}$

⠀Given that ,

⠀⠀⠀⠀▪︎⠀The points P(2,2) is equidistant from the points A(-2,k) and B(-2k,-4) .

$\qquad \therefore \:\:\sf PA \:=\: PB \:\:\\\\\\\qquad :\implies \:\:\sf PA \:=\: PB \:\:\\\\\\ \qquad :\implies \sf\: \Big\{\: PA \:\Big\}^2\:=\: \Big\{\: PB \:\Big\}^2\\\\\\ \qquad :\implies \sf\: \Big[\: \{ \:(\:2\:-\:(-2) )^2 \: + \: ( \:2 \:-\:k\:)^2 \} \:\Big]\:=\: \Big[\: \{ \:(\:2\:-\:(-2k) )^2 \: + \: ( \:2 \:-\:(-3)\:)^2 \} \:\Big]\:\\\\\\ \qquad :\implies \sf\: \Big[\: \{ \:(\:4\:)^2 \: + \: ( \:2 \:-\:k\:)^2 \} \:\Big]\:=\: \Big[\: \{ \:(\:2\:+\:2k )^2 \: + \: ( \:5\:)^2 \} \:\Big]\:\\\\\\ \qquad :\implies \sf\: \Big[\: \:16 \: + \: k^2\:+ \:4\:-\:4k \: \:\Big]\:=\: \Big[\: \:4\:+\:4k^2 \: +\:8k \: \:25 \:\Big]\:\\\\\\ \qquad :\implies \sf\: 3k^2\: +\:12k\:+ \:9\:=\:0\:\\\\\\ \qquad :\implies \sf\: k^2\: +\:4k\:+ \:3\:=\:0\:\\\\\\\qquad :\implies \sf\: k^2\: +\:3k\:+\:k\:+ \:3\:=\:0\:\\\\\\ \qquad :\implies \sf\: (k\: +\:3)\:( k \: + \: 1\;)\:=\:0\:\\\\\\ \qquad :\implies \underline {\boxed{\pmb{\frak{\: k \:=\:-\:1 \:\:or\:\:-\:3\:}}}}\:\:\bigstar \:$

⠀⠀⠀⠀∴ Hence , the value of k is –1 or –3 .

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding Length of AP [ A(-2,k) & P(2,2) ] :

⠀⠀▪︎⠀By Using the Distance Formula , we can find length of an A.P :

⠀⠀⠀⠀⠀⠀~ If Value of k will be – 1 , then Length of an AP will be ,

$\qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ \bigg( 2 - (-2) \bigg)^2 + \bigg( 2 - 1 ( - 1 )\: \bigg)^2}\\\\\\ \qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ \bigg( 2 + 2 \bigg)^2 + \bigg( 2 + 1\: \bigg)^2}\\\\\\ \qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ \bigg( 4 \bigg)^2 + \bigg( 3\: \bigg)^2}\\\\\\ \qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ 16 + 9 }\\\\\\\qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ 25 }\\\\\\ \qquad \dashrightarrow \underline {\boxed{\pmb{\frak{\: Length \:_{ \:(\:AP \: )\:} \: = \;\:5\:units\:}}}}\:\:\bigstar \:$

⠀⠀⠀⠀⠀⠀~ If Value of k will be – 3 , then Length of an AP will be ,

$\qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ \bigg( 2 - (-2) \bigg)^2 + \bigg( 2 - 1 ( - 3 )\: \bigg)^2}\\\\\\ \qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ \bigg( 2 + 2 \bigg)^2 + \bigg( 2 + 3\: \bigg)^2}\\\\\\ \qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ \bigg( 4 \bigg)^2 + \bigg( 5\: \bigg)^2}\\\\\\ \qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ 16 + 25 }\\\\\\\qquad \dashrightarrow \sf \: Length \:_{ \:(\:AP \: )\:} \: = \; \sqrt{ 41 }\\\\\\ \qquad \dashrightarrow \underline {\boxed{\pmb{\frak{\: Length \:_{ \:(\:AP \: )\:} \: = \;\:\sqrt{41}\:units\:}}}}\:\:\bigstar \:$

⠀⠀⠀⠀∴ Hence , If value of k is – 1 then , Length of an AP will be 5 units & If value of k is – 3 then , Length of an AP will be √41 units .