Question

In the figure, ∆ABC and ∆ABD are equilateral triangles, then find the coordinates of points C and D.

Answer 1

$\rm \red{Given : }$

$\sf{\Delta ABC \: and \: \Delta ABD \: are \: equilateral \: triangles}$

$\underline{ \underline{ \rm \red{Properties \: of \: Equilateral \: triangle}}} \red:$

$\sf \bullet \: All \: sides \: are \: equal \: i.e. \: AB = BC = CA$

$\sf \bullet \: Each \: and \: every \: angle \: measures \: 60°$

$\sf \: the \: coordinates \: of \: equilateral \: triangle \: are :$

$\sf \: A = ( - a,0) , \: B = (a,0)$

$\rm \red{Let : }$

$\sf \: C = (0,Y_1), \: D = (0,Y_2)$

$\underline{ \underline{ \rm \red{case \: i}}} \red{:} \sf{From \: \Delta ABC}$

$\rm{AB = BC = CA}$

$\rm{The \: distance \: between \: points \: A(-a,0), \: B(a,0)}$

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

$\rm{AB = \sqrt{ {(a + a)}^{2} + (0 - 0) ^{2} } }$

$\rm{AB \: = \sqrt{4 {a}^{2} } }$

$\rm{AB = 2a}$

$\rm{The \: distance \: between \: points \: B(a,0) \: ,C(0,y_1)}$

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

$\rm{BC = \sqrt{( {0 - a)}^{2} + {(y_1 - 0)}^{2} } }$

$\rm{BC = \sqrt{ {a}^{2} + y_1^{2} } }$

$\sf \red{From \: the \: property \: of \: Equilateral \: triangle : }$

$\rm \: AB = BC$

$\rm{2a = \sqrt{ {a}^{2} + {y_1}^{2} } }$

$\sf \red{Squaring \: on \: both \: sides : }$

$\rm \: 4 {a}^{2} = {a}^{2} + y_1^{2}$

$\rm \: {y_1 ^{2} = 3 {a}^{2} }$

$\rm \: y_1 = \pm\sqrt{3} a$

$\rm{As \: C \: is \: in \: postive \: y-axis \: we \: get \: y_1 = \sqrt{3}a}$

$\boxed {\boxed {\rm \therefore \: cordinates \: of \: C = (0, \sqrt{3}a)}}$

$\underline{ \underline{ \rm \red{case \: ii}}} \red{:} \sf{From \: \Delta ABD}$

$\rm{AB = B D= DA}$

$\rm{The \: distance \: between \: points \: A(-a,0), \: B(a,0)}$

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

$\rm{AB = \sqrt{ {(a + a)}^{2} + (0 - 0) ^{2} } }$

$\rm{AB \: = \sqrt{4 {a}^{2} } }$

$\rm{AB = 2a}$

$\rm{The \: distance \: between \: points \: B(a,0) \: ,D(0,y_2)}$

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

$\rm{BD = \sqrt{( {0 - a)}^{2} + {(y_2 - 0)}^{2} } }$

$\rm{BD = \sqrt{ {a}^{2} + y_2^{2} } }$

$\sf \red{From \: the \: property \: of \: Equilateral \: triangle : }$

$\rm \: AB = BD$

$\rm{2a = \sqrt{ {a}^{2} + {y_2}^{2} } }$

$\sf \red{Squaring \: on \: both \: sides : }$

$\rm \: 4 {a}^{2} = {a}^{2} + y_2^{2}$

$\rm \: {y_2 ^{2} = 3 {a}^{2} }$

$\rm \: y_2 = \pm\sqrt{3} a$

$\rm{As \: D \: is \: in \: negative \: y-axis \: we \: get \: y_2 = - \sqrt{3}a}$

$\boxed{ \boxed{ \rm \therefore \: cordinates \: of \: D= (0, - \sqrt{3}a)}}$

Answer 2

$\large\underline\orange{GIVEN ⤵}$

$Here, \: \: \: AC = 2a \: and \: AO = a \\ {OC}^{2} = {AC}^{2} - {AO}^{2} \\ = {4a}^{2} - {a}^{2} \\ = {3a}^{2} \\ ➡ \: OC \: = a\sqrt{3} \\ Therefore \: ,coordinates \: of \: C \: \\ are(0,a\sqrt{3}) \: and \: coordinates \: of \\ D \: are \: (0, - a\sqrt{3})$