Question

A triangle OAB is an isosceles triangle with sides OA = AB = 13 units. Find the coordinates of the vertices​

Answer 1

Answer:

Step-by-step explanation:

here, let point O be on origin

and OA be on x axis

∴ coordinates of O = (0,0)

let the above coordinates be x' and y' respectively

but O and A are on x axis

∴ coordinates of A = (13,0)      

let above coordinates  be (x,y) respectively

let coordinates of B be (x",y")  

in ,  ΔOAB

as, OA = AB

⇒OA/AB = 1

⇒tan O = OA/OB

but , tan45° = 1

⇒ tan O = tan 45°

∠O = 45°

∴, sin O = AB/ BO

sin 45° = 13/BO

1/√2 = 13/BO

⇒BO = 13√2 units

$\sqrt(x"-x')^2 + (y"-y')^2$ = BO

$\sqrt{x}"^2 + y"^2 = 13\sqrt2$

∴ x" + y" = 13√2 - - - - (i)

⇒x" = 13√2 - y"

but

$\sqrt (x -x")^2 - (y-y")^2 = AB$

$\sqrt (13-x)^2 + y^{2}$ = 13

169 -26x" +x"² + y"² = 169

x"² + y"² = 26x"

 13√2 = 26x"        ( by (i))

⇒x" = √2/2

putting the above value in eq(i) , we get ,

√2/2 + y" = 13√2

⇒√2 + 2y" + 26√2

⇒2y" = 25√2

⇒ y" = 25√2 / 2

∴ Coordinates of vertices of ΔOAB are as follows

coordinates of O = (0 , 0)

coordinates of A  = ( 13 , 0)

coordinates of B = (√2/2 , 25√2/2)

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