Find the third vertix of an equilateral triangle is two of the vertices are (0,0) and (3, root3) Share with your friends Share 0 Abhishek Jha answered this Dear student, Here is the solution of your asked query: Let A0,0; B3,3 and Cx,y be the vertices of an equilateral triangle.Since all sides of an equilateral triangle are equal.So, AB=BC=CAAB=CA⇒3-02+3-02=x-02+y-02 ⇒3-02+3-02=x-02+y-02⇒3+9=x2+y2⇒x2+y2=12 ...(i)BC=CA⇒x-32+y-32=x-02+y-02 {using the distance formula}⇒x-32+y-32=x2+y2⇒x2+9-6x+y2+3-23y=x2+y2⇒9-6x+3-23y=0⇒6x+23y=12 ⇒233x+y=12⇒3x+y=23 ⇒y=32-x ...(ii)From (i) and (ii) we get;x2+32-x 2=12⇒x2+32-x2=12⇒x2+34+x2-4x=12⇒x2+12+3x2-12x=12⇒4x2-12x=0⇒4xx-3=0⇒x=0 or x=3When x=0, y=32-0 =23And when x=3, y=32-3=-3Therefore the coordinate of C i.e. third vertex is either C0,23 or C3,-3 Regards 0 View Full Answer Bavishya Sm answered this Let the vertices be A(0,0), B(3,root3) and C(x,y). AB =BC =AC => AC=BC AC²=BC² --- (1) DISTANCE= root of {(x2 - x1)²-(y2 - y1)²} Substituting the formula in (1)... => (x-0)² + (y-0)² =(x-3)²+ (y-root 3)2 x²+y² = x² -6x + 9+ y² - 2root3 y + 3 => 0= -6x+12- 2root3 y => 3x+ root3 y= 4 X= 4- root3 y /3 Substitute the above equation in the following equation... AB²= BC² U Will get x and y values I did not do the last part bcos I want u to try on ur own!! Hope it helps u... 0
