Question

The coordinates of the centroid of a triangle are (√3,2), and two of its vertices are (2√3,-1) and (2√3,5). Find the third vertex of the triangle.

Answer 1

Answer:

GIVEN: Equilateral triangle ABC , A(x,y),

B(3,4), C(-2,3)

TO FIND: Coordinates of A, x & y

By distance formula, we calculate AB, BC & AC. Then we equate them to get x & y.

Distance between 2 points

= √[( x1-x2)² + ( y1-y2)²], where( x1, y1) &( x2,y2) are coordinates of 2 given points

So, in the given triangle ABC,

BC= √(5²+1²) = √26 ………….(1)

AB = √[(x-3)² +(y-4)²] …………(2)

AC = √[(x+2)² +(y-3)² …………..(3)

Let's equate (1) & (3) AC = BC => AC² =BC²

(x+2)² + (y-3)² =26

=> x² +4x +4 + y² -6y +9 = 26

Now, we equate(2) & (3) AB = AC => AB² = AC²

=> (x-3)² +(y-4)² = (x+2)² +(y-3)²

=> x² + 9 - 6x + y² + 16 -8y = x² +4 + 4x + y² +9-6y

=> 10x +2y -12 =0

=> 5x + y -6 =0

=> 5x +y = 6

=> y = 6–5x ……………..(5)

Now, put the value of y in equation (4)

we get, x² + (6–5x)² + 4x -6( 6–5x) -13 =0

=> x² +36 +25x² - 60x +4x -36+30x -13 =0

=> 26x² -26x -13 = 0

=> 2x² - 2x -1 =0

= x = (2 +,- √12)/4 ( by quadratic formula)

= > x = (2+,- 2√3)/4

=> x = (1+ ,- √3)/2 ……………..(6)

Now, get Y by eq (5)

y = 6 - 5{( 1+, - √3)/2}

y= 6 - ( 5+,-5√3 )/2

= y= (7 +,- 5√3)/2 …………..(7)

Now, after verifying we get the coordinates of the third vertex A

If x= (1+√3)/2 then y= (7–5√3)/2 ……….ANS★

&if x = (1-√3)/2 then y = (7+5√3)/2………..ANS★