Question

12. (a) Two vertices of an equilateral triangle are (0, 3) and (4,3). Find the third vertex of the triangle. ​

Answer 1

Answer:

(2,6.464)

Step-by-step explanation:

for any two points say, (x1,y1) and (x2,y2)

$distance \: between \: two \: points \\ = \sqrt{{(y2 - y1)}^{2} + {(x2 - x1)}^{2} }$

let the unknown vertex be (x,y)

and distance between (0,3) : (4,3)

$distance \: between (0,3) : (4,3) \: \\ = \sqrt{ {(3 - 3)}^{2} + {(4 - 0)}^{2} } \\ = \sqrt{0 + 4} = \sqrt{4} = 2 \\ so \: 4 = \sqrt{ {(y - 3)}^{2} + {(x - 0)}^{2} } which \: gives \\ {y}^{2} + 9 - 6y + {x}^{2} = 16 \\ {x}^{2} + {y}^{2} - 6y = 7 - - - - - (1) \\ and \: 4 = \sqrt{ {(3 - y)}^{2} + {(4 - x)}^{2} } \: which \: gives \\ 9 + {y}^{2} - 6y + 16 + {x}^{2} - 8x = 16 \\ {x}^{2} + {y}^{2} - 6y - 8x = - 9 - - - (2) \\ from \: equation \: (2) \: {x}^{2} + {y}^{2} - 6y = 8x - 9 \: substitute \: this \: value \: in \: equation \: (1) \\ 8x - 9 = 7 \\ 8x = 16 \\ x = \frac{16}{8} = 2 \\ substitute \: value \: of \: x \: in \: equation \: (1) \\ {2}^{2} + {y}^{2} - 6y = 7 \\ {y}^{2} - 6y - 3 = 0 \\ then \: y = 6.464 \: or \: - 0.464 \: as \: y \: should \: be \: positive \: y = 6.464$

; (0,3) : (x,y); and (x,y) : (4,3) should be equal as the triangle should be equilateral.