Question
An equilateral triangle has two vertices at the points $(3, 4)$ and $(-2, 3),$ find the coordinates of the third vertex.
✓
Answer
Let two vertices of an equilateral triangle are $A(3, 4),$ and $B(-2, 3)$ and let the third vertex be $C(x, y).$
Now, $\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(-2-3)^2+(3-4)^2}=\sqrt{(-5)^2+(-1)^2}$$\sqrt{25+1}=\sqrt{26}$
$\text{BC}=\sqrt{(\text{x}+2)^2+(\text{y}-3)^2}$
and, $\text{CA}=\sqrt{({3}-\text{x})^2+({4}-\text{y})^2}$
$\because$ The triangle is an equilateral triangle.
$\therefore$ AB = BC = CA$\because$ BC = AB
$\therefore\sqrt{(\text{x}+2)^2+(\text{y}-3)^2}=\sqrt{26}$
$\Rightarrow (x + 2)^2 + (y - 3)^2 = 26 ($Squaring$) \Rightarrow x^2 + 4x + 4 + y^2 - 6y + 9 = 26$$ \Rightarrow x^2 + y^2 + 4x - 6y + 13 = 26 $$\Rightarrow x^2 + y^2 + 4x - 6y = 26 - 13 = 13 .....(i)$ Again $CA = AB$$\therefore\sqrt{(3-\text{x})^2+(4-\text{y})^2}=\sqrt{26}$
Squaring, $(3 - x)^2 + (4 - y)^2 = 26$
$\Rightarrow 9 + x^2 - 6x + 16 + y^2 - 8y = 26$
$\Rightarrow x^2 + y^2 - 6x - 8y + 25 = 26$
$\Rightarrow x^2 + y^2 - 6x - 8y = 26 - 25 = 1 ......(ii)$ Subtracting (ii) from (i) $10x + 2y = 12$$ \Rightarrow 5x + y = 6 ......(iii)$$y = 6 - 5x$
Subtracting in $(i)x^2 + (6 - 5x)^2 + 4x - 6(6 - 5x) = 13$
$\Rightarrow x^2 + 36 + 25x^2 - 60x + 4x - 36 + 30x - 13 = 0$$ \Rightarrow 26x^2 - 26x - 13 = 0$
$\Rightarrow 2x^2 - 2x - 1 = 0 $ Here $a = 2, b = -2, C = -1$$\therefore\ \text{x}={-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-2)\pm\sqrt{(-2)^2-4\times2\times(-1)}}{2\times2}$
$=\frac{2\pm\sqrt{4+8}}{4}$
$=\frac{2\pm\sqrt{12}}{4}$
$=\frac{2\pm\sqrt{4\times3}}{4}$
$=\frac{2\pm2\sqrt{3}}{4}=\frac{1\pm\sqrt{3}}{2}$
$\text{x}=\frac{1+\sqrt{3}}{2}$ and $\frac{1-\sqrt{3}}{2}$
If $\text{x}=\frac{1+\sqrt{3}}{2},$ then $\text{y}=6-5\text{x}$
$=6-\frac{5(1+\sqrt{3})}{2}$
$=\frac{12-5-5\sqrt{3}}{2}=\frac{7-5\sqrt{3}}{2}$ and if $\text{x}=\frac{1-\sqrt{3}}{2},$ then $\text{y}=6-5\text{x}$
$=6-\frac{5(1-\sqrt{3})}{2}$
$=\frac{12-5+5\sqrt{3}}{2}=\frac{7+5\sqrt{3}}{2}$ Hence co-ordinates of the point will be $\Big(\frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2}\Big)$ or $\Big(\frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2}\Big).$
Now, $\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(-2-3)^2+(3-4)^2}=\sqrt{(-5)^2+(-1)^2}$$\sqrt{25+1}=\sqrt{26}$
$\text{BC}=\sqrt{(\text{x}+2)^2+(\text{y}-3)^2}$
and, $\text{CA}=\sqrt{({3}-\text{x})^2+({4}-\text{y})^2}$
$\because$ The triangle is an equilateral triangle.
$\therefore$ AB = BC = CA$\because$ BC = AB
$\therefore\sqrt{(\text{x}+2)^2+(\text{y}-3)^2}=\sqrt{26}$
$\Rightarrow (x + 2)^2 + (y - 3)^2 = 26 ($Squaring$) \Rightarrow x^2 + 4x + 4 + y^2 - 6y + 9 = 26$$ \Rightarrow x^2 + y^2 + 4x - 6y + 13 = 26 $$\Rightarrow x^2 + y^2 + 4x - 6y = 26 - 13 = 13 .....(i)$ Again $CA = AB$$\therefore\sqrt{(3-\text{x})^2+(4-\text{y})^2}=\sqrt{26}$
Squaring, $(3 - x)^2 + (4 - y)^2 = 26$
$\Rightarrow 9 + x^2 - 6x + 16 + y^2 - 8y = 26$
$\Rightarrow x^2 + y^2 - 6x - 8y + 25 = 26$
$\Rightarrow x^2 + y^2 - 6x - 8y = 26 - 25 = 1 ......(ii)$ Subtracting (ii) from (i) $10x + 2y = 12$$ \Rightarrow 5x + y = 6 ......(iii)$$y = 6 - 5x$
Subtracting in $(i)x^2 + (6 - 5x)^2 + 4x - 6(6 - 5x) = 13$
$\Rightarrow x^2 + 36 + 25x^2 - 60x + 4x - 36 + 30x - 13 = 0$$ \Rightarrow 26x^2 - 26x - 13 = 0$
$\Rightarrow 2x^2 - 2x - 1 = 0 $ Here $a = 2, b = -2, C = -1$$\therefore\ \text{x}={-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-2)\pm\sqrt{(-2)^2-4\times2\times(-1)}}{2\times2}$
$=\frac{2\pm\sqrt{4+8}}{4}$
$=\frac{2\pm\sqrt{12}}{4}$
$=\frac{2\pm\sqrt{4\times3}}{4}$
$=\frac{2\pm2\sqrt{3}}{4}=\frac{1\pm\sqrt{3}}{2}$
$\text{x}=\frac{1+\sqrt{3}}{2}$ and $\frac{1-\sqrt{3}}{2}$
If $\text{x}=\frac{1+\sqrt{3}}{2},$ then $\text{y}=6-5\text{x}$
$=6-\frac{5(1+\sqrt{3})}{2}$
$=\frac{12-5-5\sqrt{3}}{2}=\frac{7-5\sqrt{3}}{2}$ and if $\text{x}=\frac{1-\sqrt{3}}{2},$ then $\text{y}=6-5\text{x}$
$=6-\frac{5(1-\sqrt{3})}{2}$
$=\frac{12-5+5\sqrt{3}}{2}=\frac{7+5\sqrt{3}}{2}$ Hence co-ordinates of the point will be $\Big(\frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2}\Big)$ or $\Big(\frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2}\Big).$
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