Explanation
We have mid point formula,
$$P(X,Y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$$
Let $$\Delta ABC$$ be the equilateral triangle, where $$AB=BC=AC$$.
Vertices of this triangle be,
$$A\left( 3,2 \right),B\left( -3,2 \right)$$ and $$C\left( x,y \right)$$.
The mid-point of the side $$AB$$ is $$M\left( 0,2 \right)$$.
$$ AB=\sqrt{{{\left( 3+3 \right)}^{2}}+{{\left( 2-2 \right)}^{2}}} $$
$$ AB=6\ units $$
Therefore,
$$\Rightarrow AB=BC=AC=6\ units$$
Also,
$$\Rightarrow AM=3\ units$$
As two vertices are $$A\left( 3,2 \right)$$ and $$B\left( -3,2 \right)$$, so third vertex will be at $$y-$$axis.
$$\Rightarrow C\left( 0,y \right)$$
It will be located below the origin as the triangle contains the origin.
Now,
$$ {{y}^{2}}={{\left( AC \right)}^{2}}-{{\left( AM \right)}^{2}} $$
$$ {{y}^{2}}=36-9=27 $$
$$ y=3\sqrt{3}\ units $$
$$\Rightarrow C\left( 0,-3\sqrt{3} \right)$$
Find the value of $$x$$ such that $$AB=BC$$ where the coordinates of A, B and C are $$(2,1)$$, $$(x,0)$$ and $$(-2,-1)$$ respectively.
$${\rm{since point is an y - axis}}$$
$${\rm{its x and z coordinate is 0}}$$
$${\rm{Let pt}}{\rm{. on y - axis be A (0,a,0)}}$$
$${\rm{PA = }}\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} $$
$$5\sqrt2 = \sqrt {{{( - 3)}^2} + {{(a + 2)}^2} + {{(0 - 5)}^2}} $$
$$5\sqrt2 = \sqrt {a + {a^2} + 4 + 4a + 25} $$
$$5\sqrt2 = \sqrt {{a^2} + 4a + 38} $$
$$25 \times 2 = {a^2} + 4a + 38$$
$${a^2} + 4a - 12 = 0$$
$${a^2} + 6a - 2(a + b) = 0$$
$$a = 2,a = - 6$$
$$co - ordinate{\rm{ of point (0,2,0)and(0, - 6,0)}}$$
AC = 2 BC
$${{AC} \over {BC}} = {2 \over 1}$$
$$x = {{2{x^2} + 1 \times \left( { - 3} \right)} \over {1 + 2}} = {{4 - 3} \over 3} = {1 \over 3}$$
$$y = {{2 \times 1 + 1 \times 4} \over {2 + 1}} = {{2 + 4} \over 3} = 2$$
$$C\left( {{1 \over 3},2} \right)$$
Area of a triangle whose vertices are $$A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$$ is given as,
$$Area=(\dfrac{1}{2})[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] $$
Let $$A$$ be the required area
$$A=(\dfrac{1}{2})[(-5)(-5-2)+3(2+1)+5(-1+5)]\\ (\dfrac{1}{2})[35+9+20]\\=32sq\>unit$$
So, the correct option is (B)
$$\\(x+4)^2(y-7)^2=41\\and y=3x \\\therefore (x+4)^2+(3x-7)^2=41\\x^2+8x+16+9x^2+49-42x=41\\10x^2-34x+24=0\\5x^2-17x+12\\5x^2-5x-12x+12=0\\5x(x-1)-12(x-1)=0\\or (x-1)(5x-12)=0\\\therefore x=1 and x=(\frac{12}{5})$$
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