Explanation
We have mid point formula,
P(X,Y)=(x1+x22,y1+y22)
Let ΔABC be the equilateral triangle, where AB=BC=AC.
Vertices of this triangle be,
A(3,2),B(−3,2) and C(x,y).
The mid-point of the side AB is M(0,2).
AB=√(3+3)2+(2−2)2
AB=6 units
Therefore,
⇒AB=BC=AC=6 units
Also,
⇒AM=3 units
As two vertices are A(3,2) and B(−3,2), so third vertex will be at y−axis.
⇒C(0,y)
It will be located below the origin as the triangle contains the origin.
Now,
y2=(AC)2−(AM)2
y2=36−9=27
y=3√3 units
⇒C(0,−3√3)
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})
Please disable the adBlock and continue. Thank you.