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.
sincepointisany−axis
itsxandzcoordinateis0
Letpt.ony−axisbeA(0,a,0)
PA=√(x2−x1)2+(y2−y1)2+(z2−z1)2
5√2=√(−3)2+(a+2)2+(0−5)2
5√2=√a+a2+4+4a+25
5√2=√a2+4a+38
25×2=a2+4a+38
a2+4a−12=0
a2+6a−2(a+b)=0
a=2,a=−6
co−ordinateofpoint(0,2,0)and(0,−6,0)
AC = 2 BC
ACBC=21
x=2x2+1×(−3)1+2=4−33=13
y=2×1+1×42+1=2+43=2
C(13,2)
Area of a triangle whose vertices are A(x1,y1),B(x2,y2),C(x3,y3) is given as,
Area=(12)[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
Let A be the required area
A=(12)[(−5)(−5−2)+3(2+1)+5(−1+5)](12)[35+9+20]=32squnit
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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